Latihan Soal 12 Persiapan PAS Gasal Matematika Peminatan Kelas X (Fungsi Eksponen dan Fungsi Logaritma)

Latihan Soal 11 Persiapan PAS Gasal Matematika Peminatan Kelas X (Fungsi Eksponen dan Fungsi Logaritma)

 $\begin{array}{ll}\\ 102.&\textrm{Agar}\: \: \log \left ( x^{2}-1 \right )<0\: \: \textrm{maka}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-1<x<1\\ \textrm{b}.&-\sqrt{2}<x<\sqrt{2}\\ \textrm{c}.&x<-1\: \: \textrm{atau}\: \: x>1\\ \textrm{d}.&x<-\sqrt{2}\: \: \textrm{atau}\: \: x>\sqrt{2}\\ \color{red}\textrm{e}.&-\sqrt{2}<x<-1\: \: \textrm{atau}\: \: 1<x<\sqrt{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\log \left ( x^{2}-1 \right )<0\\ &\textrm{Diketahui}\: \: \color{red}\log f(x)<0,\: \: \color{blue}\textrm{maka}\\ &\color{purple}\textrm{Syarat (1)},\: \: \color{red}f(x)>0\\ &\Leftrightarrow x^{2}-1>0\\ &\Leftrightarrow x<-1\: \: \textrm{atau}\: \: x>1\\ &\color{purple}\textrm{Syarat (2)},\: \: \log \left ( x^{2}-1 \right )<0\\ &\log \left ( x^{2}-1 \right )<\log 1\\ &\Leftrightarrow x^{2}-1<1\\ &\Leftrightarrow x^{2}-2<0\\ &\Leftrightarrow x^{2}-\left ( \sqrt{2} \right )^{2}<0\\ &\Leftrightarrow -\sqrt{2}<x<\sqrt{2}\\ &\textrm{Jadi},\: \: \color{red}-\sqrt{2}<x<-1\: \: \color{black}\textrm{atau}\: \: \color{red}1<x<\sqrt{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 103.&\textrm{Himpunan penyelesaian dari}\\ &^{.^{\frac{1}{2}}}\log \left ( x^{2}-3 \right )>0 \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\left \{ x|-2<x<-\sqrt{3}\: \: \textrm{atau}\: \: \sqrt{3}<x<2 \right \}\\ \textrm{b}.&\left \{ x|-\sqrt{3}<x<-1\: \: \textrm{atau}\: \: \sqrt{3}<x<2 \right \}\\ \textrm{c}.&\left \{ x|-2<x<-\sqrt{3} \right \}\\ \textrm{d}.&\left \{ x|-2<x<-\sqrt{3} \right \}\\ \textrm{e}.&\left \{ x|\sqrt{3}<x<2 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&^{.^{\frac{1}{2}}}\log \left ( x^{2}-3 \right )>0\\ &\textrm{Diketahui}\: \: \color{red}^{.^{\frac{1}{2}}}\log f(x)>0,\: \: \color{blue}\textrm{maka}\\ &\color{purple}\textrm{Syarat (1)},\: \: \color{red}f(x)>0\\ &\Leftrightarrow x^{2}-3>0\\ &\Leftrightarrow x<-\sqrt{3}\: \: \textrm{atau}\: \: x>\sqrt{3}\\ &\color{purple}\textrm{Syarat (2)},\: \: ^{.^{\frac{1}{2}}}\log \left ( x^{2}-3 \right )>0\\ &^{.^{\frac{1}{2}}}\log \left ( x^{2}-3 \right )>\: ^{.^{\frac{1}{2}}}\log 1\\ &\Leftrightarrow x^{2}-3<1\quad \left (\color{black}\textrm{karena basisnya}\: \: \displaystyle \frac{1}{2}<1 \right )\\ &\Leftrightarrow x^{2}-4<0\\ &\Leftrightarrow x^{2}-2^{2}>0\\ &\Leftrightarrow -2<x<2\\ &\textrm{Jadi},\: \: \color{red}-2<x<-\sqrt{3}\: \: \color{black}\textrm{atau}\: \: \color{red}\sqrt{3}<x<2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 104.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &^{2}\log \left ( x^{2}-x \right )\leq 1 \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x<0\: \: \textrm{atau}\: \: x>1\\ \textrm{b}.&-1\leq x\leq 2,\: x\neq 1\: \: \textrm{atau}\: \: x\neq 0\\ \color{red}\textrm{c}.&-1\leq x< 0\: \: \textrm{atau}\: \: 1<x\leq 2\\ \textrm{d}.&-1< x\leq 0\: \: \textrm{atau}\: \: 1\leq x< 2\\ \textrm{e}.&-1\leq x\leq 0\: \: \textrm{atau}\: \: 1\leq x\leq 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&^{2}\log \left ( x^{2}-x \right )\leq 1\\ &\textrm{Diketahui}\: \: \color{red}^{2}\log f(x)>0,\: \: \color{blue}\textrm{maka}\\ &\color{purple}\textrm{Syarat (1)},\: \: \color{red}f(x)>0\\ &\Leftrightarrow x^{2}-x>0\Leftrightarrow x(x-1)>0\\ &\Leftrightarrow x<0\: \: \textrm{atau}\: \: x>1\\ &\color{purple}\textrm{Syarat (2)},\: \: ^{2}\log \left ( x^{2}-x \right )\leq 1\\ &^{2}\log \left ( x^{2}-x \right )\leq \: ^{2}\log 2\\ &\Leftrightarrow x^{2}-x\leq 2\\ &\Leftrightarrow x^{2}-x-2\leq 0\\ &\Leftrightarrow (x+1)(x-2)\leq 0\\ &\Leftrightarrow -1\leq x\leq 2\\ &\textrm{Jadi},\: \: \color{red}-1\leq x< 0\: \: \color{black}\textrm{atau}\: \: \color{red}1<x\leq 2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 105.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\left | \log (x+1) \right |> 1 \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x<-0,9\: \: \textrm{atau}\: \: x>9\\ \textrm{b}.&x<-9\: \: \textrm{atau}\: \: x>9\\ \color{red}\textrm{c}.&-1<x<-0,9\: \: \textrm{atau}\: \: x>9\\ \textrm{d}.&-9< x<0,9\\ \textrm{e}.&-0,9<x<9 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Ingat bahwa}\\ &\left | x \right |>A\Leftrightarrow \color{black}x<-A\: \: \textrm{atau}\: \: x>A,\: \: \color{red}A>0\\ &\Leftrightarrow \log (x+1)<-1\: \: \textrm{atau}\: \: \log (x+1)>1\\ &\color{red}\textrm{Syarat (1) buat keduanya},\: \: \color{red}f(x)>0\\ &(x+1)>0\Leftrightarrow x>-1\\ &\color{red}\textrm{Syarat (2)},\: \: \log \left ( x+1 \right )<-1\\ &\log (x+1)<\log 10^{-1}\\ &x+1<\displaystyle \frac{1}{10}\Leftrightarrow x<-\frac{9}{10}\\ &\color{red}\textrm{Syarat (3)},\: \: \log \left ( x+1 \right )> 1\\ &\log (x+1)>\log 10^{1}\\ &(x+1)>10\Leftrightarrow x>9\\ &\textrm{Jadi},\: \: \color{red}-1<x<-0,9\: \: \color{black}\textrm{atau}\: \: \color{red}x>9 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 106.&\textrm{Himpunan penyelesaian pertidaksamaan}\\ &\log 4+\log (x+3)\leq \log x^{2}\\ &\textrm{a}.\quad \left \{ x|x\geq 6,\, x\in \mathbb{R} \right \}\\ &\textrm{b}.\quad \left \{ x|-3<x\leq -2\: \: \textrm{atau}\: \: x\geq 6\, x\in \mathbb{R} \right \}\\ &\textrm{c}.\quad \left \{ x|-3<x\leq -2\: \: \textrm{atau}\: \: 0\leq x\leq 6\, x\in \mathbb{R} \right \}\\ &\textrm{d}.\quad \left \{ x|x\leq -2\: \: \textrm{atau}\: \: x\geq 6,\, x\in \mathbb{R} \right \}\\ &\textrm{e}.\quad \left \{ x|x\leq -4\: \: \textrm{atau}\: \: x\geq 4 ,\, x\in \mathbb{R} \right \}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\textrm{Diketahui}\\ &\log 4+\log (x+3)\leq \log x^{2}\\ &\textrm{adalah bentuk}\: \: \color{red}^{a}\log f(x)\leq \, ^{a}\log g(x)\\ &\color{blue}\textrm{Syarat penyelesaian ada 2}\\ &\bullet \quad \textrm{basis}:\: \: a=10>0,\, \neq 1\\ &\bullet \quad \textrm{numerus}:\begin{cases} (1) & x+3>0\Rightarrow x>-3 \\ (2) & x^{2}>0\Rightarrow x\neq 0 \end{cases}\\ &\color{blue}\textrm{Proses penyelesaian}\\ &\log 4(x+3)\leq \log x^{2}\\ &4(x+3)\leq x^{2}\Leftrightarrow x^{2}\geq 4x+12\\ &\Leftrightarrow x^{2}-4x-12\geq 0\\ &\Leftrightarrow (x+2)(x-6)\geq 0\\ &x\leq -2\: \: \textrm{atau}\: \: x\geq 6\\ &\textrm{Jadi},\, \textrm{HP}=\color{red}\left \{ x|-3<x\leq -2\: \: \textrm{atau}\: \: x\geq 6\,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$.

$.\qquad\: \begin{aligned}&\textrm{Proses penyelesaian}\\ &^{6}\log \left (x^{2}-x-6 \right )>1\\ &\Leftrightarrow \, ^{6}\log \left (x^{2}-x-6 \right )>1.\, ^{6}\log 6\\ &\Leftrightarrow \, ^{6}\log \left (x^{2}-x-6 \right )>\, ^{6}\log 6\\ &\Leftrightarrow \, ^{a}\log f(x)>\, ^{a}\log p,\\ &\textrm{Karena}\: \color{blue}\textrm{basis}=a=6, \, \color{black}\textrm{maka tanda}\\ &\textrm{pertidaksamaan tetap. Selanjutnya}\\ &f(x)>p)\\ &\Leftrightarrow x^{2}-x-6>6\\ &\Leftrightarrow x^{2}-x-12>0\\ &\Leftrightarrow (x+3)(x-4)<0\\ &\Leftrightarrow \color{red}x<-3\: \: \color{black}\textrm{atau}\: \: \color{red}x>4\\ &\textrm{Karena}\: \: -2<x\: \: \textrm{atau}\: \: x>3,\: \textrm{maka}\\ &\textrm{HP}=\left \{ \color{red}x<-3\: \: \color{black}\textrm{atau}\: \: \color{red}x>4 \right \} \end{aligned}$.

$\begin{array}{ll} 108.&\textrm{Himpunan penyelesaian dari}\\ &\textrm{pertidaksamaan bentuk}\\ &\log x^{2}< \log (x+3)+2\log 2\\ &\textrm{adalah}\: ....\\ &\begin{array}{lll} \textrm{a}.&\left \{ -3<x<0\: \: \textrm{atau}\: \: 0<x<6 \right \}\\ \textrm{b}.&\color{red}\left \{ -2<x<0\: \: \textrm{atau}\: \: 0<x<6 \right \}\\ \textrm{c}.&\left \{ -1<x<0\: \: \textrm{atau}\: \: 0<x<6 \right \}\\ \textrm{d}.&\left \{ -2<x<0\: \: \textrm{atau}\: \: 0<x<7 \right \}\\ \textrm{e}.&\left \{ -1<x<0\: \: \textrm{atau}\: \: 0<x<8 \right \}\\ \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{array}{|c|c|}\hline \textrm{Syarat Numerus}&\textrm{Syarat Numerus}\\\hline \begin{aligned}f(x)>0&\\ x^{2}>0&\\ x\neq 0& \end{aligned}&\begin{aligned}g(x)&>0\\ x+3&>0\\ x&>-3 \end{aligned}\\\hline \textrm{Kita pilih}&\begin{cases} & -3<x<0 \\ & \textrm{atau}\\ & x>0 \end{cases}\\\hline \end{array}\\ &\begin{aligned}&\log x^{2}< \log (x+3)+2\log 2\\ &\Leftrightarrow \log x^{2}< \log (x+3)+\log 2^{2}\\ &\Leftrightarrow \log x^{2}< \log (x+3). 2^{2}\\ &\Leftrightarrow \, ^{a}\log f(x)<\, ^{a}\log g(x),\\ &\textrm{Karena}\: \color{blue}\textrm{basis}=a=10, \, \color{black}\textrm{maka tanda}\\ &\textrm{pertidaksamaan tetap. Selanjutnya}\\ &f(x)<g(x)\\ &\Leftrightarrow x^{2}<(x+3).2^{2}\\ &\Leftrightarrow x^{2}<(x+3).4\\ &\Leftrightarrow x^{2}<4x+12\\ &\Leftrightarrow x^{2}-4x-12<0\\ &\Leftrightarrow (x+2)(x-6)<0\\ &\Leftrightarrow \color{red}-2<x<6\\ &\textrm{Karena}\: \: -3<x<0\: \: \textrm{atau}\: \: x>0,\: \textrm{maka}\\ &\textrm{HP}=\left \{ -2<x<0\: \: \textrm{atau}\: \: 0<x<6 \right \} \end{aligned} \end{array}$.

$\begin{array}{ll} 109.&\textrm{Suatu larutan memiliki konsentrasi}\\ &\textrm{ion}\: \: \textrm{H}^{+}\: \: \textrm{sebesar}\: \: 2\times 10^{-6}.\\ &\textrm{PH dari larutan tersebut adalah}\: ....\\ &(\log 2=0,3010)\\ &\begin{array}{lllllll} \textrm{a}.&\color{red}4.3&&&\textrm{d}.&5,7\\ \textrm{b}.&4,7\quad &\textrm{c}.&5,3\quad&\textrm{e}.&6,3 \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned} \textrm{pH}&=-\log \left [\textrm{ H}^{+} \right ]\\ &=-\log \left ( 2\times 10^{-6} \right )\\ &=-\log 2-\log 10^{-6}\\ &=-0,3010-(-6)\log 10\\ &=-0,3010+6.1\\ &=-0,3010+6\\ &=6-0,3010\\ &=5,699\quad \textrm{dibulatkan}\\ &=\color{red}5,7 \end{aligned} \end{array}$

Latihan Soal 10 Persiapan PAS Gasal Matematika Peminatan Kelas X (Fungsi Eksponen dan Fungsi Logaritma)

