$\begin{array}{ll}\\ 93.&\textrm{Persamaan}\: \: ^{x}\log 2+\: ^{x}\log (3x-4)=2\\ &\textrm{mempunyai akar}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2},\: \: \textrm{maka}\\ &\textrm{nilai}\: \: x_{1}+x_{2}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \textrm{b}.&3\\ \textrm{c}.&4\\ \color{red}\textrm{d}.&6\\ \textrm{e}.&8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\color{purple}\textrm{Alternatif 1}\\ &^{x}\log 2+\: ^{x}\log (3x-4)=2\\ &\Leftrightarrow \: ^{x}\log 2(3x-4)=2\\ &\Leftrightarrow \: ^{x}\log 6x-8=2\\ &\Leftrightarrow \: \color{black}6x-8=x^{2}\\ &\Leftrightarrow \: \color{black}x^{2}-6x+8=0,\: \: \color{purple}\textrm{dengan}\: \begin{cases} a &=1 \\ b &=-6 \\ c &=8 \end{cases}\\ &\Leftrightarrow \: x_{1}+x_{2}=-\displaystyle \frac{b}{a}\\ &\Leftrightarrow \: x_{1}+x_{2}=\color{red}-\displaystyle \frac{-6}{1}=6\\ &\color{purple}\textrm{Alternatif 2}\\ &\Leftrightarrow \: \color{black}x^{2}-6x+8=0\\ &\Leftrightarrow \: (x-2)(x-4)\\ &\Leftrightarrow \: x_{1}=2\: \: \textrm{atau}\: \: x_{2}=4\\ &\Leftrightarrow \: x_{1}+x_{2}=\color{red}2+4=6 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 94.&\textrm{Jika}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\: \: \textrm{memenuhi}\\ &(\log x)(2\log x-3)=\log 100\\ &\textrm{maka}\: \: x_{1}\times x_{2}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&100\\ \color{red}\textrm{b}.&10\sqrt{10}\\ \textrm{c}.&\sqrt{10}\\ \textrm{d}.&-\sqrt{10}\\ \textrm{e}.&-10\sqrt{10} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&(\log x)(2\log x-3)=\color{red}\log 100\\ &\Leftrightarrow (\log x)\left ( 2\log x-3 \right )=\color{red}2\\ &\color{black}\Leftrightarrow 2\log ^{2}x-3\log x-2=0\: \color{purple}\begin{cases} a &=2 \\ b &=-3 \\ c &=-2 \end{cases}\\ &\Leftrightarrow \log x_{1}+\log x_{2}=-\displaystyle \frac{-3}{2}=\frac{3}{2}\\ &\Leftrightarrow \log \left ( x_{1}\times x_{2} \right )=1\displaystyle \frac{1}{2}\\ &\Leftrightarrow \color{red}\left ( x_{1}\times x_{2} \right )=10^{1\frac{1}{2}}=10\sqrt{10} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 95.&\textrm{Persamaan}\\ & 10^{\,^{\, ^{2}}\log x^{2} }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ & \textrm{mempunyai dua akar yaitu}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\\ &\textrm{Nilai}\: \: x_{1}\times x_{2}=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \textrm{b}.&-5\\ \color{red}\textrm{c}.&2\\ \textrm{d}.&5\\ \textrm{e}.&10 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&10^{\,^{\, ^{2}}\log x^{2} }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ &10^{\, 2^{\, ^{2}}\log x }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ &\color{black}\textrm{adalah persamaan kuadrat dalam}\: \: \color{red}10^{\,^{\, ^{2}}\log x }\\ &\color{black}\textrm{Misalkan}\: \: \color{red}p=10^{\,^{\, ^{2}}\log x },\: \: \textrm{maka persamaan}\\ &\color{black}\textrm{menjadi}\: \: \color{purple}p^{2}-7p+10=0\: \begin{cases} a & =1 \\ b & =-7 \\ c & =10 \end{cases}\\ & \color{red}\textrm{Karena nilai}\: \: \color{black}p_{1}\times p_{2}=\displaystyle \frac{c}{a}\: \: \textrm{maka}\\ &10^{\,^{\, ^{2}}\log x_{1} }\times 10^{\,^{\, ^{2}}\log x_{2} }=\displaystyle \frac{10}{1}=10\\ &10^{\,^{\, ^{2}}\log x_{1}\: +\: ^{2}\log x_{2} }=10\\ &10^{\,^{\, ^{2}}\log x_{1}\: +\: ^{2}\log x_{2} }=10^{1}\\ &\Leftrightarrow \: ^{2}\log x_{1}\: +\: ^{2}\log x_{2}=1\\ &\Leftrightarrow \: ^{2}\log x_{1}\times x_{2}=1\\ &\Leftrightarrow \: \color{red}x_{1}\times x_{2}=2^{1}=2\\ \end{aligned} \end{array}$
