Latihan Soal 1 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

 $\begin{array}{l}\\ 1.& \text{Nilai}{75}^{\circ }\text{jika dinyatakan ke radian}\\ & \text{adalah}....\text{radian}\\ \\ & \text{a}.{\displaystyle \frac{1}{3}\pi }\\ \\ & \text{b}.{\displaystyle \frac{5}{6}\pi }\\ \\ & \text{c}.{\displaystyle \frac{5}{12}\pi }\\ \\ & \text{d}.{\displaystyle \frac{7}{12}\pi }\\ \\ & \text{e}.{\displaystyle \frac{9}{12}\pi }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diketah}& \text{ui bahwa}\\ {180}^{\circ }& =\pi radian\\ 1^{\circ }& ={\displaystyle \frac{\pi }{180}radian}\\ 75\times 1^{\circ }& =75\times {\displaystyle \frac{\pi }{180}radian}\\ {75}^{\circ }& ={\displaystyle \frac{5}{12}\pi radian}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Jika}\tan \theta ={\displaystyle \frac{5}{12}\text{untuk}{0}^{\circ }\le \theta \le {90}^{\circ }}\\ & \text{maka}\cos \theta \text{adalah}....\\ \\ & \text{a}.{\displaystyle \frac{5}{13}}\\ \\ & \text{b}.{\displaystyle \frac{12}{13}}\\ \\ & \text{c}.{\displaystyle \frac{13}{5}}\\ \\ & \text{d}.{\displaystyle \frac{13}{12}}\\ \\ & \text{e}.{\displaystyle \frac{12}{5}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikanlah gambar segitiga berikut}\end{array}$.

$.\begin{array}{rl}\text{Diketah}& \text{ui bahwa}\\ \tan \theta & ={\displaystyle \frac{5}{12},\text{untuk}{0}^{\circ }\le \theta \le {90}^{\circ }}\\ & \text{lihat gambar di atas}\\ & \text{dengan dalil Pythagoras akan}\\ & \text{didapatkan sisimiringnya}=13\\ \text{jadi}& ,\text{nilai dari}\\ \cos \theta & ={\displaystyle \frac{12}{13}}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Perhatikanlah gambar berikut}\end{array}$.