 $\begin{array}{ll}\\ 93.&\textrm{Persamaan}\: \: ^{x}\log 2+\: ^{x}\log (3x-4)=2\\ &\textrm{mempunyai akar}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2},\: \: \textrm{maka}\\ &\textrm{nilai}\: \: x_{1}+x_{2}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \textrm{b}.&3\\ \textrm{c}.&4\\ \color{red}\textrm{d}.&6\\ \textrm{e}.&8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\color{purple}\textrm{Alternatif 1}\\ &^{x}\log 2+\: ^{x}\log (3x-4)=2\\ &\Leftrightarrow \: ^{x}\log 2(3x-4)=2\\ &\Leftrightarrow \: ^{x}\log 6x-8=2\\ &\Leftrightarrow \: \color{black}6x-8=x^{2}\\ &\Leftrightarrow \: \color{black}x^{2}-6x+8=0,\: \: \color{purple}\textrm{dengan}\: \begin{cases} a &=1 \\ b &=-6 \\ c &=8 \end{cases}\\ &\Leftrightarrow \: x_{1}+x_{2}=-\displaystyle \frac{b}{a}\\ &\Leftrightarrow \: x_{1}+x_{2}=\color{red}-\displaystyle \frac{-6}{1}=6\\ &\color{purple}\textrm{Alternatif 2}\\ &\Leftrightarrow \: \color{black}x^{2}-6x+8=0\\ &\Leftrightarrow \: (x-2)(x-4)\\ &\Leftrightarrow \: x_{1}=2\: \: \textrm{atau}\: \: x_{2}=4\\ &\Leftrightarrow \: x_{1}+x_{2}=\color{red}2+4=6 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 94.&\textrm{Jika}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\: \: \textrm{memenuhi}\\ &(\log x)(2\log x-3)=\log 100\\ &\textrm{maka}\: \: x_{1}\times x_{2}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&100\\ \color{red}\textrm{b}.&10\sqrt{10}\\ \textrm{c}.&\sqrt{10}\\ \textrm{d}.&-\sqrt{10}\\ \textrm{e}.&-10\sqrt{10} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&(\log x)(2\log x-3)=\color{red}\log 100\\ &\Leftrightarrow (\log x)\left ( 2\log x-3 \right )=\color{red}2\\ &\color{black}\Leftrightarrow 2\log ^{2}x-3\log x-2=0\: \color{purple}\begin{cases} a &=2 \\ b &=-3 \\ c &=-2 \end{cases}\\ &\Leftrightarrow \log x_{1}+\log x_{2}=-\displaystyle \frac{-3}{2}=\frac{3}{2}\\ &\Leftrightarrow \log \left ( x_{1}\times x_{2} \right )=1\displaystyle \frac{1}{2}\\ &\Leftrightarrow \color{red}\left ( x_{1}\times x_{2} \right )=10^{1\frac{1}{2}}=10\sqrt{10} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 95.&\textrm{Persamaan}\\ & 10^{\,^{\, ^{2}}\log x^{2} }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ & \textrm{mempunyai dua akar yaitu}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\\ &\textrm{Nilai}\: \: x_{1}\times x_{2}=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \textrm{b}.&-5\\ \color{red}\textrm{c}.&2\\ \textrm{d}.&5\\ \textrm{e}.&10 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&10^{\,^{\, ^{2}}\log x^{2} }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ &10^{\, 2^{\, ^{2}}\log x }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ &\color{black}\textrm{adalah persamaan kuadrat dalam}\: \: \color{red}10^{\,^{\, ^{2}}\log x }\\ &\color{black}\textrm{Misalkan}\: \: \color{red}p=10^{\,^{\, ^{2}}\log x },\: \: \textrm{maka persamaan}\\ &\color{black}\textrm{menjadi}\: \: \color{purple}p^{2}-7p+10=0\: \begin{cases} a & =1 \\ b & =-7 \\ c & =10 \end{cases}\\ & \color{red}\textrm{Karena nilai}\: \: \color{black}p_{1}\times p_{2}=\displaystyle \frac{c}{a}\: \: \textrm{maka}\\ &10^{\,^{\, ^{2}}\log x_{1} }\times 10^{\,^{\, ^{2}}\log x_{2} }=\displaystyle \frac{10}{1}=10\\ &10^{\,^{\, ^{2}}\log x_{1}\: +\: ^{2}\log x_{2} }=10\\ &10^{\,^{\, ^{2}}\log x_{1}\: +\: ^{2}\log x_{2} }=10^{1}\\ &\Leftrightarrow \: ^{2}\log x_{1}\: +\: ^{2}\log x_{2}=1\\ &\Leftrightarrow \: ^{2}\log x_{1}\times x_{2}=1\\ &\Leftrightarrow \: \color{red}x_{1}\times x_{2}=2^{1}=2\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 96.&\textrm{Nilai}\: \: x\: \: \: \: \textrm{yang memenuhi}\\ &^{x}\log (x+12)-3.\: ^{x}\log 4+1=0\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\\ \textrm{b}.&2\\ \color{red}\textrm{c}.&4\\ \textrm{d}.&8\\ \textrm{e}.&16 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&^{x}\log (x+12)-3.\: ^{x}\log 4+1=0\\ &^{x}\log (x+12)-\: ^{x}\log 4^{3}=-1\\ &^{x}\log \displaystyle \frac{x+12}{64}=-1\\ &\displaystyle \frac{x+12}{64}=x^{-1}=\frac{1}{x}\\ &x+12=\displaystyle \frac{64}{x}\\ &\color{purple}x^{2}+12x-64=0\\ &\color{purple}(x+16)(x-4)=0\\ &x=-16\: \: \color{black}\textrm{atau}\: \: \color{red}x=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 97.&\textrm{Nilai}\: \: x\: \: \: \: \textrm{yang memenuhi}\\ &\displaystyle \frac{^{2}\log (2x-3)}{^{2}\log x}-\: ^{x}\log (x+6)+\displaystyle \frac{1}{^{(x+2)}\log x}=1\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-1\: \: \textrm{dan}\: \: 6\\ \textrm{b}.&-2\: \: \textrm{dan}\: \: 6\\ \textrm{c}.&-1\\ \textrm{d}.&-2\\ \color{red}\textrm{e}.&6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\displaystyle \frac{^{2}\log (2x-3)}{^{2}\log x}-\: ^{x}\log (x+6)+\displaystyle \frac{1}{^{(x+2)}\log x}=1\\ &^{x}\log (2x-3)-\: ^{x}\log (x+6)+\: ^{x}\log (x+2)=1\\ &^{x}\log (2x-3)(x+2)=1+\log (x+6)\\ &^{x}\log \displaystyle \frac{\left ( 2x^{2}+x-6 \right )}{(x+6)}=1\\ &\displaystyle \frac{\left ( 2x^{2}+x-6 \right )}{(x+6)}=x^{1}\\ &\left ( 2x^{2}+x-6 \right )=x^{2}+6x\\ &\color{purple}x^{2}-5x-6=0\\ &\color{purple}(x+1)(x-6)=0\\ &x=-1\: \: \color{black}\textrm{atau}\: \: \color{red}x=6 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 98.&\textrm{Himpunan penyelesaian dari persamaan}\\ & ^{3}\log \left ( x^{2}-5x+7 \right )=0\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\left \{ 2,3 \right \}&\quad&\textrm{d}.&\left \{ 3,4 \right \}\\ \textrm{b}.&\left \{ 2,4 \right \}&&\textrm{e}.&\left \{ 3,5 \right \}\\ \textrm{c}.&\left \{ 2,5 \right \}&&& \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Diketahui bahwa}:\\ & ^{3}\log \left ( x^{2}-5x+7 \right )=0\\ &\Leftrightarrow \, ^{3}\log \left ( x^{2}-5x+7 \right )=\, ^{3}\log 3^{0}\\ &\Leftrightarrow \, ^{3}\log \left ( x^{2}-5x+7 \right )=\, ^{3}\log 1\\ &\textrm{bersesuaian dengan rumus}\: ^{a}\log f(x)=\, ^{a}\log p\\ &\underline{\textrm{Syarat numerus}}\\ &f(x)>0\Leftrightarrow x^{2}-5x+7>0\\ &\textrm{adalah definit positif}\\ &\textrm{sehingga semua nilai}\: \: x\: \: \textrm{memenuhi}\\ &\underline{\textrm{Langkah berikutnya}}\\ &f(x)=p\\ &\Leftrightarrow x^{2}-5x+7=1\\ &\Leftrightarrow x^{2}-5x+6=0\\ &\Leftrightarrow (x-2)(x-3)=0\\ &\Leftrightarrow x=2\: \: \textrm{atau}\: \: x=3\\ &\textrm{Jadi, HP}=\color{red}\left \{ 2,3 \right \} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 99.&\textrm{Himpunan penyelesaian dari}\\ & \log x^{2}=\log 4+\log (x+3)\\ &\textrm{adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\left \{ 1,4 \right \}&&&\textrm{d}.&\left \{ 2,6 \right \}\\ \textrm{b}.&\left \{ 1,6 \right \}\quad&\textrm{c}.&\displaystyle \left \{ 2,4 \right \}\quad&\textrm{e}.&\left \{ 4,6 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textrm{tidak ada}\\ &\textrm{ingat formula}:\: ^{a}\log f(x)=\, ^{a}\log g(x)\\ &\begin{array}{|c|c|}\hline \textbf{Syarat Numerus}&\textbf{Syarat Numerus}\\\hline f(x)&g(x)\\\hline \begin{aligned}x^{2}>0&\\ x<0\: \textrm{atau}\: x&>0 \end{aligned}&\begin{aligned}x+3&>0\\ x&>-3 \end{aligned}\\\hline \begin{aligned}&\textrm{Yang digunakan}\\ &\textrm{adalah yang }\\ &\textrm{memenuhi}\\ &\textrm{keduanya}\\ &\textrm{yaitu}:\: \end{aligned}&\begin{aligned}&-3<x<0\\ &\textrm{atau}\\ &x>0 \end{aligned}\\\hline \end{array}\\ &\begin{aligned}&\textrm{Syarat Penyelesaian}\\ & \log x^{2}=\log 4+\log (x+3)\\ &\Leftrightarrow \log x^{2}=\log 4(x+3)\\ &\textrm{maka},\: \: f(x)=g(x)\\ &x^{2}=4(x+3)\\ &\Leftrightarrow x^{2}=4x+12\\ &\Leftrightarrow x^{2}-4x-12=0\\ &\Leftrightarrow (x+2)(x-6)=0\\ &\Leftrightarrow x+2=0\: \: \textrm{atau}\: \: x-6=0\\ &\Leftrightarrow x=-2\: \: \textrm{atau}\: \: x=6\\ &\textrm{karena syaratnya},\: -3<x<0\: \: \textrm{atau}\: x>0,\\ & \textrm{maka keduanya memenuhi}\\ &\textrm{HP}=\color{red}\left \{ -2,6 \right \} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 100.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ & \log x^{2}=\log x\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&1&\qquad&\textrm{d}.&4\\ \textrm{b}.&2&&\textrm{e}.&5\\ \textrm{c}.&3&&& \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Diketahui}\: \: \log x^{2}=\log x\\ &\textrm{bersesuaian rumus}\: \: ^{a}\log f(x)=\, ^{a}\log g(x)\\ &\underline{\textrm{Syarat numerus}}\\ &\begin{array}{l|l} \begin{aligned}&f(x)>0\\ &x^{2}>0\\ &x>0\: \: \textrm{atau}\: \: x>0 \end{aligned}&\begin{aligned}&g(x)>0\\ &x>0\\ & \end{aligned} \end{array}\\ &\textrm{Sehingga syarat numerusnya}\: \: x>0\\ &\underline{\textrm{Syarat berikutnya}}\\ &\begin{aligned}f(x)&=g(x)\\ x^{2}&=x\\ x^{2}-x&=0\\ x(x-1)&=0\\ x=0\: \: \textrm{atau}\: \: x&=1 \end{aligned}\\ &\textrm{Jadi, HP}=\color{red}\left \{ 1 \right \} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 101.&\textrm{Salah satu nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ & 2\log^{2} x-9\log x+4=0\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\displaystyle \sqrt{10}&\qquad&\textrm{d}.&100\\ \textrm{b}.&\displaystyle 1&&\textrm{e}.&1000\\ \textrm{c}.&10&&& \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Diketahui}\: \: 2\log^{2} x-9\log x+4=0\\ &\textrm{bersesuaian rumus}: \\ &A\left ( ^{a}\log f(x) \right )^{2}+B\left ( ^{a}\log f(x) \right )+C=0\\ &\underline{\textrm{Langkah pengerjaan}}\\ &\begin{aligned}&2\log^{2} x-9\log x+4=0\\ &\Leftrightarrow \left ( 2\log x-1 \right )\left ( \log x-4 \right )=0\\ &\Leftrightarrow 2\log x-1=0\: \: \textrm{atau}\: \: \log x-4=0\\ &\Leftrightarrow \log x=\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: \log x=4\\ &\Leftrightarrow x=10^{.^{ \frac{1}{2}}}\: \: \textrm{atau}\: \: x=10^{4}\\ &\Leftrightarrow x=\sqrt{10}\: \: \textrm{atau}\: \: x=10000 \end{aligned}\\ &\textrm{Jadi, HP}=\color{red}\left \{ \sqrt{10},10000 \right \} \end{aligned} \end{array}$.

Latihan Soal 9 Persiapan PAS Gasal Matematika Peminatan Kelas X (Fungsi Eksponen dan Fungsi Logaritma)

 $\begin{array}{ll}\\ 85.&\textrm{Perhatikan pernyataan berikut}\\ &(1).\quad ^{2}\log 7+\, ^{2}\log 2=\, ^{2}\log 14\\ &(2).\quad ^{2}\log 12-\, ^{2}\log 4=\, ^{2}\log 8\\ &(3).\quad ^{2}\log \displaystyle \frac{1}{2}+\, ^{2}\log 16=3\\ &(4).\quad ^{2}\log \displaystyle \frac{1}{2}\times \, ^{2}\log 16=3\\ &\textrm{Pernyataan di atas yang benar adalah}\\ &\textrm{a}.\quad (1)\: \: \textrm{dan}\: \: (2)\\ &\textrm{b}.\quad (1)\: \: \textrm{dan}\: \: (3)\\ &\textrm{c}.\quad (2)\: \: \textrm{dan}\: \: (3)\\ &\textrm{d}.\quad (2)\: \: \textrm{dan}\: \: (4)\\ &\textrm{e}.\quad (3)\: \: \textrm{dan}\: \: (4)\\\\ &\begin{aligned}&\textbf{Jawab}:\quad \textbf{b}\\ &\textrm{Perhatikan pernyataan (1)}\\ &\color{red}^{2}\log 7+\, ^{2}\log 2=\, ^{2}\log 14\\ &\textrm{benar karena},\\ &^{a}\log b+\, ^{a}\log c=\, ^{a}\log bc\\ &\textrm{Perhatikan pernyataan (2)}\\ &\color{red}^{2}\log 12-\, ^{2}\log 4=\, ^{2}\log 8\\ &\textrm{adalah salah, seharusnya}\\ &^{2}\log 12-\, ^{2}\log 4=\, \color{blue}^{2}\log \frac{12}{4}=\, ^{2}\log 3\\ &\textrm{ingat sifat berikut}\\ &^{a}\log b-\, ^{a}\log c=\, ^{a}\log \displaystyle \frac{b}{c}\\ &\textrm{Perhatikan pernyataan (3)}\\ & ^{2}\log \displaystyle \frac{1}{2}+\, ^{2}\log 16=3\\ &\textrm{Benar, karena sama dengan sifat no (1) di atas}\\ &\textrm{yaitu}:\\ &^{2}\log \displaystyle \frac{1}{2}+\, ^{2}\log 16=\, ^{2}\log \displaystyle \frac{1}{2}.16\\ &\qquad =\, ^{2}\log 8=\, ^{2}\log 2^{3}=\color{red}3 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 86.&\textrm{Perhatikan pernyataan berikut}\\ &(1).\quad ^{2}\log 10+\, ^{2}\log 3=\, ^{2}\log 13\\ &(2).\quad ^{2}\log 20-\, ^{2}\log 4=\, ^{2}\log 5\\ &(3).\quad ^{2}\log \displaystyle \frac{1}{2}+\, ^{2}\log 16=-4\\ &(4).\quad ^{2}\log \displaystyle \frac{1}{2}\times \, ^{2}\log 16=-4\\ &\textrm{Pernyataan di atas yang benar adalah}\\ &\textrm{a}.\quad (1)\: \: \textrm{dan}\: \: (2)\\ &\textrm{b}.\quad (1)\: \: \textrm{dan}\: \: (3)\\ &\textrm{c}.\quad (2)\: \: \textrm{dan}\: \: (3)\\ &\textrm{d}.\quad (2)\: \: \textrm{dan}\: \: (4)\\ &\textrm{e}.\quad (3)\: \: \textrm{dan}\: \: (4)\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned} &\textrm{Perhatikan pernyataan (1)}\\ &^{2}\log 10+\, ^{2}\log 3=\, ^{2}\log 13\\ &\textrm{adalah salah karena},\\ &^{a}\log b+\, ^{a}\log c=\, ^{a}\log bc\\ &\color{red}\textrm{Perhatikan pernyataan (2)}\\ &^{2}\log 20-\, ^{2}\log 4=\, ^{2}\log 5\\ &\textrm{benar, karena}\\ &^{2}\log 20-\, ^{2}\log 4=\, ^{2}\log \frac{20}{4}=\, ^{2}\log 5\\ &\textrm{ingat sifat berikut}\\ &^{a}\log b-\, ^{a}\log c=\, ^{a}\log \displaystyle \frac{b}{c}\\ &\color{red}\textrm{Perhatikan pernyataan (3)}\\ & ^{2}\log \displaystyle \frac{1}{2}+\, ^{2}\log 16=-4\\ &\textrm{salah, karena sama dengan sifat no (1) di atas}\\ &\textrm{yaitu}:\\ &^{2}\log \displaystyle \frac{1}{2}+\, ^{2}\log 16=\, ^{2}\log \displaystyle \frac{1}{2}.16\\ &\qquad =\, ^{2}\log 8=\, ^{2}\log 2^{3}=3\\ &^{2}\log \displaystyle \frac{1}{2}\times \, ^{2}\log 16=^{2}\log 2^{-1}\times \, ^{2}\log 16\\ &\qquad =\, ^{2}\log \displaystyle 2^{-1}\times \, ^{2}\log 2^{4}=(-1).(4)=\color{red}-4 \end{aligned} \end{array}$.