$\begin{array}{ll}\\ 96.&\textrm{Nilai}\: \: x\: \: \: \: \textrm{yang memenuhi}\\ &^{x}\log (x+12)-3.\: ^{x}\log 4+1=0\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\\ \textrm{b}.&2\\ \color{red}\textrm{c}.&4\\ \textrm{d}.&8\\ \textrm{e}.&16 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&^{x}\log (x+12)-3.\: ^{x}\log 4+1=0\\ &^{x}\log (x+12)-\: ^{x}\log 4^{3}=-1\\ &^{x}\log \displaystyle \frac{x+12}{64}=-1\\ &\displaystyle \frac{x+12}{64}=x^{-1}=\frac{1}{x}\\ &x+12=\displaystyle \frac{64}{x}\\ &\color{purple}x^{2}+12x-64=0\\ &\color{purple}(x+16)(x-4)=0\\ &x=-16\: \: \color{black}\textrm{atau}\: \: \color{red}x=4 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 97.&\textrm{Nilai}\: \: x\: \: \: \: \textrm{yang memenuhi}\\ &\displaystyle \frac{^{2}\log (2x-3)}{^{2}\log x}-\: ^{x}\log (x+6)+\displaystyle \frac{1}{^{(x+2)}\log x}=1\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-1\: \: \textrm{dan}\: \: 6\\ \textrm{b}.&-2\: \: \textrm{dan}\: \: 6\\ \textrm{c}.&-1\\ \textrm{d}.&-2\\ \color{red}\textrm{e}.&6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\displaystyle \frac{^{2}\log (2x-3)}{^{2}\log x}-\: ^{x}\log (x+6)+\displaystyle \frac{1}{^{(x+2)}\log x}=1\\ &^{x}\log (2x-3)-\: ^{x}\log (x+6)+\: ^{x}\log (x+2)=1\\ &^{x}\log (2x-3)(x+2)=1+\log (x+6)\\ &^{x}\log \displaystyle \frac{\left ( 2x^{2}+x-6 \right )}{(x+6)}=1\\ &\displaystyle \frac{\left ( 2x^{2}+x-6 \right )}{(x+6)}=x^{1}\\ &\left ( 2x^{2}+x-6 \right )=x^{2}+6x\\ &\color{purple}x^{2}-5x-6=0\\ &\color{purple}(x+1)(x-6)=0\\ &x=-1\: \: \color{black}\textrm{atau}\: \: \color{red}x=6 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 98.&\textrm{Himpunan penyelesaian dari persamaan}\\ & ^{3}\log \left ( x^{2}-5x+7 \right )=0\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\left \{ 2,3 \right \}&\quad&\textrm{d}.&\left \{ 3,4 \right \}\\ \textrm{b}.&\left \{ 2,4 \right \}&&\textrm{e}.&\left \{ 3,5 \right \}\\ \textrm{c}.&\left \{ 2,5 \right \}&&& \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Diketahui bahwa}:\\ & ^{3}\log \left ( x^{2}-5x+7 \right )=0\\ &\Leftrightarrow \, ^{3}\log \left ( x^{2}-5x+7 \right )=\, ^{3}\log 3^{0}\\ &\Leftrightarrow \, ^{3}\log \left ( x^{2}-5x+7 \right )=\, ^{3}\log 1\\ &\textrm{bersesuaian dengan rumus}\: ^{a}\log f(x)=\, ^{a}\log p\\ &\underline{\textrm{Syarat numerus}}\\ &f(x)>0\Leftrightarrow x^{2}-5x+7>0\\ &\textrm{adalah definit positif}\\ &\textrm{sehingga semua nilai}\: \: x\: \: \textrm{memenuhi}\\ &\underline{\textrm{Langkah berikutnya}}\\ &f(x)=p\\ &\Leftrightarrow x^{2}-5x+7=1\\ &\Leftrightarrow x^{2}-5x+6=0\\ &\Leftrightarrow (x-2)(x-3)=0\\ &\Leftrightarrow x=2\: \: \textrm{atau}\: \: x=3\\ &\textrm{Jadi, HP}=\color{red}\left \{ 2,3 \right \} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 99.&\textrm{Himpunan penyelesaian dari}\\ & \log x^{2}=\log 4+\log (x+3)\\ &\textrm{adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\left \{ 1,4 \right \}&&&\textrm{d}.