$.\begin{array}{l}\\ & \text{Panjang BC adalah}....\\ & \text{a}.{\displaystyle 20\sin {36}^{\circ }}\\ & \text{b}.{\displaystyle 20\cos {36}^{\circ }}\\ & \text{c}.{\displaystyle 20\tan {36}^{\circ }}\\ & \text{d}.{\displaystyle 15}\\ & \text{e}.{\displaystyle 16}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diketah}& \text{ui bahwa}\\ \tan {36}^{\circ }& ={\displaystyle \frac{BC}{20}}\\ \iff & BC=20\tan {36}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Nilai}\tan {300}^{\circ }\text{adalah}....\\ \\ & \text{a}.{\displaystyle \sqrt{3}}\\ \\ & \text{b}.{\displaystyle \frac{1}{3}\sqrt{3}}\\ \\ & \text{c}.{\displaystyle -\frac{1}{3}\sqrt{3}}\\ \\ & \text{d}.{\displaystyle \frac{1}{2}\sqrt{3}}\\ \\ & \text{e}.{\displaystyle -\sqrt{3}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\tan {300}^{\circ }& =\tan ({360}^{\circ }-{60}^{\circ })\\ & =-\tan {60}^{\circ }\\ & =-\sqrt{3}\\ \mathbf{\text{catatan}}& :\text{ingat sudut berelasi}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Nilai}\tan {60}^{\circ }-\sin {120}^{\circ }-\tan {210}^{\circ }\text{adalah}....\\ \\ & \text{a}.{\displaystyle \frac{1}{6}\sqrt{6}}\\ \\ & \text{b}.{\displaystyle \frac{1}{3}\sqrt{6}}\\ \\ & \text{c}.{\displaystyle -\frac{1}{2}\sqrt{6}}\\ \\ & \text{d}.{\displaystyle \frac{1}{3}\sqrt{3}}\\ \\ & \text{e}.{\displaystyle \frac{1}{6}\sqrt{3}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \tan {60}^{\circ }-\sin {120}^{\circ }-\tan {210}^{\circ }\\ & =\tan {60}^{\circ }-\sin ({180}^{\circ }-{60}^{\circ })-\tan ({180}^{\circ }+{30}^{\circ })\\ & =\tan {60}^{\circ }-\sin {60}^{\circ }-\tan {30}^{\circ }\\ & =\sqrt{3}-{\displaystyle \frac{1}{2}\sqrt{3}-{\displaystyle \frac{1}{3}\sqrt{3}}}\\ & =(1-{\displaystyle \frac{1}{2}-\frac{1}{3}})\sqrt{3}\\ & ={\displaystyle \frac{1}{6}\sqrt{3}}\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Nilai}x\text{positif terkecil yang memenuhi}\\ & \sin x=-{\displaystyle \frac{1}{2}\sqrt{3}\text{adalah}....}\\ \\ & \text{a}.{\displaystyle {30}^{\circ }}\\ \\ & \text{b}.{\displaystyle {60}^{\circ }}\\ \\ & \text{c}.{\displaystyle {120}^{\circ }}\\ \\ & \text{d}.{\displaystyle {240}^{\circ }}\\ \\ & \text{e}.{\displaystyle {300}^{\circ }}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\sin x& =-\frac{1}{2}\sqrt{3}\\ \text{Gun}& \text{akan rumus persamaan}\\ & \text{sederhana, yaitu}:\\ \sin x& =-\sin {60}^{\circ }\\ & =\sin ({180}^{\circ }+{60}^{\circ })\\ & =\sin {240}^{\circ }\\ x& ={240}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Jika}\cos x={\displaystyle \frac{2\sqrt{5}}{5}\text{maka nilai}}\\ & \cot x\left({\displaystyle \frac{\pi }{2}-x}\right)\text{adalah}....\\ \\ & \text{a}.{\displaystyle \frac{1}{2}}\\ \\ & \text{b}.{\displaystyle \frac{1}{3}}\\ \\ & \text{c}.{\displaystyle \frac{1}{6}}\\ \\ & \text{d}.{\displaystyle \frac{1}{7}}\\ \\ & \text{e}.{\displaystyle \frac{1}{8}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\cos x& =\frac{2\sqrt{5}}{5},\text{maka}\\ {\sin }^{2}x& +{\cos }^{2}x=1\\ \sin x& =1-{\cos }^{2}x\\ & =\sqrt{1-{\cos }^{2}x}=\sqrt{1-{\left({\displaystyle \frac{2\sqrt{5}}{5}}\right)}^2}\\ & =\sqrt{1-{\displaystyle \frac{20}{25}}}=\sqrt{{\displaystyle \frac{5}{25}}}={\displaystyle \frac{\sqrt{5}}{5}}\\ \cot & \left({\displaystyle \frac{\pi }{2}-x}\right)=\tan x,\text{maka}\\ \tan x& ={\displaystyle \frac{\sin x}{\cos x}}\\ & ={\displaystyle \frac{\sqrt{5}}{2\sqrt{5}}}\\ & ={\displaystyle \frac{1}{2}}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Pada setiap}\alpha \text{berlaku}\\ & \tan \alpha +\cos +\tan (-\alpha )+\cos (-\alpha )=....\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 2\tan \alpha \\ \text{c}.& 2\cos \alpha \\ \text{d}.& 2(\tan \alpha +\cos \alpha )\\ \text{e}.& 2\\ \\ & & \mathbf{\text{SAT Subjeck Test}}\end{array}\\ & \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\tan \alpha & +\cos +\tan (-\alpha )+\cos (-\alpha )\\ & =\tan \alpha +\cos -\tan \alpha +\cos \alpha \\ & =2\cos \alpha \end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Jika}\sin x=-\sin {35}^{\circ }\text{untuk}{90}^{\circ }\le x\le {270}^{\circ }\\ & \text{maka nilai}x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {215}^{\circ }\\ \text{b}.& {235}^{\circ }\\ \text{c}.& {240}^{\circ }\\ \text{d}.& {255}^{\circ }\\ \text{e}.& {270}^{\circ }\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Diketahui}\text{bahwa}\sin x=-\sin {35}^{\circ }\\ & \text{untuk}{90}^{\circ }\le x\le {270}^{\circ },\text{dengan}\\ & \sin x=-\sin {35}^{\circ }\\ & \text{(hanya terjadi dikuadran III dan IV)}\\ & \text{karena ba}\text{tasnya}\text{hanya untuk kuadran III saja, }\\ & \text{maka}\\ & =\sin ({180}^{\circ }+{35}^{\circ })\\ & =\sin {215}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Jika}\tan y=\tan {83}^{\circ }\text{untuk}{90}^{\circ }<y<{270}^{\circ }\\ & \text{maka nilai}x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {173}^{\circ }\\ \text{b}.& {187}^{\circ }\\ \text{c}.& {263}^{\circ }\\ \text{d}.& {268}^{\circ }\\ \text{e}.& {293}^{\circ }\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa nilai}\\ \tan y& =\tan {83}^{\circ }\text{untuk}{90}^{\circ }<y<{270}^{\circ }\\ \tan y& =\tan ({180}^{\circ }+{83}^{\circ }),\\ & \text{karena berada dikuadran III}\\ & =\tan {263}^{\circ }\end{array}\end{array}$.