$\begin{array}{ll} 87.&\textrm{Jika}\: \: \log 2=0,301\: \: \log 3=0,477\\ &\textrm{maka nilai}\: \: \log \sqrt[3]{225}\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\displaystyle 0,714&&&\textrm{d}.&0,778\\ \textrm{b}.&\displaystyle 0,734\quad&\textrm{c}.&0,756\quad&\textrm{e}.&\color{red}0,784 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}\log \sqrt[3]{225}&=\log 225^{.^{\frac{1}{3}}}\\ &=\log (15^{2})^{.^{\frac{1}{3}}}\\ &=\log 15^{.^{\frac{2}{3}}}\\ &=\displaystyle \frac{2}{3}\log 15\\ &=\displaystyle \frac{2}{3}\log 3\times 5\\ &=\displaystyle \frac{2}{3}\left ( \log 3+\log 5 \right )\\ &=\displaystyle \frac{2}{3}\left ( \log 3+\log \displaystyle \frac{10}{2} \right )\\ &=\displaystyle \frac{2}{3}\left ( \log 3+\log 10-\log 2 \right )\\ &=\displaystyle \frac{2}{3}\left ( \log 3+1-\log 2 \right )\\ &=\displaystyle \frac{2}{3}\left ( 0,477+1,000+0,301 \right )\\ &=\displaystyle \frac{2}{3}\left ( 1,176 \right )\\ &=\displaystyle \frac{2,352}{3}\\ &=\color{red}0,784 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 88.&\textrm{Fungsi invers dari}\: \: f(x)=5^{x}\\ &\begin{array}{lllllll} \textrm{a}.&f^{-1}(x)=5^{-x}&\\ \textrm{b}.&f^{-1}(x)=\left ( \displaystyle \frac{1}{5} \right )^{x}\\ \textrm{c}.&f^{-1}(x)=\left ( \displaystyle \frac{1}{5} \right )^{-x}&\\ \textrm{d}.&f^{-1}(x)=\, ^{5}\log x\\ \textrm{e}.&f^{-1}(x)=\, ^{x}\log 5\\ \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Diketahui bahwa}:\: \: f(x)=5^{x},\: \textrm{maka inversnya}\\ &\textrm{adalah}:\\ &\textrm{Langkah mula-mula dilogkan}\\ &\textrm{masing-masing ruas untuk mencari nilai}\\ &x,\: \: \textrm{yaitu}:\\ &\log f(x)=\log 5^{x}\Leftrightarrow \log f(x)=x\log 5\\ &\Leftrightarrow x=\displaystyle \frac{\log f(x)}{\log 5}=\, ^{5}\log f(x)\\ &\textrm{Selanjutnya kita ganti}\: \: x\: \: \textrm{dengan}\: \: f^{-1}(x),\: \: \textrm{dan}\\ & f(x)\: \: \textrm{dengan}\: \: x,\: \: \textrm{sehingga menjadi bentuk}\\ &\Leftrightarrow f^{-1}(x)=\, \color{red}^{5}\log x \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 89.&\textrm{Fungsi invers dari}\: \: y=\left ( \displaystyle \frac{1}{3} \right )^{x}\\ &\begin{array}{lllllll} \textrm{a}.&y^{-1}=\, ^{.^{\frac{1}{3}}}\log x&\quad&\textrm{d}.&y^{-1}=\left (\displaystyle \frac{1}{3} \right )^{-x}\\ \textrm{b}.&y^{-1}=\, ^{-3}\log x&&\textrm{e}.&y^{-1}=x^{.^{\frac{1}{3}}}\\ \textrm{c}.&y^{-1}=3^{-x}&&& \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Diketahui bahwa}:\: \: y=\left ( \displaystyle \frac{1}{3} \right )^{x},\: \textrm{maka inversnya}\\ &\textrm{sama seperti pada nomor sebelumnya, yaitu}:\\ &\textrm{dilogkan masing-masing ruas untuk mencari nilai}\\ &x,\: \: \textrm{yaitu}:\\ &\log y=\log \left ( \displaystyle \frac{1}{3} \right )^{x}\Leftrightarrow \log f(x)=x\log \left ( \displaystyle \frac{1}{3} \right )\\ &\Leftrightarrow x=\displaystyle \frac{\log y}{\log \left ( \displaystyle \frac{1}{3} \right )}=\, ^{.^{\frac{1}{3}}}\log y\\ &\textrm{Selanjutnya kita ganti}\: \: x\: \: \textrm{dengan}\: \: y^{-1},\: \: \textrm{dan}\\ & y\: \: \textrm{dengan}\: \: x,\: \: \textrm{sehingga menjadi bentuk}\\ &\Leftrightarrow y^{-1}=\, \color{red}^{.^{\frac{1}{3}}}\log x \end{aligned} \end{array}$.

$\begin{array}{ll} 90.&\textrm{Diketahui}\: \: f(x)=\, ^{a}\log (x+1).\: \textrm{Nilai}\: \: f(7)\\ &\textrm{jika nilai}\: \: f(1)=1\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\displaystyle \frac{1}{4}&&&\textrm{d}.&3\\\\ \textrm{b}.&\displaystyle \frac{1}{2}\qquad&\textrm{c}.&\color{red}2\qquad&\textrm{e}.&4 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&f(x)=\, ^{a}\log (x+1)\\ &f(1)=\, ^{a}\log (1+1)=1\\ &\qquad\qquad\Leftrightarrow \: ^{a}\log 2=1\\ &\qquad\qquad\Leftrightarrow \: 2=a^{1}\\ &\qquad\qquad\Leftrightarrow \: \color{red}2\color{black}=a \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 91.&\textrm{Jika}\: \: x=\: ^{15}\log 75\: \: \textrm{dan}\: \: y=\: ^{\displaystyle ^{\frac{3}{5}}}\log \displaystyle \frac{9}{125},\\ &\textrm{maka nilai}\: \: 5x+3y-2xy\: \: \textrm{adalah}\: ....\\\\ &\color{blue}\textrm{KOMPETISI HARDIKNAS ONLINE}\\ &\color{blue}\textrm{POSI}(\textrm{Pelatihan Olimpiade Sain Indonesia})\\ &\color{blue}\textrm{Bidang Matematika 2020}\\ &\begin{array}{llll}\\ \textrm{a}.&-1\\ \textrm{b}.&1\\ \textrm{c}.&3\\ \textrm{d}.&5\\ \color{red}\textrm{e}.&7 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\color{black}5x+3y-2xy\\ &=5\left ( ^{15}\log 75 \right )+3\left ( ^{\displaystyle ^{\frac{3}{5}}}\log \displaystyle \frac{9}{125} \right )\\ &\qquad -2\left ( ^{15}\log 75 \right )\left ( ^{\displaystyle ^{\frac{3}{5}}}\log \displaystyle \frac{9}{125} \right )\\ &=5\left ( \displaystyle \frac{\log 75}{\log 15} \right )+3\left ( \displaystyle \frac{\log \displaystyle \frac{9}{125}}{\log \displaystyle \frac{3}{5}} \right )\\ &\qquad -2\left ( \displaystyle \frac{\log 75}{\log 15} \right )\left ( \displaystyle \frac{\log \displaystyle \frac{9}{125}}{\log \displaystyle \frac{3}{5}} \right )\\ &=5\left ( \displaystyle \frac{\log 3.5^{2}}{\log 3.5} \right )+3\left ( \displaystyle \frac{\log -\log 125}{\log 3-\log 5} \right )\\ &\qquad -2\left ( \displaystyle \frac{\log 3.5^{2}}{\log 3.5} \right )\left ( \displaystyle \frac{\log 9-\log 125}{\log 3-\log 5} \right )\\ &=5\left ( \displaystyle \frac{\log 3+\log 5^{2}}{\log 3-\log 5} \right )+3\left ( \displaystyle \frac{\log 3^{2}-\log 5^{3}}{\log 3-\log 5} \right )\\ &\qquad -2\left ( \displaystyle \frac{\log 3+\log 5^{2}}{\log 3+\log 5} \right )\left ( \displaystyle \frac{\log 3^{2}-\log 5^{3}}{\log 3-\log 5} \right )\\ &=5\left ( \displaystyle \frac{\log 3+2\log 5}{\log 3-\log 5} \right )+3\left ( \displaystyle \frac{2\log 3-3\log 5}{\log 3-\log 5} \right )\\ &\qquad -2\left ( \displaystyle \frac{\log 3+2\log 5}{\log 3+\log 5} \right )\left ( \displaystyle \frac{2\log 3-3\log 5}{\log 3-\log 5} \right )\\\\ &\color{red}\textrm{Misalkan}\: \: \color{black}\log 3=A,\: \: \log 5=B \end{aligned} \end{array}$

$.\qquad\begin{aligned} &\color{red}\textrm{Selanjutnya}\\ &=5\left ( \displaystyle \frac{A+2B}{A+B} \right )+3\left ( \displaystyle \frac{2A-3B}{A-B} \right )\\ &\qquad -2\left ( \displaystyle \frac{A+2B}{A+B} \right )\left ( \displaystyle \frac{2A-3B}{A-B} \right )\\ &=\left ( \displaystyle \frac{5A+10B}{A+B} \right )+\left ( \displaystyle \frac{6A-9B}{A-B} \right )\\ &\qquad -\left ( \displaystyle \frac{2A+4B}{A+B} \right )\left ( \displaystyle \frac{2A-3B}{A-B} \right )\\ &=\displaystyle \frac{(5A+10B)(A-B)+(6A-9B)(A+B)}{A^{2}-B^{2}}\\ &\qquad -\left (\displaystyle \frac{4A^{2}-6AB+8AB-12B^{2}}{A^{2}-B^{2}} \right )\\ &=\displaystyle \frac{5A^{2}-5AB+10AB-10B^{2}}{A^{2}-B^{2}}\\ &\quad +\displaystyle \frac{6A^{2}+6AB-9AB-9B^{2}}{A^{2}-B^{2}}\\ &\qquad -\left (\displaystyle \frac{4A^{2}-6AB+8AB-12B^{2}}{A^{2}-B^{2}} \right )\\ &=\displaystyle \frac{7A^{2}-7B^{2}}{A^{2}-B^{2}}\\ &=\displaystyle \frac{7\left ( A^{2}-B^{2} \right )}{A^{2}-B^{2}}\\ &=\color{red}7 \end{aligned}$

$\begin{array}{ll}\\ 92.&\textrm{Diberikan}\: \: A=\: ^{6}\log 16\: \: \textrm{dan}\: \: B=\: ^{12}\log 27\\ &\textrm{Terdapat bilangan-bilangan bulat positif}\\ &a,\: b,\: \: \textrm{dan}\: \: c\: \: \textrm{sehingga}\: \: (A+a)(B+b)=c\\ &\textrm{Nilai dari}\: \: a+b+c\: \: \textrm{adalah}\: ....\\\\ &\color{blue}\textrm{KOMPETISI HARDIKNAS ONLINE}\\ &\color{blue}\textrm{POSI}(\textrm{Pelatihan Olimpiade Sain Indonesia})\\ &\color{blue}\textrm{Bidang Matematika 2020}\\ &\begin{array}{llll}\\ \textrm{a}.&23\\ \textrm{b}.&24\\ \textrm{c}.&27\\ \textrm{d}.&30\\ \textrm{e}.&34 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{....}\\ &\begin{aligned}&\color{black}\textrm{Diketahui}\\ &A=\: ^{6}\log 16=\displaystyle \frac{\log 16}{\log 6}=\displaystyle \frac{\log 2^{4}}{\log 2.3}=\frac{4\log 2}{\log 2+\log 3}\\ &\Leftrightarrow \color{black}\log 2+\log 3=\displaystyle \frac{4\log 2}{A}\: ...........\color{red}(1)\\ &B=\: ^{12}\log 27=\displaystyle \frac{\log 27}{\log 12}=\frac{\log 3^{3}}{\log 2^{2}.3}=\frac{3\log 3}{2\log 2+\log 3}\\ &\Leftrightarrow \color{black}2\log 2+\log 3=\displaystyle \frac{3\log 3}{B}\: .........\color{red}(2)\\ &\color{black}\textrm{ELIMINASI}\\ &\textrm{Dari persamaan (1) dan (2) diperoleh}:\\ &\bullet \quad \log 2=\displaystyle \frac{3\log 3}{B}-\displaystyle \frac{4\log 2}{A}\\ &\qquad \Leftrightarrow \log 2=\displaystyle \frac{3A\log 3-4B\log 2}{AB}\\ &\qquad \Leftrightarrow AB\log 2=3A\log 3-4B\log 2\\ &\qquad \Leftrightarrow AB\log 2+4B\log 2=3A\log 3\\ &\qquad \Leftrightarrow (AB+4B)\log 2=3A\log 3\\ &\qquad \Leftrightarrow \displaystyle \frac{\log 2}{\log 3}=\color{purple}\displaystyle \frac{3A}{AB+4B}\: ..........\color{red}(3)\\ &\bullet \quad \log 3=\displaystyle \frac{8\log 2}{A}-\displaystyle \frac{3\log 3}{B}\\ &\qquad \Leftrightarrow \log 3=\displaystyle \frac{8B\log 2-3A\log 3}{AB}\\ &\qquad \Leftrightarrow AB\log 3=8B\log 2-3A\log 3\\ &\qquad \Leftrightarrow AB\log 3+3A\log 3=8B\log 2\\ &\qquad \Leftrightarrow (AB+3A)\log 3=8B\log 2\\ &\qquad \Leftrightarrow \frac{\log 2}{\log 3}=\color{purple}\displaystyle \frac{AB+3A}{8B}...........\color{red}(4)\\ &\color{black}\textrm{KESAMAAN}\\ &\qquad\quad \frac{\log 2}{\log 3}=\frac{\log 2}{\log 3}\\ &\color{purple}\displaystyle \frac{AB+3A}{8B}=\color{purple}\displaystyle \frac{3A}{AB+4B}\\ &\qquad \Leftrightarrow (AB+3A)(AB+4B)=(8B).(3A)\\ &\qquad \Leftrightarrow (B+3)(A+4)=24\\ &\qquad \Leftrightarrow (A+4)(B+3)=24\\ &\color{black}\textrm{KESIMPULAN}\\ &a=4,\: b=3,\: \: \textrm{dan}\: \: c=24,\\ &\color{purple}\textrm{maka}\: \: \color{black}a+b+c=4+3+24=\color{red}31 \end{aligned} \end{array}$.