&\left \{ 2,6 \right \}\\ \textrm{b}.&\left \{ 1,6 \right \}\quad&\textrm{c}.&\displaystyle \left \{ 2,4 \right \}\quad&\textrm{e}.&\left \{ 4,6 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textrm{tidak ada}\\ &\textrm{ingat formula}:\: ^{a}\log f(x)=\, ^{a}\log g(x)\\ &\begin{array}{|c|c|}\hline \textbf{Syarat Numerus}&\textbf{Syarat Numerus}\\\hline f(x)&g(x)\\\hline \begin{aligned}x^{2}>0&\\ x<0\: \textrm{atau}\: x&>0 \end{aligned}&\begin{aligned}x+3&>0\\ x&>-3 \end{aligned}\\\hline \begin{aligned}&\textrm{Yang digunakan}\\ &\textrm{adalah yang }\\ &\textrm{memenuhi}\\ &\textrm{keduanya}\\ &\textrm{yaitu}:\: \end{aligned}&\begin{aligned}&-3<x<0\\ &\textrm{atau}\\ &x>0 \end{aligned}\\\hline \end{array}\\ &\begin{aligned}&\textrm{Syarat Penyelesaian}\\ & \log x^{2}=\log 4+\log (x+3)\\ &\Leftrightarrow \log x^{2}=\log 4(x+3)\\ &\textrm{maka},\: \: f(x)=g(x)\\ &x^{2}=4(x+3)\\ &\Leftrightarrow x^{2}=4x+12\\ &\Leftrightarrow x^{2}-4x-12=0\\ &\Leftrightarrow (x+2)(x-6)=0\\ &\Leftrightarrow x+2=0\: \: \textrm{atau}\: \: x-6=0\\ &\Leftrightarrow x=-2\: \: \textrm{atau}\: \: x=6\\ &\textrm{karena syaratnya},\: -3<x<0\: \: \textrm{atau}\: x>0,\\ & \textrm{maka keduanya memenuhi}\\ &\textrm{HP}=\color{red}\left \{ -2,6 \right \} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 100.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ & \log x^{2}=\log x\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&1&\qquad&\textrm{d}.&4\\ \textrm{b}.&2&&\textrm{e}.&5\\ \textrm{c}.&3&&& \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Diketahui}\: \: \log x^{2}=\log x\\ &\textrm{bersesuaian rumus}\: \: ^{a}\log f(x)=\, ^{a}\log g(x)\\ &\underline{\textrm{Syarat numerus}}\\ &\begin{array}{l|l} \begin{aligned}&f(x)>0\\ &x^{2}>0\\ &x>0\: \: \textrm{atau}\: \: x>0 \end{aligned}&\begin{aligned}&g(x)>0\\ &x>0\\ & \end{aligned} \end{array}\\ &\textrm{Sehingga syarat numerusnya}\: \: x>0\\ &\underline{\textrm{Syarat berikutnya}}\\ &\begin{aligned}f(x)&=g(x)\\ x^{2}&=x\\ x^{2}-x&=0\\ x(x-1)&=0\\ x=0\: \: \textrm{atau}\: \: x&=1 \end{aligned}\\ &\textrm{Jadi, HP}=\color{red}\left \{ 1 \right \} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 101.&\textrm{Salah satu nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ & 2\log^{2} x-9\log x+4=0\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\displaystyle \sqrt{10}&\qquad&\textrm{d}.&100\\ \textrm{b}.&\displaystyle 1&&\textrm{e}.&1000\\ \textrm{c}.&10&&& \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Diketahui}\: \: 2\log^{2} x-9\log x+4=0\\ &\textrm{bersesuaian rumus}: \\ &A\left ( ^{a}\log f(x) \right )^{2}+B\left ( ^{a}\log f(x) \right )+C=0\\ &\underline{\textrm{Langkah pengerjaan}}\\ &\begin{aligned}&2\log^{2} x-9\log x+4=0\\ &\Leftrightarrow \left ( 2\log x-1 \right )\left ( \log x-4 \right )=0\\ &\Leftrightarrow 2\log x-1=0\: \: \textrm{atau}\: \: \log x-4=0\\ &\Leftrightarrow \log x=\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: \log x=4\\ &\Leftrightarrow x=10^{.^{ \frac{1}{2}}}\: \: \textrm{atau}\: \: x=10^{4}\\ &\Leftrightarrow x=\sqrt{10}\: \: \textrm{atau}\: \: x=10000 \end{aligned}\\ &\textrm{Jadi, HP}=\color{red}\left \{ \sqrt{10},10000 \right \} \end{aligned} \end{array}$.
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