Latihan Soal 8 Persiapan PAS Gasal Matematika Peminatan Kelas X (Fungsi Eksponen dan Fungsi Logaritma)

 $\begin{array}{ll}\\ 71.&\textbf{(SPMB '05)}\\ &\textrm{Jika}\: \: ^{3}\log 2=p\: \: \textrm{dan}\: \: ^{2}\log 7=q,\\ &\textrm{maka}\: \: ^{14}\log 54=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{p+3}{p+q}\\ \textrm{b}.&\displaystyle \frac{2p}{p+q}\\ \color{red}\textrm{c}.&\displaystyle \frac{p+3}{p(q+1)}\\ \textrm{d}.&\displaystyle \frac{p+q}{p(q+1)}\\ \textrm{e}.&\displaystyle \frac{p(q+1)}{p+q} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&^{14}\log 54\\ &=\displaystyle \frac{^{...}\log 54}{^{...}\log 14}=\frac{^{...}\log (2.27)}{^{...}\log (2.7)},\: \: \color{red}\textrm{pilih basis 2}\\ &\color{blue}\textrm{mengapa tidak pilih basis selain 2}\\ &\color{blue}\textrm{lihat penyebut, di sana terdapat numerus 7}\\ &\color{blue}\textrm{pada soal, pasangan numerus 7 adalah 2},\\ &\textrm{makanya basis 2 dipilih, bukan yang lain}\\ &=\frac{^{2}\log (2.27)}{^{2}\log (2.7)}=\displaystyle \frac{^{2}\log 2+\: ^{2}\log 27}{^{2}\log 2+\: ^{2}\log 7}\\ &=\displaystyle \frac{^{2}\log 2+\: ^{2}\log 3^{3}}{^{2}\log 2+\: ^{2}\log 7}=\displaystyle \frac{^{2}\log 2+\left (3\times \: \displaystyle \frac{1}{^{3}\log 2} \right )}{^{2}\log 2+\: ^{2}\log 7}\\ &=\displaystyle \frac{1+\displaystyle \frac{3}{p}}{1+q}\\ &=\color{red}\displaystyle \frac{p+3}{p(q+1)} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 72.&\textbf{(SPMB '04)}\\ &\textrm{Jika}\: \: a>1\: ,\: \textrm{maka penyelesaian untuk}\\ &\left (^{a}\log (2x+1) \right )\left ( ^{3}\log \sqrt{a} \right )=1\: \: \textrm{adalah}\: ....\\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 1\\ \textrm{b}.&\displaystyle 2\\ \textrm{c}.&\displaystyle 3\\ \color{red}\textrm{d}.&\displaystyle 4\\ \textrm{e}.&\displaystyle 5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\left (^{a}\log (2x+1) \right )\left ( ^{3}\log \sqrt{a} \right )&=1\\ \left ( ^{3}\log \sqrt{a} \right )\left (^{a}\log (2x+1) \right )&=1\\ \left ( ^{3}\log a^{.^{^{\frac{1}{2}}}} \right )\left (^{a}\log (2x+1) \right )&=1\\ \displaystyle \frac{1}{2}\left ( ^{3}\log a \right )\left (^{a}\log (2x+1) \right )&=1\\ ^{3}\log (2x+1)&=2\\ 2x+1&=3^{2}\\ 2x&=9-1\\ 2x&=8\\ x&=\color{red}4 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 73.&\textbf{(SPMB '04)}\\ &\textrm{Nilai}\: \: \displaystyle \frac{\left ( ^{5}\log 10 \right )^{2}-\left ( ^{5}\log 2 \right )^{2}}{^{5}\log \sqrt{20}}\: \: \textrm{adalah}\: ....\\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\\ \textrm{b}.&\displaystyle 1\\ \color{red}\textrm{c}.&\displaystyle 2\\ \textrm{d}.&\displaystyle 4\\ \textrm{e}.&\displaystyle 5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\displaystyle \frac{\left ( ^{5}\log 10 \right )^{2}-\left ( ^{5}\log 2 \right )^{2}}{^{5}\log \sqrt{20}}\\ &=\displaystyle \frac{\left ( ^{5}\log 10+\: ^{5}\log 2 \right )\left ( ^{5}\log 10-\: ^{5}\log 2 \right )}{^{5}\log (20)^{.^{\frac{1}{2}}}}\\ &=\displaystyle \frac{^{5}\log (10.2)\times ^{5}\log \left (\frac{10}{2} \right )}{\displaystyle \frac{1}{2}\times \: ^{5}\log 20}\\ &=2\times \left ( \displaystyle \frac{^{5}\log 20}{^{5}\log 20} \right )\times \: ^{5}\log 5\\ &=2.1.1\\ &=\color{red}2 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 74.&\textbf{(SPMB '03)}\\ &\textrm{Jika diketahui bahwa}\\ &^{4}\log ^{4}\log x-\: ^{4}\log ^{4}\log ^{4}\log 16=2\\ & \textrm{maka}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle ^{2}\log x=8\\ \textrm{b}.&\displaystyle ^{2}\log x=4\\ \color{red}\textrm{c}.&\displaystyle ^{4}\log x=8\\ \textrm{d}.&\displaystyle ^{4}\log x=16\\ \textrm{e}.&\displaystyle ^{16}\log x=8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{4}\log \: ^{4}\log 16=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{4}\log \: ^{4}\log 4^{2}=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{4}\log 2=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{2^{2}}\log 2^{1}=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{4}\log \left ( \displaystyle \frac{1}{2} \right )=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{2^{2}}\log 2^{-1}=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\left ( -\displaystyle \frac{1}{2} \right )=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x+\frac{1}{2}=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x=2-\displaystyle \frac{1}{2}=\frac{3}{2}\\ &\Leftrightarrow \: ^{4}\log x=4^{.\frac{3}{2}}\\ &\Leftrightarrow \: ^{4}\log x=\left (2^{2} \right )^{.^{\frac{3}{2}}}\\ &\Leftrightarrow \: ^{4}\log x=2^{3}\\ &\Leftrightarrow \: ^{4}\log x=\color{red}8\\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 75.&\textbf{(UMPTN '92)}\\ &\textrm{Jika}\: \: x\: \: \textrm{memenuhi persamaan}\\ &^{4}\log ^{4}\log x-\: ^{4}\log ^{4}\log ^{4}\log 16=2\\ & \textrm{maka nilai}\: \: ^{16}\log x=\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle 4\\ \textrm{b}.&\displaystyle 2\\ \textrm{c}.&\displaystyle 1\\ \textrm{d}.&\displaystyle -2\\ \textrm{e}.&\displaystyle -4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{4}\log \: ^{4}\log 16=2\\ &\textrm{menyebabkan}\\ & ^{4}\log x=8\Rightarrow x=4^{8}\\ &(\color{purple}\textrm{lihat pembahasan no.23})\\ &\textrm{maka},\\ &\: ^{16}\log x=\: ^{16}\log 4^{8}=\: ^{4^{2}}\log 4^{8}\\ &=\displaystyle \frac{8}{2}\\ &=\color{red}4 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 78.&\textbf{(UMPTN '94)}\\ &\textrm{Hasil kali akar-akar persamaan}\\ &^{3}\log x^{.\left (^{2+\: ^{3}\log x} \right )}=15\\ & \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{1}{9}\\ \textrm{b}.&\displaystyle \frac{1}{3}\\ \textrm{c}.&\displaystyle 1\\ \textrm{d}.&\displaystyle 3\\ \textrm{e}.&\displaystyle 9 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&^{3}\log x^{.\left (^{2+\: ^{3}\log x} \right )}=15\\ &\Leftrightarrow \: \left ({2+\: ^{3}\log x} \right )^{3}\log x-15=0\\ &\Leftrightarrow 2\: ^{3}\log x+\: \left ( ^{3}\log x \right )^{2}-15=0\\ &\Leftrightarrow \: \left ( ^{3}\log x \right )^{2}+2\: ^{3}\log x-15=0\\ &\Leftrightarrow \: \left (^{3}\log x_{1}+5 \right )\left ( ^{3}\log x_{2}-3 \right )=0\\ &\Leftrightarrow \: ^{3}\log x_{1}+5=0\: \: \textrm{atau}\: \: ^{3}\log x_{1}-3=0\\ &\Leftrightarrow \: ^{3}\log x_{1}=-5\: \: \textrm{atau}\: \: ^{3}\log x_{2}=3\\ &\Leftrightarrow \: x_{1}=3^{-5}\: \: \textrm{atau}\: \: x_{2}=3^{3}\\ &\qquad \textrm{maka}\\ &\Leftrightarrow \: x_{1}\times x_{2}=3^{-5}\times 3^{3}=3^{-5+3}=3^{-2}\\ &\Leftrightarrow \qquad =\displaystyle \frac{1}{3^{2}}\\ &\Leftrightarrow \qquad =\color{red}\frac{1}{9} \end{aligned} \end{array}$.

$\begin{array}{l}\\ 79.&\textrm{Diketahui bahwa}\\ & ^{^{2}}\log 3=p \: \: \textrm{dan}\: \: ^{^{3}}\log 11=q,\\ &\textrm{maka nilai}\: \: ^{^{44}}\log 66=....\\\\ &\textrm{Jawab}:\\ &\begin{aligned}^{^{44}}\log 66&=\displaystyle \frac{^{^{^{...}}}\log 66}{^{^{^{...}}}\log 44}\\ &=\displaystyle \frac{^{^{^{...}}}\log \left (2\times 3\times 11 \right )}{^{^{^{...}}}\log \left (2^{2}\times 11 \right )}\\ &=\displaystyle \frac{^{^{^{3}}}\log 2+\: ^{^{^{3}}}\log 3+\:^{^{^{3}}}\log 11}{^{^{^{3}}}\log 2^{2}+\: ^{^{^{3}}}\log 11}\\ &=\displaystyle \frac{\frac{1}{^{^{^{2}}}\log 3}+\: ^{^{^{3}}}\log 3+\:^{^{^{3}}}\log 11}{\frac{2}{^{^{^{3}}}\log 2^{2}}+\: ^{^{^{3}}}\log 11}\\ &=\displaystyle \frac{\frac{1}{p}+1+q}{\frac{2}{p}+q}\\ &=\displaystyle \frac{\frac{1}{p}+1+q}{\frac{2}{p}+q}\times \displaystyle \frac{p}{p}\\ &=\color{red}\frac{1+p+pq}{2+pq} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 80.&\textbf{(AIME 1984)}\\ &\textrm{Diketahui bahwa}\: \: x\: \: \textrm{dan}\: \: y\\ & \textrm{adalah bilangan real yang memenuhi}\\\\ &\left\{\begin{matrix} ^{^{8}}\log x+\: ^{^{4}}\log y^{2}=5\\ \\ ^{^{8}}\log y+\: ^{^{4}}\log x^{2}=7 \end{matrix}\right.\\\\ &\textrm{Tentukanlah nilai dari}\: \: xy\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&^{^{^{8}}}\log x+\: ^{^{4}}\log y^{2}=5\\ &\Leftrightarrow \: ^{^{^{2^{3}}}}\log x+\: ^{^{^{2^{2}}}}\log y^{2}=5....(1)\\ &^{^{^{8}}}\log y+\: ^{^{4}}\log x^{2}=7\\ &\Leftrightarrow \: ^{^{^{2^{3}}}}\log y+\: ^{^{^{2^{2}}}}\log x^{2}=7....(2)\\ &\textrm{selanjutnya},&\\ &\frac{1}{3}\:. ^{^{^{2}}}\log x+\: ^{^{^{2}}}\log y=5\: \: |\times \frac{1}{3}|\\ &\Rightarrow \frac{1}{9}\: .^{^{^{2}}}\log x+\: \frac{1}{3}.\: ^{^{^{2}}}\log y=\frac{5}{3}....(3)\\ &^{^{^{2}}}\log x+\:\frac{1}{3}\: . ^{^{^{2}}}\log y=7\: \: |\times 1|\\ &\Rightarrow ^{^{^{2}}}\log x+\:\frac{1}{3}\: . ^{^{^{2}}}\log y=7....(4)\\ &\textrm{saat persamaan}\: \: (3)-(4)\\ &=-\frac{8}{9}.\: ^{^{^{2}}}\log x=\frac{5}{3}-7=-\frac{16}{3}\\ &\color{purple}\textrm{maka}\\ &^{^{^{2}}}\log x=\left ( -\frac{16}{3} \right )\left ( -\frac{9}{8} \right )\\ &^{^{^{2}}}\log x=6\Leftrightarrow x=2^{6}\Leftrightarrow x=64\\ &\color{purple}\textrm{Pada persamaan 1 selanjutnya}\\ &\frac{1}{3}.\: ^{^{^{2}}}\log x+\: ^{^{^{2}}}\log y=5\\ &\Leftrightarrow \: \: \frac{1}{3}.\: ^{^{^{2}}}\log 2^{6}+\: ^{^{^{2}}}\log y=5\\ &\Leftrightarrow \: \: \frac{1}{3}.6+\: ^{^{^{2}}}\log y=5\\ &\Leftrightarrow \: \: 2+\: ^{^{^{2}}}\log y=5\\ &\Leftrightarrow \: \: ^{^{^{2}}}\log y=5-2=3\Leftrightarrow y=2^{3}=8\\ &\textrm{Jadi},\: \: x.y=64.8=\color{red}512 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 81.&\textrm{Tentukanlah nilai dari}\\ &\textrm{a}.\quad \left ( 2^{\: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: ^{^{^{9}}}\log 5} \right )\left ( 5^{\: ^{^{^{\frac{1}{5}}}}\log 2} \right )\\ &\textrm{b}.\quad \left ( ^{27}\log 125 \right )\left ( ^{25}\log \displaystyle \frac{1}{64} \right )\left ( ^{64}\log \frac{1}{9} \right )\\ &\textrm{c}.\quad \left ( ^{625}\log 19 \right )\left ( ^{7}\log \displaystyle \frac{1}{25} \right )\left ( ^{19}\log 7 \right )\\\\ &\textrm{Jawab}:\\ &\begin{array}{|l|}\hline \begin{aligned}\textrm{a}.\quad &\left ( 2^{\: \: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: ^{^{^{9}}}\log 5} \right )\left ( 5^{\: ^{^{^{\frac{1}{5}}}}\log 2} \right )\\ &=\left ( 2^{\: \: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: \: ^{^{^{3^{2}}}}\log 5} \right )\left ( 5^{\: \: ^{^{^{5^{-1}}}}\log 2} \right )\\ &=\left ( 2^{\: \: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: \: ^{^{^{3}}}\log 5^{^{\frac{1}{2}}}} \right )\left ( 5^{\: \: ^{^{^{5}}}\log 2^{-1}} \right )\\ &=\left ( 2^{\: ^{^{2}}\log 6} \right )\left ( 3^{\: ^{^{3}}\log \sqrt{5}} \right )\left ( 5^{\: ^{^{5}}\log \frac{1}{2}} \right )\\ &=6\times \sqrt{5}\times \frac{1}{2}\\ &=\color{red}3\sqrt{5} \end{aligned}\\\hline \end{array}\\ &\begin{array}{|l|}\hline \begin{aligned}\textrm{b}.\quad &\left ( ^{27}\log 125 \right )\left ( ^{25}\log \displaystyle \frac{1}{64} \right )\left ( ^{64}\log \frac{1}{9} \right )\\ &=\left ( ^{^{3^{3}}}\log 5^{3} \right )\left ( ^{^{5^{2}}}\log 4^{-3} \right )\left ( ^{^{4^{3}}}\log 3^{-2} \right )\\ &=\displaystyle \frac{3}{3}.\left ( -\frac{3}{2} \right ).\left ( -\frac{2}{3} \right ).\: ^{^{3}}\log 5.\: ^{^{5}}\log 4.\: ^{^{4}}\log 3\\ &=\color{red}1\end{aligned}\\\hline \end{array}\\ &\color{blue}\textrm{Pembahasan diserahkan kepada}\\ &\color{blue}\textrm{Pembaca yang budiman untuk poin c} \end{array}$

$\begin{array}{ll}\\ 82.&\textrm{Tentukanlah nilai}\: \: a+b\: \: \textrm{dimana}\: \: a\: \: \textrm{dan}\: \: b\\ &\textrm{adalah bilangan riil positif}.\\ &^{7}\log \left ( 1+a^{2} \right )-\: ^{7}\log 25=\: ^{7}\log \left ( 2ab-15 \right )-\: ^{7}\log \left ( 25+b^{2} \right )\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&^{7}\log \displaystyle \frac{\left ( 1+a^{2} \right )}{25}=\: ^{7}\log \displaystyle \frac{\left ( 2ab-15 \right )}{\left ( 25+b^{2} \right )}\\ &\textrm{diambil}\: \textrm{persamaannya, maka}\\ &\displaystyle \frac{\left ( 1+a^{2} \right )}{25}=\displaystyle \frac{\left ( 2ab-15 \right )}{\left ( 25+b^{2} \right )}\\ &\displaystyle \left ( 1+a^{2} \right )\left ( 25+b^{2} \right )=25\left ( 2ab-15 \right )\\ &\begin{cases} \left ( 1+a^{2} \right ) & \text{ faktor dari} \: \: 25,\: \: a> 0,\: a\in \mathbb{R} \\ &\textrm{atau}\\ \left ( 25+b^{2} \right ) & \text{ faktor dari }\: \: 25,\: \: b> 0,\: b\in \mathbb{R}\: \: \: \textrm{juga} \end{cases}\\ \end{aligned}\\\\ &\color{red}\begin{array}{|c|l|c|l|c|c|c|}\hline \textrm{No}&a&\left ( 1+a^{2} \right )&b&\left ( 25+b^{2} \right )&\textrm{Keterangan}&a+b\\\hline 1&2&5&10&125&\textrm{Memenuhi}&12\\\hline 2&7&50&\cdots &\cdots &\textrm{tidak}&\cdots \\\hline 3&\cdots &\cdots &5&50&\textrm{tidak}&\cdots \\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 83.&\textrm{Jika}\: \: 60^{a}=3\: \: \textrm{dan}\: \: 60^{b}=5\\ &\textrm{maka hasil dari}\: \: 12^{^{\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )}}\\\\ &\textrm{Jawab}:\\ &\textrm{Perhatikanlah bahwa}\\ &\color{blue}\begin{aligned}&\begin{cases} 60^{a}=3 & \Rightarrow \: ^{60}\log 3=a \\ 60^{b}=5 & \Rightarrow \: ^{60}\log 5=b \end{cases} \end{aligned} \\ &\color{blue}\textrm{Selanjutnya}\\ &\color{blue}\textrm{Untuk}\: \: (1-a-b),\: \textrm{maka}\\ &\begin{aligned}1-a-b&=1-\: ^{60}\log 3-\: ^{60}\log 5\\ &=\: ^{60}\log 60-\: ^{60}\log 3-\: ^{60}\log 5\\ &=\: ^{60}\log \displaystyle \frac{60}{3\times 5}\\ &=\: ^{60}\log 4\\ &=\: ^{60}\log 2^{2} \end{aligned} \\ &\textrm{Untuk}\: \: 2(1-b),\: \textrm{maka}\\ &\begin{aligned}2(1-b)&=2\left ( 1-\: ^{60}\log 5 \right )\\ &=2\left ( ^{60}\log 60-\: ^{60}\log 5 \right )\\ &=2\left ( ^{60}\log \displaystyle \frac{60}{5} \right )\\ &=2\left ( ^{60}\log 12 \right )\\ &=\: ^{60}\log 12^{2} \end{aligned} \\ &\textrm{Untuk}\: \: \left ( \displaystyle \frac{1-x-y}{2(1-b)} \right ),\: \textrm{maka}\\ &\begin{aligned}\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )&=\displaystyle \frac{\: ^{60}\log 2^{2}}{\: ^{60}\log 12^{2}}\\ &=\displaystyle \frac{2\: ^{60}\log 2}{2\: ^{60}\log 12}\\ &=\displaystyle \frac{\: ^{60}\log 2}{\: ^{60}\log 12}\\ &=\: ^{12}\log 2 \end{aligned}\\ &\color{blue}\textrm{Jadi},\: \: 12^{^{\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )}}=12^{^{\: ^{12}\log 2}}=\color{red}2 \end{array}$

$\begin{array}{ll}\\ 84.&\textrm{Diberikan bilangan riil positif}\: \: x,\: y,\: \textrm{dan}\: z\\ & \textrm{yang memenuhi persamaan}\\ &2\: ^{x}\log (2y)=2\: ^{2x}\log (4z)=2\: ^{4x}\log (8yz)\neq 0.\\ &\textrm{Jika nilai}\: \: xy^{5}z\: \: \textrm{dapat dinyatakan dengan}\: \: \displaystyle \frac{1}{2^{\displaystyle \frac{p}{q}}}\\ & \textrm{dengan}\: \: p\: \: \textrm{dan}\: \: q\: \: \textrm{bilangan asli yang saling prima},\\ &\textrm{maka nilai dari}\: \: p+q=....\\\\ &\textrm{Jawab}:\\\\ &\begin{aligned}&2\: ^{x}\log (2y)=2\: ^{2x}\log (4z)=2\: ^{4x}\log (8yz)\neq 0\\ &\textrm{maka}\\ &^{x}\log (2y)=\: ^{2x}\log (4y)\\ &\Rightarrow \quad \log (2y)\times \log (2x)=\log x\times \log (4y)...(1)\\ &^{x}\log (2y)=\: ^{4x}\log (8yz)\\ &\Rightarrow \quad \log (2y)\times \log (4x)=\log x\times \log (8yz)....(2)\\ &^{2x}\log (4y)=\: ^{4x}\log (8yz)\\ &\Rightarrow \quad \log (4y)\times \log (4x)=\log (2x)\times \log (8yz)....(3) \end{aligned}\\\\ &\begin{aligned}&\textrm{Perhatikan persamaan}\: \: (2),\: \textrm{yaitu}:\\ &\log (2y)\times \log (4x)=\log x\times \log (8yz)\\ &\log (2y)\times \left (\log (2x)+\log 2 \right )=\log x\times \log (8yz)\\ &\log (2y)\times \log (2x)+\log (2y)\times \log 2=\log x\times \log (8yz)\\ &\log x\times \log (4y)+\log (2y)\times \log 2=\log x\times \log (8yz)\\ &\quad\textrm{persamaan di atas, persamaan}\: \: (1)\: \: \textrm{disubstitusikan}\\ &\log (2y)=\displaystyle \frac{\log x\times \log (8yz)-\log x\times \log (4y)}{\log 2}\\ &\log (2y)=\displaystyle \frac{\log x\times \left ( \log \displaystyle \frac{8yz}{4y} \right )}{\log 2}\\ &\log (2y)=\color{red}\displaystyle \frac{\log x\times \log (2z)}{\log 2}\: \color{black}......(4) \end{aligned}\\\\ &\begin{aligned}&\textrm{Perhatikan juga persamaan}\: \: (3),\: \textrm{yaitu}:\\ &\log (4y)\times \log (4x)=\log (2x)\times \log (8yz)\\ &\left (\log (2y)+\log 2 \right )\times \log (4x)=\log (2x)\times \log (8yz)\\ &\log (2y)\times \log (4x)+\log 2\times \log (4x)=\log (2x)\times \log (8yz)\\ &\log x\times \log (8yz)+\log 2\times \log (4x)=\log (2x)\times \log (8yz)\\ &\quad\textrm{di atas, persamaan}\: \: (2)\: \: \textrm{disubstitusikan}\\ &\log 2\times \log (4x)=\log (2x)\times \log (8yz)-\log x\times \log (8yz)\\ &\log 2\times \log (4x)=\log (8yz)\times \left ( \log (2x)-\log x \right )\\ &\log 2\times \log (4x)=\log (8yz)\times \left ( \log \displaystyle \frac{2x}{x} \right )\\ &\log 2\times \log (4x)=\log (8yz)\times \log 2\\ &\log 4x=\log (8yz)\\ &4x=8yz\\ &\displaystyle \frac{x}{z}=\color{red}2y\: \color{black}....(5) \end{aligned} \end{array}$

$.\: \: \qquad\begin{aligned}&\textrm{dari persamaan}\: \: (4)\: \: \textrm{dan}\: \: (5)\\ &\log (2y)=\displaystyle \frac{\log x\times \log (2z)}{\log 2}\\ &\log \left ( \displaystyle \frac{x}{z} \right )=\displaystyle \frac{\log x\times \log (2z)}{\log 2}\\ &\log 2\left ( \log x-\log z \right )=\log x\times \log (2z)\\ &\log 2\times \log x-\log 2\times \log z=\log x\times \left ( \log 2+\log z \right )\\ &\log 2\times \log x-\log 2\times \log z=\log x\times \log 2+\log x\times \log z\\ &-\log 2\times \log z=\log x\times \log z\\ &\log 2^{-1}=\log x\\ &\displaystyle \frac{1}{2}=\color{red}x\: \color{black}.....(6) \end{aligned}$

$.\: \: \qquad\begin{array}{|c|c|}\hline \textrm{persamaan}\: \: (2)&\textrm{Menentukan nilai}\: \: z\\\hline \begin{aligned} \log 2y\times \log (4x)&=\log x\times \log (8yz)\\ \log 2y\times \log (4(2yz))&=\log x\times \log (8yz)\\ \log 2y\times \log (8yz)&=\log x\times \log (8yz)\\ \log (2y)&=\log x\\ 2y&=x\\ y&=\displaystyle \frac{1}{2}x\\ &=\displaystyle \frac{1}{2}\times \frac{1}{2}\\ &=\color{red}\displaystyle \frac{1}{4}\: \color{black}.....(7)\\ & \end{aligned}&\begin{aligned}&\\ x&=2yz\\ \displaystyle \frac{1}{2}&=2\left ( \displaystyle \frac{1}{4} \right )z\\ 1&=z\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array}$

$.\: \: \qquad\begin{aligned}&\textrm{maka nilai untuk}\: \: xy^{5}z\: \: \textrm{adalah}\\ &xy^{5}z=\left ( \displaystyle \frac{1}{2} \right ).\left ( \displaystyle \frac{1}{4} \right )^{5}.1\\ &=\displaystyle \frac{1}{2\times 4^{5}}\\ &=\displaystyle \frac{1}{2\times \left ( 2^{2} \right )^{5}}=\displaystyle \frac{1}{2^{1+10}}\\ &=\displaystyle \frac{1}{2^{11}}=\displaystyle \frac{1}{2^{^{\frac{11}{1}}}}=\displaystyle \frac{1}{2^{^{\frac{p}{q}}}}\\ &\begin{cases} p & =11 \\ q & =1 \end{cases}\quad \textrm{dan jelas bahwa} \: \: p\: \: \textrm{dan}\: \: q\: \: \textrm{saling prima}\\ &\textrm{Jadi},\\ &p+q=11+1=\color{red}12 \end{aligned}$

DAFTAR PUSTAKA

1. Idris, M., Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
2. Sembiring, S. 2002. Olimpiade Matematika untuk SMU. Bandung: YRAMA WIDYA.

 

Latihan Soal 7 Persiapan PAS Gasal Matematika Peminatan Kelas X (Fungsi Eksponen dan Fungsi Logaritma)

 $\begin{array}{ll}\\ 61.&\textrm{Nilai dari}\\ & ^{2}\log \displaystyle \frac{4}{3}+\: 2.\, ^{2}\log \sqrt{12}\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 6\\ \color{red}\textrm{b}.&\displaystyle 4\\ \textrm{c}.&3 \displaystyle \frac{1}{2}\\ \textrm{d}.& 2\displaystyle \frac{1}{2}\\ \textrm{e}.& \displaystyle 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&=\: ^{2}\log \displaystyle \frac{4}{3}+\: 2.\, ^{2}\log \sqrt{12}\\ &=\: ^{2}\log \frac{4}{3}+\: ^{2}\log \left ( \sqrt{12} \right )^{2}\\ &=\: ^{2}\log \frac{4}{3}+\: ^{2}\log 12\\ &=\: ^{2}\log \frac{4}{3}\times 12\\ &=\: ^{2}\log 16\\ &=\: ^{2}\log 2^{4}\\ &=\color{red}4 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 62.&\textrm{Nilai dari}\\ & \displaystyle \frac{1}{6}.\, ^{2}\log 25-\: \frac{1}{3}.\, ^{2}\log 10\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -3\\ \textrm{b}.&\displaystyle -2\\ \textrm{c}.&-2 \displaystyle \frac{1}{2}\\ \textrm{d}.& -\displaystyle \frac{1}{2}\\ \color{red}\textrm{e}.&- \displaystyle \frac{1}{3} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&=\: \displaystyle \frac{1}{6}.\, ^{2}\log 25-\: \frac{1}{3}.\, ^{2}\log 10\\ &=\displaystyle \frac{1}{3}.\frac{1}{2}.\, ^{2}\log 25-\: \frac{1}{3}.\, ^{2}\log 10\\ &=\frac{1}{3}\left ( ^{2}\log 25^{\frac{1}{2}} -\: ^{2}\log 10\right )\\ &=\frac{1}{3}\left ( ^{2}\log 5-\: ^{2}\log 10 \right )\\ &=\frac{1}{3}\left ( ^{2}\log \frac{5}{10} \right )\\ &=\frac{1}{3}\left ( ^{2}\log \frac{1}{2} \right )\\ &=\frac{1}{3}\left ( ^{2}\log 2^{-1} \right )\\ &=\color{red}-\frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 63.&\textrm{Nilai dari}\\ & ^{2}\log \left ( ^{2}\log \sqrt{\sqrt{2}} \right )\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -4\\ \color{red}\textrm{b}.&\displaystyle -2\\ \textrm{c}.&-1 \displaystyle \frac{1}{2}\\ \textrm{d}.& -\displaystyle \frac{1}{2}\\ \textrm{e}.&- \displaystyle \frac{1}{4} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&=\: ^{2}\log \left ( ^{2}\log \sqrt{\sqrt{2}} \right )\\ &=\: ^{2}\log \left ( ^{2}\log 2^{\frac{1}{4}} \right )\\ &=\: ^{2}\log \frac{1}{4}\\ &=\: ^{2}\log \left (2 \right )^{-2}\\ &=\color{red}-2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 64.&\textbf{(UMPTN '99)}\\ &\textrm{Diketahui}\: \: \log 2=0,3010\: \: \textrm{dan}\: \: \log 3=0,4771\\ &\textrm{maka}\: \: \log \left ( \sqrt[3]{2}\times \sqrt{3} \right )\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 0,1505\\ \textrm{b}.&\displaystyle 0,1590\\ \textrm{c}.&\displaystyle 0,2007\\ \color{red}\textrm{d}.&\displaystyle 0,3389\\ \textrm{e}.&\displaystyle 0,3891 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\log \left ( \sqrt[3]{2}\times \sqrt{3} \right )\\ &=\log \sqrt[3]{2}+\log \sqrt{3}\\ &=\log 2^{\frac{1}{3}}+\log 3^{\frac{1}{2}}\\ &=\displaystyle \frac{1}{3}\log 2+\displaystyle \frac{1}{2}\log 3\\ &=\displaystyle \frac{1}{3}(0,3010)+\displaystyle \frac{1}{2}(0,4771)\\ &=0,1003+0,2386\\ &=\color{red}0,3389 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 65.&\textbf{(UMPTN '98)}\\ &\textrm{Nilai}\: \: ^{a}\log \displaystyle \frac{1}{b}\times \: ^{b}\log \displaystyle \frac{1}{c^{2}}\times \: ^{c}\log \displaystyle \frac{1}{a^{3}}\: =\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle -6\\ \textrm{b}.&\displaystyle 6\\ \textrm{c}.&\displaystyle \frac{b}{a^{2}c}\\ \textrm{d}.&\displaystyle \frac{a^{2}c}{b}\\ \textrm{e}.&\displaystyle -\frac{1}{6} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&^{a}\log \displaystyle \frac{1}{b}\times \: ^{b}\log \displaystyle \frac{1}{c^{2}}\times \: ^{c}\log \displaystyle \frac{1}{a^{3}}\\ &=\: ^{a}\log \displaystyle b^{-1}\times \: ^{b}\log \displaystyle c^{-2}\times \: ^{c}\log \displaystyle a^{-3}\\ &=(-1).(-2).(-3)\times \: ^{a}\log \displaystyle a\times \: ^{b}\log \displaystyle c\times \: ^{c}\log \displaystyle a\\ &=-6\times \: ^{a}\log a\\ &=-6\times 1\\ &=\color{red}-6 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 66.&\textbf{(UMPTN '01)}\\ &\textrm{Jika}\: \: \displaystyle \frac{^{2}\log a}{^{3}\log b}=m\: \: \textrm{dan}\: \: \displaystyle \frac{^{3}\log a}{^{2}\log b}=n\\ &\textrm{dengan}\: \: a> 1,\: b> 1,\: \textrm{maka}\: \: \displaystyle \frac{m}{n}=....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle ^{2}\log 3\\ \textrm{b}.&\displaystyle ^{3}\log 2\\ \textrm{c}.&\displaystyle ^{4}\log 9\\ \textrm{d}.&\displaystyle \left ( ^{3}\log 2 \right )^{2}\\ \color{red}\textrm{e}.&\displaystyle \left ( ^{2}\log 3 \right )^{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}\displaystyle \frac{m}{n}&=\displaystyle \frac{\displaystyle \frac{^{2}\log a}{^{3}\log b}}{\displaystyle \frac{^{3}\log a}{^{2}\log b}}\\ &=\displaystyle \frac{^{2}\log a\times \: ^{2}\log b}{^{3}\log b\times \: ^{3}\log a}\\ &=\displaystyle \frac{^{2}\log a\times \: \displaystyle \frac{1}{^{b}\log 2}}{^{3}\log b\times \: \displaystyle \frac{1}{^{a}\log 3}}\\ &=\displaystyle \frac{^{2}\log a\times \: ^{a}\log 3}{^{3}\log b\times \: ^{b}\log 2}\\ &=\displaystyle \frac{^{2}\log 3}{^{3}\log 2}=\displaystyle \frac{^{2}\log 3}{\displaystyle \frac{1}{^{2}\log 3}}\\ &=\color{red}\left ( ^{2}\log 3 \right )^{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 67.&\textbf{(UMPTN '01)}\\ &\textrm{Jika}\: \: ^{10}\log x=b\: ,\: \textrm{maka}\: \: ^{10x}\log 100=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{b+1}\\ \color{red}\textrm{b}.&\displaystyle \frac{2}{b+1}\\ \textrm{c}.&\displaystyle \frac{1}{b}\\ \textrm{d}.&\displaystyle \frac{2}{b}\\ \textrm{e}.&\displaystyle \frac{2}{10b} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&^{10x}\log 100\\ &=\displaystyle \frac{\log 100}{\log 10x}\\ &=\displaystyle \frac{^{10}\log 100}{^{10}\log 10x},\quad \color{magenta}\textrm{pilih basis 10}\\ &\color{purple}\textrm{alasannya: supaya sama dengan soal}\\ &=\displaystyle \frac{^{10}\log 10^{2}}{^{10}\log 10+\: ^{10}\log x}\\ &=\displaystyle \frac{2}{1+b}\: \: \textrm{atau}\\ &=\color{red}\frac{2}{b+1} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 68.&\textbf{(UM UGM '03)}\\ &\textrm{Jika}\: \: ^{4}\log 6=m+1\: ,\: \textrm{maka}\: \: ^{9}\log 8=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{3}{4m-2}\\ \color{red}\textrm{b}.&\displaystyle \frac{3}{4m+2}\\ \textrm{c}.&\displaystyle \frac{3}{2m+4}\\ \textrm{d}.&\displaystyle \frac{3}{2m-4}\\ \textrm{e}.&\displaystyle \frac{3}{2m+2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\color{purple}\textrm{Sebelumnya perhatikanlah}\\ &^{4}\log 6=m+1\\ &\Leftrightarrow \: ^{2^{2}}\log (2.3)^{1}=m+1\\ &\Leftrightarrow \: \displaystyle \frac{1}{2}\times \: ^{2}\log (2.3)=m+1\\ &\Leftrightarrow \: \displaystyle \frac{1}{2}\times \: \left (^{2}\log 2 +\: ^{2}\log 3 \right )=m+1\\ &\Leftrightarrow \: \displaystyle \frac{1}{2}\times \: \left (1 +\: ^{2}\log 3 \right )=m+1\\ &\Leftrightarrow \: 1 +\: ^{2}\log 3=2m+2\\ &\Leftrightarrow \: ^{2}\log 3=2m+1\\ &\color{red}\textrm{Selanjutnya adalah}:\\ &^{9}\log 8=\: \displaystyle \frac{1}{^{8}\log 9}\\ &=\: \displaystyle \frac{1}{^{2^{3}}\log 3^{2}}\\ &=\: \displaystyle \frac{1}{\displaystyle \frac{2}{3}\: ^{2}\log 3}\\ &=\: \displaystyle \frac{3}{2\: ^{2}\log 3}\\ &=\: \displaystyle \frac{3}{2(2m+1)}\\ &=\: \color{red}\displaystyle \frac{3}{4m+2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 69.&\textbf{(UMPTN '00)}\\ &\textrm{Jika}\: \: ^{3}\log 5=p\: \: \textrm{dan}\: \: ^{3}\log 4=q,\\ &\textrm{maka}\: \: ^{4}\log 15=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{pq}{1+p}\\ \textrm{b}.&\displaystyle \frac{p+q}{pq}\\ \color{red}\textrm{c}.&\displaystyle \frac{p+1}{pq}\\ \textrm{d}.&\displaystyle \frac{p+1}{q+1}\\ \textrm{e}.&\displaystyle \frac{pq}{1-p} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&^{4}\log 15\\ &=\displaystyle \frac{^{...}\log 15}{^{...}\log 4},\: \: \color{blue}\textrm{pilih basis 5}\\ &\color{blue}\textrm{mengapa tidak pilih basis selain 5}\\ &\color{blue}\textrm{lihat penyebut, di sana terdapat numerus 4}\\ &\color{blue}\textrm{pada soal, pasangan numerus 4 adalah 5},\\ &\textrm{makanya basis 5 dipilih, bukan yang lain}\\ &=\displaystyle \frac{^{5}\log 15}{^{5}\log 4}=\displaystyle \frac{^{5}\log (3.5)}{^{5}\log 4}\\ &=\displaystyle \frac{^{5}\log 3+\: ^{5}\log 5}{^{5}\log 4}=\displaystyle \frac{\displaystyle \frac{1}{^{3}\log 5}+\: ^{5}\log 5}{^{5}\log 4}\\ &=\displaystyle \frac{\displaystyle \frac{1}{p}+1}{q}=\displaystyle \frac{1+p}{pq},\: \: \textrm{atau}\\ &=\color{red}\displaystyle \frac{p+1}{pq} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 70.&\textbf{(UMPTN '94)}\\ &\textrm{Jika}\: \: ^{6}\log 5=a\: \: \textrm{dan}\: \: ^{5}\log 4=b,\\ &\textrm{maka}\: \: ^{4}\log 0,24=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{a+2}{ab}\\ \textrm{b}.&\displaystyle \frac{2a+1}{ab}\\ \textrm{c}.&\displaystyle \frac{a-2}{ab}\\ \textrm{d}.&\displaystyle \frac{2a+1}{2ab}\\ \color{red}\textrm{e}.&\displaystyle \frac{1-2a}{ab} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&^{4}\log 0,24\\ &=\displaystyle \frac{^{...}\log 0,24}{^{...}\log 4}=\frac{^{...}\log \displaystyle \frac{6}{25}}{^{...}\log 4},\: \: \color{purple}\textrm{pilih basis 5}\\ &\color{red}\textrm{mengapa tidak pilih basis selain 5}\\ &\color{red}\textrm{lihat penyebut, di sana terdapat numerus 4}\\ &\color{red}\textrm{pada soal, pasangan numerus 4 adalah 5},\\ &\textrm{makanya basis 5 dipilih, bukan yang lain}\\ &=\frac{^{5}\log \displaystyle \frac{6}{25}}{^{5}\log 4}=\displaystyle \frac{^{5}\log 6-\: ^{5}\log 25}{^{5}\log 4}\\ &=\displaystyle \frac{\displaystyle \frac{1}{^{6}\log 5}-\: ^{5}\log 5^{2}}{^{5}\log 4}=\displaystyle \frac{\displaystyle \frac{1}{a}-2}{b}=\color{red}\frac{1-2a}{ab} \end{aligned} \end{array}$.

Latihan Soal 6 Persiapan PAS Gasal Matematika Peminatan Kelas X (Fungsi Eksponen dan Fungsi Logaritma)

 $\begin{array}{ll}\\ 52.&\textrm{Nilai dari}\: \: ^{2}\log \displaystyle \frac{1}{32}\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -7\\ \color{red}\textrm{b}.&\displaystyle -5\\ \textrm{c}.& \displaystyle -3\\ \textrm{d}.& \displaystyle -2\\ \textrm{e}.& 5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&=\: ^{2}\log \displaystyle \frac{1}{32}\\ &=\: ^{2^{1}}\log 2^{-5}\\ &=\displaystyle \frac{-5}{1}\times \: ^{2}\log 2\\ &=\color{red}-5\\ &\\ & \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 53.&\textrm{Nilai dari}\: \: ^{0,333...}\log 0,111....\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{3}\\ \textrm{b}.&\displaystyle \frac{1}{2}\\ \color{red}\textrm{c}.& 2\\ \textrm{d}.& 3\\ \textrm{e}.& 6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&=\: ^{0,333...}\log 0,111...\\ &=\: ^{\frac{1}{3}}\log \frac{1}{9}\\ &=\: ^{\frac{1}{3}}\log \left (\frac{1}{3} \right )^{2}\\ &=\color{red}2 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 54.&\textrm{Nilai dari}\: \: ^{5}\log 25\sqrt{5}\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{5}{2}\\ \textrm{b}.&\displaystyle \frac{3}{2}\\ \textrm{c}.& \displaystyle \frac{1}{2}\\ \textrm{d}.& \displaystyle 2\\ \textrm{e}.& 3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&=\: ^{5}\log 25\sqrt{5}\\ &=\: ^{5^{1}}\log 5^{2}.5^{\frac{1}{2}}\\ &=\: ^{5^{1}}\log 5^{\frac{5}{2}}\\ &=\displaystyle \frac{\frac{5}{2}}{1}\times \: ^{5}\log 5\\ &=\color{red}\displaystyle \frac{5}{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 55.&\textrm{Nilai dari}\: \: ^{\sqrt{3}}\log 81\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 12\\ \textrm{b}.&\displaystyle 10\\ \textrm{c}.& \displaystyle 9\\ \color{red}\textrm{d}.& \displaystyle 8\\ \textrm{e}.& 6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&=\: ^{\sqrt{3}}\log 81\\ &=\: ^{\displaystyle 3^{\frac{1}{2}}}\log 3^{4}\\ &=\displaystyle \frac{4}{\frac{1}{2}}\times \: ^{3}\log 3\\ &=\color{red}8 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 56.&\textrm{Nilai dari}\: \: ^{\frac{1}{3}}\log \displaystyle \frac{1}{243}\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 6\\ \color{red}\textrm{b}.&\displaystyle 5\\ \textrm{c}.& \displaystyle 4\\ \textrm{d}.& \displaystyle 3\\ \textrm{e}.& \displaystyle 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&=\: ^{\frac{1}{3}}\log \displaystyle \frac{1}{243}\\ &=\: ^{\left (\frac{1}{3} \right )^{1}}\log \displaystyle \left (\frac{1}{3} \right )^{5}\\ &=\displaystyle \frac{5}{1}\times \: ^{\frac{1}{3}}\log \frac{1}{3}\\ &=\color{red}5 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 57.&\textrm{Nilai dari}\: \: ^{\sqrt{2}}\log 16\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 10\\ \textrm{b}.&\displaystyle 9\\ \color{red}\textrm{c}.& \displaystyle 8\\ \textrm{d}.& \displaystyle 6\\ \textrm{e}.& \displaystyle 4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&=\: ^{\sqrt{2}}\log 16\\ &=\: ^{\displaystyle 2^{\frac{1}{2}}}\log 2^{4}\\ &=\displaystyle \frac{4}{\frac{1}{2}}\times \: ^{2}\log 2\\ &=\color{red}8\end{aligned} \end{array}$

$\begin{array}{ll}\\ 58.&\textrm{Nilai dari}\: \: ^{\sqrt{5}}\log \sqrt{125}\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -3\\ \textrm{b}.&\displaystyle -2\\ \textrm{c}.& \displaystyle 2\\ \color{red}\textrm{d}.& \displaystyle 3\\ \textrm{e}.& \displaystyle 5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&=\: ^{\sqrt{5}}\log \sqrt{125}\\ &=\: ^{\sqrt{5}^{1}}\log \left ( \sqrt{5} \right )^{3}\\ &=\displaystyle \frac{3}{1}\times \: ^{\sqrt{5}}\log \sqrt{5}\\ &=\color{red}3\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 59.&\textrm{Nilai dari}\: \: ^{\sqrt{\sqrt{2}}}\log \sqrt{8\sqrt{8}}\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 4\\ \textrm{b}.&\displaystyle 6\\ \textrm{c}.& \displaystyle 8\\ \color{red}\textrm{d}.& \displaystyle 9\\ \textrm{e}.& \displaystyle 12 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&=\: ^{\sqrt{\sqrt{2}}}\log \sqrt{8\sqrt{8}}\\ &=\: ^{\sqrt[4]{2}}\log \left ( 8\left ( 8 \right )^{\frac{1}{2}} \right )^{\frac{1}{2}}\\ &=\: ^{2^{\frac{1}{4}}}\log 8^{\left (\frac{1}{2}+\frac{1}{4} \right )}\\ &=\: ^{2^{\frac{1}{4}}}\log 2^{3\left ( \frac{3}{4} \right )}\\ &=\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}\times \: ^{2}\log 2\\ &=\color{red}9 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 60.&\textrm{Nilai dari}\\ & ^{6}\log 8 +\: ^{6}\log \displaystyle \frac{9}{2}\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 4\\ \textrm{b}.&\displaystyle 3\\ \textrm{c}.&3 \displaystyle \frac{1}{2}\\ \textrm{d}.& 2\displaystyle \frac{1}{2}\\ \color{red}\textrm{e}.& \displaystyle 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&=\: ^{6}\log 8 +\: ^{6}\log \displaystyle \frac{9}{2}\\ &=\: ^{6}\log 8\times \frac{9}{2}\\ &=\: ^{6}\log 36\\ &=\: ^{6}\log 6^{2}\\ &=\color{red}2\\ &\\ &\\ & \end{aligned} \end{array}$

Latihan Soal 5 Persiapan PAS Gasal Matematika Peminatan Kelas X (Fungsi Eksponen dan Fungsi Logaritma)

 $\begin{array}{ll}\\ 41.&\textrm{Penyelesaian pertidaksamaan eksponen}\\ &\left ( \displaystyle \frac{1}{3} \right )^{2x+1}>\sqrt{\displaystyle \frac{27}{3^{x-1}}} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&x>\displaystyle \frac{6}{5}\\ \textrm{b}.&x<-\displaystyle \frac{6}{5}\\ \textrm{c}.&x>\displaystyle \frac{5}{6}\\ \color{red}\textrm{d}.&x<-2\\ \textrm{e}.&x<2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\left ( \displaystyle \frac{1}{3} \right )^{2x+1}&>\sqrt{\displaystyle \frac{27}{3^{x-1}}}\\ 3^{-(2x+1)}&>3^{\frac{1}{2}(3-(x-1))}\\ -(2x+1)&>\displaystyle \frac{1}{2}(3-(x-1))\\ -4x-2&>4-x\\ -4x+x&>4+2\\ -3x&>6\\ 3x&<-6\\ x&\color{red}<-2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 42.&(\textbf{UMPTN 01})\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &4^{x^{2}-x-2}.2^{x^{2}+3x-10}<\displaystyle \frac{1}{16} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&x<-5\: \: \textrm{atau}\: \: x>-2\\ \textrm{b}.&x<-2\: \: \textrm{atau}\: \: x>\displaystyle \frac{5}{3}\\ \textrm{c}.&-2<x<-1\\ \color{red}\textrm{d}.&-2<x<\displaystyle \frac{5}{3}\\ \textrm{e}.&-5<x<2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}4^{x^{2}-x-2}.2^{x^{2}+3x-10}&<\displaystyle \frac{1}{16}\\ 2^{2\left ( x^{2}-x-2 \right )+\left ( x^{2}+3x-10 \right )}&<2^{-4}\\ 2\left ( x^{2}-x-2 \right )+\left ( x^{2}+3x-10 \right )&<-4\\ 3x^{2}-2x+3x-4-10+4&<0\\ 3x^{2}+x-10+&<0\\ (x+2)(3x-5)&<0\\ \therefore \qquad-2<x&<\displaystyle \frac{5}{3} \end{aligned} \end{array}$

$\begin{array}{l}\\ 43.&(\textbf{SPMB 04 Mat IPA})\textrm{Himpunan Penyelesaian}\\ & \textrm{pertidaksamaan eksponen}\\ &2\sqrt{4^{x^{2}-3x+2}}<\sqrt[3]{\left (\displaystyle \frac{1}{2} \right )^{3-6x}} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x|x> 4 \right \}\\ \textrm{b}.&\left \{ x|x> 2 \right \}\\ \textrm{c}.&\left \{ x|x<1 \right \}\\ \color{red}\textrm{d}.&\left \{ x|1<x<4 \right \}\\ \textrm{e}.&\left \{ x|2\leq x\leq 3 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}2\sqrt{4^{x^{2}-3x+2}}&<\sqrt[3]{\left (\displaystyle \frac{1}{2} \right )^{3-6x}}\\ 2^{1+\frac{2}{2}\left ( x^{2}-3x+2 \right )}&<2^{- \frac{3-6x}{3}}\\ 1+\left ( x^{2}-3x+2 \right )&<-\displaystyle \frac{3-6x}{3}\\ x^{2}-3x+3&<-1+2x\\ x^{2}-5x+4&<0\\ (x-1)(x-4)&<0\\ 1<x&<4 \end{aligned} \end{array}$

$\begin{array}{l}\\ 44.&(\textbf{SBMPTN 2015 Mat IPA})\\ &\textrm{Nilai}\: \: c\: \: \textrm{yang memenuhi}\\ &(0,12)^{4x^{2}+8x+c}<(0,0144)^{x^{2}+4x+4} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&c>0\\ \textrm{b}.&c>2\\ \textrm{c}.&c>4\\ \textrm{d}.&c>6\\ \color{red}\textrm{e}.&c>8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}(0,12)^{4x^{2}+8x+c}&<(0,0144)^{x^{2}+4x+4}\\ (0,12)^{4x^{2}+8x+c}&<(0,12)^{2\left (x^{2}+4x+4 \right )}\\ 4x^{2}+8x+c&>2\left (x^{2}+4x+4 \right )\\ 2x^{2}+c-8&>0\quad \color{magenta}\textrm{haruslah definit positif}\\ \textrm{Syaratnya}&\begin{cases} a &=2>0 \\ D &=b^{2}-4ac<0\\ \end{cases}\\ \textrm{Maka}\quad D&=b^{2}-4ac<0\\ \textbf{ambil dari}&\: \: \: 2x^{2}-c-8=0\begin{cases} a &=2 \\ b &=0 \\ c &=c-8 \end{cases}\\ &=0^{2}-4.2(c-8)<0\\ -8c&+64<0\\ -8c&<-64\\ 8c&>64\\ c&\color{red}>8 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 45.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &9^{2x}-10.9^{x}+9>0, \: \: x\in \mathbb{R}\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&x<1\: \: \textrm{atau}\: \: x>0\\ \color{red}\textrm{b}.&x<0\: \: \textrm{atau}\: \: x>1\\ \textrm{c}.&x<-1\: \: \textrm{atau}\: \: x>2\\ \textrm{d}.&x<1\: \: \textrm{atau}\: \: x>2\\ \textrm{e}.&x<-1\: \: \textrm{atau}\: \: x>1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}9^{2x}-10.9^{x}+9&>0\\ \left (9^{x} \right )^{2}-10.\left ( 9^{x} \right )+9&>0\\ \left ( 9^{x}-1 \right )\left ( 9^{x}-9 \right )&>0\\ 9^{x}<1\: \: \textrm{atau}\: \: 9^{x}&>9\\ 9^{x}<9^{0}\: \: \textrm{atau}\: \: 9^{x}&>9^{1}\\ x<0\: \: \textrm{atau}\: \: x&\color{red}>1 \end{aligned} \end{array}$

$\begin{array}{l}\\ 46.&\textrm{Jumlah akar-akar persamaan}\\ & 5^{x+1}+5^{2-x}-30=0\: \: \textrm{adalah}\: ....\\\\ &\begin{array}{lllllllll}\\ \textrm{a}.&-2&&&\\ \textrm{b}.&-1\\ \textrm{c}.&0\\ \color{red}\textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}5^{x+1}+5^{2-x}-30&=0\\ \left (5^{x} \right ).5^{1}+\displaystyle \frac{5^{2}}{5^{x}}-30&=0\\ 5\left ( 5^{x} \right )^{2}+25-30\left ( 5^{x} \right )&=0\\ \textrm{Persamaan kuadrat}&\: \textrm{dalam}\: \: 5^{x},\: \textrm{maka}\\ 5(5^{x})^{2}-30(5^{x})+25&=0\begin{cases} a & =5 \\ b & =-30 \\ c & =25 \end{cases}\\ (5^{x_{1}}).\left ( 5^{x_{2}} \right )&=\displaystyle \frac{c}{a}\\ 5^{x_{1}+x_{2}}&=\displaystyle \frac{25}{5}=5\\ 5^{x_{1}+x_{2}}&=5^{1}\\ x_{1}+x_{2}&=\color{red}1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 47.&\textrm{Jumlah akar-akar persamaan}\\ &2020^{x^{2}-7x+7}=2021^{x^{2}-7x+7}\: \: \textrm{adalah}\: ....\\\\ &\begin{array}{lllllllll}\\ \textrm{a}.&-7\\ \textrm{b}.&-5\\ \textrm{c}.&-3\\ \textrm{d}.&5\\ \color{red}\textrm{e}.&7 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}2020^{x^{2}-7x+7}&=2021^{x^{2}-7x+7}\\ \textrm{Karena basis}&\: \textrm{tidak sama},\\ \textrm{maka harusl}&\textrm{ah pangkatnya}=0,\\ x^{2}-7x+7&=0\\ \textrm{dan jumlah}\: &\textrm{akar-akarnya adalah}:\\ x_{1}+x_{2}&=-\displaystyle \frac{b}{a}, \: \: \textrm{dari persamaan}\\ x^{2}-7x+7&=0\begin{cases} a &=1 \\ b &=-7 \\ c &=7 \end{cases}\\ \textrm{maka}\: \: x_{1}+x_{2}&=-\displaystyle \frac{b}{a}=-\frac{-7}{1}=\color{red}7 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 48.&\textrm{Nilai dari}\: \: \displaystyle \frac{2^{2020}+2^{2018}}{2^{2018}+2^{2016}} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \color{red}\textrm{b}.&5\\ \textrm{c}.&10\\ \textrm{d}.&20\\ \textrm{e}.&40 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\displaystyle \frac{2^{2020}+2^{2018}}{2^{2018}+2^{2016}}&=\displaystyle \frac{2^{4}.2^{2016}+2^{2}.2^{2016}}{2^{2}.2^{2018}+2^{2016}}\\ &=\displaystyle \frac{2^{2016}\left ( 2^{4}+2^{2} \right )}{2^{2016}\left ( 2^{2}+1 \right )}\\ &=\displaystyle \frac{16+4}{4+1}\\ &=\displaystyle \frac{20}{5}\\ &=\color{red}4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 49.&(\textbf{UM IPB})\textrm{Jika}\: \: ab=a^{b} \: \: \textrm{dan}\: \: \displaystyle \frac{a}{b}=a^{3b}\\ &\textrm{maka nilai}\: \: a\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&0\\ \textrm{b}.&0,5\\ \textrm{c}.&1\\ \textrm{d}.&0,25\\ \textrm{e}.&0,75 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\textrm{Diketahui}&\\ ab&=a^{b}\\ b&=\displaystyle \frac{a^{b}}{a}=a^{b-1}.....\textbf{1}\\ \textrm{maka}&\\ \displaystyle \frac{a}{b}&=a^{3b}...............\textbf{2}\\ \textbf{1}&\: \: ke\: \: \textbf{2}\\ \displaystyle \frac{a}{a^{b-1}}&=a^{3b}\\ a^{2-b}&=a^{3b}\\ 2-b&=3b\\ -4b&=-2\\ b&=\displaystyle \frac{1}{2}................\textbf{3}\\ \textbf{3}&\: \: ke\: \: \textbf{1}\\ a\left ( \displaystyle \frac{1}{2} \right )&=a^{\frac{1}{2}}\\ \displaystyle \frac{1}{4}a^{2}&=a\\ a^{2}-4a&=0\\ a(a-4)&=0\\ a=0\: \: &\textrm{atau}\: \: \color{red}a=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 50.&\textrm{Jika}\: \: \displaystyle 3.x^{^{^{^{ \frac{3}{2}}}}}=4\: ,\: \textrm{maka}\: \: x=\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle 1,1\\ \textrm{b}.&\displaystyle 1,2\\ \color{red}\textrm{c}.&1,3\\ \textrm{d}.&\displaystyle 1,4\\ \textrm{e}.&1,5\\\\ &&&(\textbf{SAT Test Math Level 2})\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\displaystyle 3.x^{^{^{^{ \frac{3}{2}}}}}&=4\\ \left (3.x^{^{^{^{ \frac{3}{2}}}}} \right )^{2}&=4^{2}\\ 3^{2}.x^{3}&=4^{2}\\ x^{3}&=\displaystyle \frac{4^{2}}{3^{2}}\\ x^{3}&=\displaystyle \frac{4^{2}}{3^{2}}\times \frac{3}{3}\\ x^{3}&\leq \displaystyle \frac{4^{2}}{3^{2}}\times \frac{4}{3}\\ x^{3}&\leq \left ( \displaystyle \frac{4^{3}}{3^{3}} \right )\\ x^{3}&\leq \left ( \displaystyle \frac{4}{3} \right )^{3}\\ x&\leq \displaystyle \frac{4}{3}\\ x&\leq 1,\overline{333}\\ x&\color{red}\approx 1,3 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 51.&\textrm{Hitunglah}\:\\\\ &\quad\quad\qquad \displaystyle \sqrt[8]{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{...}}}}}\\\\ &\textrm{nyatakan jawabannya dalam bentuk }\: \displaystyle \frac{a\pm b\sqrt{c}}{d}\\ &\textrm{dengan a, b, c, dan d bilangan-bilangan bulat}\\ \end{array}$

Pembahasan:

$\begin{aligned}x^{8}&=2207-\displaystyle \underset{x^{8}}{\underbrace{\displaystyle \frac{1}{2207-\frac{1}{2207-\frac{1}{2207-...}}}}}\\ x^{8}&=2207-\displaystyle \frac{1}{x^{8}}\\ x^{8}+\displaystyle \frac{1}{x^{8}}&=2207\\ \left ( x^{4}+\displaystyle \frac{1}{x^{4}} \right )^{2}&=2207+2\\ \left ( x^{4}+\displaystyle \frac{1}{x^{4}} \right )&=\sqrt{2209}=\color{red}47 \end{aligned}$

$\begin{aligned}x^{4}+\displaystyle \frac{1}{x^{4}}&=47\\ \left ( x^{2}+\displaystyle \frac{1}{x^{2}} \right )^{2}&=47+2\\ x^{2}+\displaystyle \frac{1}{x^{2}}&=\sqrt{49}=7\\ \left ( x+\displaystyle \frac{1}{x} \right )^{2}&=7+2\\ x+\displaystyle \frac{1}{x}&=\sqrt{9}=3\\ x^{2}-3x+1&=0,\\ &\textrm{persamaan kuadrat dalam x,}\\ & \textbf{gunakan rumus abc}\\ x_{1,2}=&\displaystyle \frac{3\pm \sqrt{5}}{2}=\displaystyle \frac{3\pm 1\sqrt{5}}{2}=\displaystyle \frac{a\pm b\sqrt{c}}{d}\\ &\textbf{Sehingga},\quad \begin{cases} & a=3 \\ & b=1 \\ & c=5 \\ & d=\color{red}2 \end{cases} \end{aligned}$

DAFTAR PUSTAKA

1. Enung, S., Untung, W. 2009. Mandiri Matematika SMA Jilid I untuk Kelas X. Jakarta: ERLANGGA.
2. Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA
3. Kanginan, M., Terzalgi, Y. 2013. Matematika untuk SMA-MA/SMK Kelas X Wajib. Bandung: SEWU.
4. Susianto, B. 2011. Olimpiade Matematika dengan Proses Berpikir Aljabar dan Bilangan. Jakarta: GRASINDO. 

Latihan Soal 4 Persiapan PAS Gasal Matematika Peminatan Kelas X (Fungsi Eksponen dan Fungsi Logaritma)

$\begin{array}{ll}\\ 31.&\textrm{Jika bentuk}\: \: \displaystyle \frac{ab^{-1}}{a^{-1}-b^{-1}}\\ & \textrm{dinyatakan dalam pangkat positif}=\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle \frac{a^{2}}{a-b}&&&\\\\ \textrm{b}.&\displaystyle \frac{a^{2}}{a-1}\\\\ \textrm{c}.&\displaystyle \frac{b-a}{ab}\\\\ \color{red}\textrm{d}.&\displaystyle \frac{a^{2}}{b-a}\\\\ \textrm{e}.&\displaystyle \frac{1}{a-b}\\\\ &&&&(\textbf{SAT Test Math Level 2})\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\displaystyle \frac{ab^{-1}}{a^{-1}-b^{-1}}&=\displaystyle \frac{ab^{-1}}{a^{-1}-b^{-1}}\times \frac{b}{b}\\ &=\displaystyle \frac{a}{a^{-1}b-1}\\ &=\displaystyle \frac{a}{a^{-1}b-1}\times \frac{a}{a}\\ &=\color{red}\displaystyle \frac{a^{2}}{b-a} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 32.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ & \sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}=3 \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&3\\ \color{red}\textrm{b}.&6\\ \textrm{c}.&7\\ \textrm{d}.&8\\ \textrm{e}.&9 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\textrm{Misalkan}\quad A&=\sqrt{+\sqrt{x+\sqrt{+\cdots }}}\\ \sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}&=3\\ \textrm{dikuadratkan}&\\ x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}&=9\\ x+3&=9\\ x&=9-3\\ x&=\color{red}6 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 33.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &x=\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&7\sqrt[7]{7}\\ \color{red}\textrm{b}.&7\\ \textrm{c}.&14\\ \textrm{d}.&49\\ \textrm{e}.&\sqrt[3]{81} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}x&=\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\\ x^{3}&=49\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\\ x^{3}&=49x\\ x^{2}&=49\\ x&=\sqrt{49}\\ &=\color{red}7 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 34.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &x^{x^{x^{x^{x^{\cdots }}}}}=2020 \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\sqrt{2020}\\ \color{red}\textrm{b}.&\sqrt[2020]{2020}\\ \textrm{c}.&2020^{\sqrt{2020}}\\ \textrm{d}.&\sqrt{2020}^{\sqrt{2020}}\\ \textrm{e}.&\sqrt{2020\sqrt{2020}} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}x^{x^{x^{x^{x^{\cdots }}}}}&=2020\\ x^{2020}&=2020\\ x&=\color{red}\sqrt[2020]{2020} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 35.&\textrm{Nilai dari}\\ &\displaystyle \frac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}}\\ &\textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \textrm{b}.&\sqrt[3]{2}+1\\ \textrm{c}.&\sqrt[3]{2}-1\\ \textrm{d}.&\sqrt[3]{4}+1\\ \color{red}\textrm{e}.&\sqrt[3]{4}-1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\displaystyle \frac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}}\times \frac{\sqrt[3]{2}-1}{\sqrt[3]{2}-1}\\ &=\displaystyle \frac{\left ( \sqrt[3]{2} \right )^{2}-1}{\sqrt[3]{2}+\sqrt[3]{4}+\sqrt[3]{8}-1-\sqrt[3]{2}-\sqrt[3]{4}}\\ &=\displaystyle \frac{\sqrt[3]{4}-1}{\sqrt[3]{8}-1}\\ &=\displaystyle \frac{\sqrt[3]{4}-1}{2-1}\\ &=\color{red}\sqrt[3]{4}-1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 36.&\textrm{Nilai dari}\\ &\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\\ &\textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&1\\ \textrm{b}.&2\sqrt{2}-1\\ \textrm{c}.&\displaystyle \frac{1}{2}\sqrt{2}\\ \textrm{d}.&\sqrt{\displaystyle \frac{5}{3}}\\ \textrm{e}.&\sqrt{\displaystyle \frac{2}{5}} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\\ &=\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\times \frac{\sqrt{\sqrt{5}+1}}{\sqrt{\sqrt{5}+1}} -\sqrt{3-2\sqrt{2}}\\ &=\displaystyle \frac{\sqrt{7+3\sqrt{5}}+\sqrt{3-\sqrt{5}}}{\sqrt{5}+1}-\left ( \sqrt{2}-1 \right )\\ &=\displaystyle \frac{\left ( \displaystyle \frac{3+\sqrt{5}}{\sqrt{2}} \right )+\left ( \displaystyle \frac{\sqrt{5}-1}{\sqrt{2}} \right )}{\sqrt{5}+1}+1-\sqrt{2}\\ &=\displaystyle \frac{\displaystyle \frac{2+2\sqrt{5}}{\sqrt{2}}}{1+\sqrt{5}}+1-\sqrt{2}\\ &=\displaystyle \frac{2}{\sqrt{2}}+1-\sqrt{2}\\ &=\sqrt{2}+1-\sqrt{2}\\ &=\color{red}1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 37.&\textrm{Bentuk sederhana dari}\\ &\sqrt{33+\sqrt{800}}-\sqrt{27-2\sqrt{162}}=....\\\\ &\qquad\qquad\qquad(\textbf{SIMAK UI 2012 Mat IPA})\\ &\begin{array}{llllllll}\\ \textrm{a}.&2-\sqrt{2}\\ \color{red}\textrm{b}.&8-\sqrt{2}\\ \textrm{c}.&-2+\sqrt{2}\\ \textrm{d}.&2+5\sqrt{2}\\ \textrm{e}.&8+5\sqrt{2} \end{array}\\\\ &\textrm{Jawab}:\qquad \color{red}\textbf{b}\\ &\textrm{misalkan},\\ &\begin{aligned}x&=\sqrt{33+\sqrt{800}}-\sqrt{27-2\sqrt{162}}\\ &=\left ( \sqrt{33+20\sqrt{2}}-\sqrt{27-2.9\sqrt{2}}\: \right )\\ &=\sqrt{33+20\sqrt{2}}-\sqrt{27-18\sqrt{2}}\\ x^{2}&=33+20\sqrt{2}+27-18\sqrt{2}-2\sqrt{\left ( 33+20\sqrt{2} \right )\left ( 27-18\sqrt{2} \right )}\\ &=60+2\sqrt{2}-2\sqrt{33.27-33.18\sqrt{2}+27.20\sqrt{2}-20.18.2}\\ &=60+2\sqrt{2}-2\sqrt{891-720+540\sqrt{2}-594\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-54\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-2.27\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-2\sqrt{27.27}\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-2\sqrt{162.9}}\\ &=60+2\sqrt{2}-2\sqrt{162+9-2\sqrt{162.9}}\\ &=60+2\sqrt{2}-2\left ( \sqrt{162}-\sqrt{9} \right )\\ &=60+2\sqrt{2}-2\left ( 9\sqrt{2}-3 \right )\\ x^{2}&=66-16\sqrt{2}\\ x&=\sqrt{66-2.8\sqrt{2}}\\ &=\sqrt{64+2-2\sqrt{64.2}}\\ &=\sqrt{64}-\sqrt{2}\\ &=\color{red}8-\sqrt{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 38.&\textrm{Jika}\: \: \displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\displaystyle \frac{a\sqrt{2}+b\sqrt{3}+c\sqrt{5}}{12}\: ,\\\\ &\textrm{maka}\: \: a+b+c=....\\ &\qquad\qquad\qquad\qquad (\textbf{UM UGM 2016 Mat Das})\\ &\begin{array}{llllllll}\\ \textrm{a}.&0\\ \textrm{b}.&1\\ \textrm{c}.&2\\ \textrm{d}.&3\\ \color{red}\textrm{e}.&4 \end{array}\\\\ &\textrm{Jawab}:\qquad \color{red}\textbf{e}\\ &\begin{aligned}&\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\\ &=\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\times \displaystyle \frac{\left ( \sqrt{2}+\sqrt{3}-\sqrt{5} \right )}{\left ( \sqrt{2}+\sqrt{3}-\sqrt{5} \right )}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\left ( \sqrt{2}+\sqrt{3} \right )^{2}-\left ( \sqrt{5} \right )^{2}}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2+3+2\sqrt{2.3}-5}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2\sqrt{6}}\times \displaystyle \frac{\sqrt{6}}{\sqrt{6}}\\ &=\displaystyle \frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{12}\\ &=\displaystyle \frac{2\sqrt{3}+3\sqrt{2}-\sqrt{30}}{12}\\ &=\displaystyle \frac{3\sqrt{2}+2\sqrt{3}-\sqrt{30}}{12}\\ &\quad \begin{cases} a &=3 \\ b &=2 \\ c &=-1 \end{cases}\\ a+b+c&=3+2+(-1)\\ &=\color{red}4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 39.&\textrm{Tunjukkan bahwa}\\ &\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+\cdots }}}}}=3\\\\ &\textrm{Bukti}\\ &\begin{aligned}x^{2}&=x^{2}\\ x^{2}&=1+\left ( x^{2}-1 \right )\\ &=1+\left ( x-1 \right )\left ( x+1 \right )\\ &=1+\left ( x-1 \right )\sqrt{\left ( x+1 \right )^{2}}\\ &=1+\left ( x-1 \right )\sqrt{1+\left ( x+1 \right )^{2}-1}\\ &=1+\left ( x-1 \right )\sqrt{1+\left ( x+1-1 \right )\left ( x+1+1 \right )}\\ &=1+\left ( x-1 \right )\sqrt{1+x\left ( x+2 \right )}\\ &=1+\left ( x-1 \right )\sqrt{1+x\sqrt{\left ( x+2 \right )^{2}}}\\ &=1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+2 \right )^{2}-1}}\\ &=1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+2-1 \right )\left ( x+2+1 \right )}}\\ &=1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+1 \right )\left ( x+3 \right )}}\\ &=1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+1 \right )\sqrt{\left ( x+3 \right )^{2}}}}\\ x^{2}&=1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+1 \right )\sqrt{...}}}\\ x&=\color{red}\sqrt{1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+1 \right )\sqrt{...}}}} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 40.&\textrm{Jika terdapat hubungan berikut}\\ &\textrm{a}.\quad 2^{p}=3^{q}=6^{r},\: \: \textrm{tunjukkan bahwa}\: \: pr+qr-pq=0\\ &\textrm{b}.\quad 2^{x}=3^{2y}=6^{z},\: \: \textrm{tunjukkan bahwa }\: \: 2xy-2yz-xz=0\\ &\textrm{c}.\quad 3^{15a}=5^{5b}=15^{3c},\: \: \textrm{tunjukkan bahwa }\: \: 5ab-bc-3ac=0\\ \end{array}$

$\textbf{bukti}$

Yang akan ditunjukkan adalah no. 40 yang poin c, yaitu:

$\begin{aligned}3^{15a}&=5^{5b}=15^{3c}\begin{cases} 3=5^{\frac{5b}{15a}} & \\ 3^{\frac{15a}{5b}}=b &\left ( a^{b}=c^{d}\rightarrow a=c^{\frac{d}{b}}\: \: \textrm{atau}\: \: a^{\frac{b}{d}}=c \right ) \end{cases}\\ 3^{15a}&=15^{3c}\\ 3^{15a}&=(3\times 5)^{3c}\\ 3^{15a}&=(3\times 3^{\frac{15a}{5b}})^{3c}\\ 3^{15a}&=3^{3c+\frac{9c}{b}}\\ a^{f(x)}&=a^{g(x)}\\ f(x)&=g(x)\\ 15a&=3c+\frac{9ac}{b}\\ 15ab&=3bc+9ac\\ 5ab&=bc+3ac\\ 5ab-bc-3ac&=\color{red}0\quad \color{black}\blacksquare \end{aligned}$

Latihan Soal 3 Persiapan PAS Gasal Matematika Peminatan Kelas X (Fungsi Eksponen dan Fungsi Logaritma)

 $\begin{array}{ll}\\ 21.&\textrm{Jika}\: \: f(x)=b^{x},\: \: \textrm{di mana konstan positif},\\\\ &\displaystyle \frac{f\left ( x^{2}+x \right )}{f(x+1)}= ....\\ &\begin{array}{lllllllll}\\ \textrm{a}.&f\left ( x^{2} \right )&&&\\ \textrm{b}.&f(x+1)f(x-1)\\ \textrm{c}.&f(x+1)+f(x-1)\\ \textrm{d}.&f(x+1)-f(x-1)\\ \color{red}\textrm{e}.&f\left ( x^{2}-1 \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}\displaystyle \frac{f\left ( x^{2}+x \right )}{f(x+1)}&=\frac{b^{x^{2}+x}}{b^{x+1}}\\ &=b^{x^{2}+x-(x+1)}\\ &=b^{x^{2}-1}\\ &=\color{red}f\left ( x^{2}-1 \right ) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 22.&\textrm{Daerah hasil dari fungsi eksponen}\\ &y\: =x^{- \frac{2}{3}}\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\\ \textrm{a}.&y< 0\\ \color{red}\textrm{b}.&y> 0\\ \textrm{c}.&y\geq 0\\ \textrm{d}.&y\leq 0\\ \textrm{e}.&\textrm{Semua bilangan real} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\textrm{Perhatikanlah gambar berikut} \end{array}$

$.\quad\: \, \begin{aligned}\textrm{diketahui}&\\ y\: &=x^{-\displaystyle \frac{2}{3}}\\ y^{3}\: &=x^{-2}\\ y^{3}\: &=\displaystyle \frac{1}{x^{2}},\: \textrm{atau}\\ y^{3}\times x^{2}\: &=1,\\ \textrm{sehingga}&\: \: y\: \: \textrm{tidak mungkin berharga}\: \: \color{red}0 \end{aligned}$.

$\begin{array}{ll}\\ 23.&\textrm{Nilai}\: \: p-q^{p-q}\: \: \textrm{untuk}\: \:p=2\: \: \textrm{dan}\: \: q=-2\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-18\\ \textrm{b}.&-14\\ \textrm{c}.&1\\ \textrm{d}.&18\\ \color{red}\textrm{e}.&256 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}p-q^{p-q}&=(2-(-2))^{2-(-2)}\\ &=4^{4}\\ &=\color{red}256 \end{aligned} \end{array}$

$\begin{array}{l}\\ 24.&\textrm{Bentuk sederhana dari}\\ &\displaystyle \frac{3^{x+8}\left (7^{3x+1}\times 3^{4x-1} \right )^{2}}{(7^{2x}\times 3^{3x+2})^{3}}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \color{red}\textrm{a}.&49\\ \textrm{b}.&9\\ \textrm{c}.&7\\ \textrm{d}.&7^{2x+2}\\ \textrm{e}.&3^{2x-1} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\displaystyle \frac{3^{x+8}\left (7^{3x+1}\times 3^{4x-1} \right )^{2}}{(7^{2x}\times 3^{3x+2})^{3}}\\ &=\displaystyle \frac{3^{x+8+2(4x-1)}\times 7^{2(3x+1)}}{7^{3.2x}\times 3^{3(3x+2)}}\\ &=\displaystyle \frac{3^{x+8x+8-2}\times 7^{6x+2}}{3^{9x+6}\times 7^{6x}}\\ &=3^{0}\times 7^{2}\\ &=1\times 49\\ &=\color{red}49 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 25.&\textrm{Jika}\: \: \displaystyle \frac{(0,24)^{3}(0,243)^{5}}{(3,6)^{7}}=\frac{2^{p}3^{q}}{5^{r}},\\ & \textrm{maka nilai}\: \: p+q+r\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&7\\ \textrm{b}.&8\\ \color{red}\textrm{c}.&9\\ \textrm{d}.&10\\ \textrm{e}.&11 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\displaystyle \frac{(0,24)^{3}(0,243)^{5}}{(3,6)^{7}}\\ &=\displaystyle \frac{\left (\frac{24}{100} \right )^{3}\times \left ( \frac{243}{1000} \right )^{5}}{\left (\frac{36}{10} \right )^{7}}\\ &=\displaystyle \frac{(8\times 3)^{3}\times \left (3^{5} \right )^{5}}{(2^{2}\times 3^{2})^{7}}\times \displaystyle \frac{10^{7}}{100^{3}\times 1000^{5}}\\ &=\displaystyle \frac{(2^{3}\times 3)^{3}\times 3^{25}}{(2^{14}\times 3^{14})}\times \displaystyle \frac{10^{7}}{\left ( 10^{2} \right )^{3}\times \left ( 10^{3} \right )^{5}}\\ &=2^{9-14}.3^{3+25-14}.10^{7-6-15}\\ &=2^{-5}.3^{14}.10^{-14}=2^{-5}.3^{14}.(2.5)^{-14}\\ &=2^{-5-14}.3^{21}.5^{-14}\\ &=\displaystyle \frac{2^{-19}.3^{14}}{5^{14}}\\ \textrm{Sehingga}\: \: &p+q+r=-19+14+14=\color{red}9 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 26.&(\textbf{UM UGM 05})\textrm{Hasil dari}\\ &\sqrt{0,3+\sqrt{0,08}}=\sqrt{a}+\sqrt{b}\: ,\: \textrm{maka}\: \: \displaystyle \frac{1}{a}+\frac{1}{b}=....\\ &\begin{array}{llll}\\ \textrm{a}.&25\\ \textrm{b}.&20\\ \color{red}\textrm{c}.&15\\ \textrm{d}.&10\\ \textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\sqrt{0,3+\sqrt{0,08}}&=\sqrt{0,2+0,1+\sqrt{4\times 0,2\times 0,1}}\\ &=\sqrt{0,2+0,1+2\sqrt{\times 0,2\times 0,1}}\\ &=\sqrt{0,2}+\sqrt{0,1}\\ \textrm{maka},\: \: a=0,2&,\: \: b=0,1\\ \textrm{sehingga}\: \displaystyle \frac{1}{a}+\frac{1}{b}&=\displaystyle \frac{1}{0,2}+\frac{1}{0,1}=5+10=\color{red}15\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 27.&(\textbf{SPMB 06})\textrm{Jika bilangan bulat}\: \: a\: \: \: \textrm{dan}\: \: b\: \: \textrm{memenuhi}\\ &\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}=a+b\sqrt{30}\: ,\: \textrm{maka}\: \: ab=....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&-22\\ \textrm{b}.&-11\\ \textrm{c}.&-9\\ \textrm{d}.&2\\ \textrm{e}.&13 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}&=\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}\times \displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}-\sqrt{6}}\\ &=\displaystyle \frac{5-2\sqrt{30}+6}{5-6}\\ &=\displaystyle \frac{11-2\sqrt{30}}{-1}\\ &=-11+2\sqrt{30}\\ \textrm{sehingga}&\: \: \: a=-11,\: \: b=2,\: \: \textrm{maka}\\ ab&=(-11)\times 2\\ &=\color{red}-22 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 28.&(\textbf{OSK 2013})\textrm{Misal}\: \: a\: \: \textrm{dan}\: \: b\: \: \textrm{bilangan asli}\\ &\textrm{dengan}\: \: a>b.\: \: \textrm{Jika} \: \: \sqrt{94+2\sqrt{2013}}=\sqrt{a}+\sqrt{b}\\ &\textrm{maka nilai} \: \: a-b\: \: \textrm{adalah... .}\\\\ &\textrm{Jawab}:\\ &\begin{aligned} \sqrt{94+2\sqrt{2013}}&=\sqrt{61+33+2\sqrt{61\times 33}}\\ &=\sqrt{61}+\sqrt{33}\\ &=\sqrt{a}+\sqrt{b}\\ \textrm{Sehingga}\: \: a&=61,\: \: b=33,\: \: \textrm{maka}\\ a-b&=61-33\\ &=\color{red}28 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 29.&\textrm{Nilai dari}\\ &\sqrt{54+14\sqrt{5}}+\sqrt{12-2\sqrt{35}}+\sqrt{32-10\sqrt{7}}\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&10\\ \textrm{b}.&11\\ \color{red}\textrm{c}.&12\\ \textrm{d}.&5\sqrt{6}\\ \textrm{e}.&6\sqrt{6} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\textrm{misal diketah}&\textrm{ui}\\ x&=\sqrt{54+14\sqrt{5}}+\sqrt{12-2\sqrt{35}}+\sqrt{32-10\sqrt{7}}\\ \textrm{untuk}&\\ \sqrt{54+14\sqrt{5}}&=\sqrt{49+5+2.7\sqrt{5}}=7+\sqrt{5}\\ \sqrt{12-2\sqrt{35}}&=\sqrt{7+5-2\sqrt{7.5}}=\sqrt{7}-\sqrt{5}\\ \sqrt{32-10\sqrt{7}}&=\sqrt{25+7-2.5\sqrt{7}}=5-\sqrt{7}\qquad +\\ &---------------\\ &\qquad\qquad\qquad\quad\qquad=7+5\\ &\qquad\qquad\qquad\quad\qquad=\color{red}12 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 30.&\textrm{Nilai eksak dari}\\ &\displaystyle \frac{1}{10^{-2020}+1}+\frac{1}{10^{-2019}+1}+...+\displaystyle \frac{1}{10^{2019}+1}+\frac{1}{10^{2020}+1} \\ &\begin{array}{lllllllll}\\ \textrm{a}.&2020&&&\\ \color{red}\textrm{b}.&2020,5\\ \textrm{c}.&2021\\ \textrm{d}.&2021,5\\ \textrm{e}.&2022 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\textrm{Misal},\: \: x&=\frac{1}{10^{-2020}+1}+\frac{1}{10^{-2019}+1}+...+\frac{1}{10^{2019}+1}+\frac{1}{10^{2020}+1}\\ \textrm{maka},\, \: x&=\frac{1}{\frac{1}{10^{2020}}+1}+\frac{1}{\frac{1}{10^{2019}}+1}+...+\frac{1}{10^{2019}+1}+\frac{1}{10^{2020}+1}\\ &=\frac{10^{2020}}{1+10^{2020}}+\frac{10^{2019}}{1+10^{2019}}+...+\frac{1}{10^{2019}+1}+\frac{1}{10^{2020}+1}\\ &=\frac{10^{2020}}{1+10^{2020}}+\frac{1}{10^{2020}+1}+\frac{10^{2019}}{1+10^{2019}}+\frac{1}{10^{2019}+1}+...+\frac{1}{1^{0}+1}\\ &=\frac{10^{2020}+1}{10^{2020}+1}+\frac{10^{2019}+1}{10^{2019}+1}+\frac{10^{2018}+1}{10^{2018}+1}+...+\frac{10^{1}+1}{10^{1}+1}+\frac{1}{1^{0}+1}\\ &=\underset{\textrm{sebanyak}\: 2020}{\underbrace{1+1+1+1+1+...+1}}+\frac{1}{2}\\ &=\color{red}2020,5 \end{aligned} \end{array}$