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)

 $\begin{array}{l}\\ 102.& \text{Agar}\log (x^2-1)<0\text{maka}....\\ & \begin{array}{l}\\ \text{a}.& -1<x<1\\ \text{b}.& -\sqrt{2}<x<\sqrt{2}\\ \text{c}.& x<-1\text{atau}x>1\\ \text{d}.& x<-\sqrt{2}\text{atau}x>\sqrt{2}\\ \text{e}.& -\sqrt{2}<x<-1\text{atau}1<x<\sqrt{2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \log (x^2-1)<0\\ & \text{Diketahui}\log f(x)<0,\text{maka}\\ & \text{Syarat (1)},f(x)>0\\ & \iff x^2-1>0\\ & \iff x<-1\text{atau}x>1\\ & \text{Syarat (2)},\log (x^2-1)<0\\ & \log (x^2-1)<\log 1\\ & \iff x^2-1<1\\ & \iff x^2-2<0\\ & \iff x^2-{\left(\sqrt{2}\right)}^2<0\\ & \iff -\sqrt{2}<x<\sqrt{2}\\ & \text{Jadi},-\sqrt{2}<x<-1\text{atau}1<x<\sqrt{2}\end{array}\end{array}$

$\begin{array}{l}\\ 103.& \text{Himpunan penyelesaian dari}\\ & {}^{{.}^{\frac{1}{2}}}\log (x^2-3)>0\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{x|-2<x<-\sqrt{3}\text{atau}\sqrt{3}<x<2\}\\ \text{b}.& \{x|-\sqrt{3}<x<-1\text{atau}\sqrt{3}<x<2\}\\ \text{c}.& \{x|-2<x<-\sqrt{3}\}\\ \text{d}.& \{x|-2<x<-\sqrt{3}\}\\ \text{e}.& \{x|\sqrt{3}<x<2\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {}^{{.}^{\frac{1}{2}}}\log (x^2-3)>0\\ & \text{Diketahui}{}^{{.}^{\frac{1}{2}}}\log f(x)>0,\text{maka}\\ & \text{Syarat (1)},f(x)>0\\ & \iff x^2-3>0\\ & \iff x<-\sqrt{3}\text{atau}x>\sqrt{3}\\ & \text{Syarat (2)},{}^{{.}^{\frac{1}{2}}}\log (x^2-3)>0\\ & {}^{{.}^{\frac{1}{2}}}\log (x^2-3)>{}^{{.}^{\frac{1}{2}}}\log 1\\ & \iff x^2-3<1\left(\text{karena basisnya}{\displaystyle \frac{1}{2}<1}\right)\\ & \iff x^2-4<0\\ & \iff x^2-2^2>0\\ & \iff -2<x<2\\ & \text{Jadi},-2<x<-\sqrt{3}\text{atau}\sqrt{3}<x<2\end{array}\end{array}$

$\begin{array}{l}\\ 104.& \text{Nilai}x\text{yang memenuhi}\\ & {}^2\log (x^2-x)\le 1\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& x<0\text{atau}x>1\\ \text{b}.& -1\le x\le 2,x\ne 1\text{atau}x\ne 0\\ \text{c}.& -1\le x<0\text{atau}1<x\le 2\\ \text{d}.& -1<x\le 0\text{atau}1\le x<2\\ \text{e}.& -1\le x\le 0\text{atau}1\le x\le 2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {}^2\log (x^2-x)\le 1\\ & \text{Diketahui}{}^2\log f(x)>0,\text{maka}\\ & \text{Syarat (1)},f(x)>0\\ & \iff x^2-x>0\iff x(x-1)>0\\ & \iff x<0\text{atau}x>1\\ & \text{Syarat (2)},{}^2\log (x^2-x)\le 1\\ & {}^2\log (x^2-x)\le {}^2\log 2\\ & \iff x^2-x\le 2\\ & \iff x^2-x-2\le 0\\ & \iff (x+1)(x-2)\le 0\\ & \iff -1\le x\le 2\\ & \text{Jadi},-1\le x<0\text{atau}1<x\le 2\end{array}\end{array}$

$\begin{array}{l}\\ 105.& \text{Nilai}x\text{yang memenuhi}\\ & |\log (x+1)|>1\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& x<-0,9\text{atau}x>9\\ \text{b}.& x<-9\text{atau}x>9\\ \text{c}.& -1<x<-0,9\text{atau}x>9\\ \text{d}.& -9<x<0,9\\ \text{e}.& -0,9<x<9\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Ingat bahwa}\\ & \left|x\right|>A\iff x<-A\text{atau}x>A,A>0\\ & \iff \log (x+1)<-1\text{atau}\log (x+1)>1\\ & \text{Syarat (1) buat keduanya},f(x)>0\\ & (x+1)>0\iff x>-1\\ & \text{Syarat (2)},\log (x+1)<-1\\ & \log (x+1)<\log {10}^{-1}\\ & x+1<{\displaystyle \frac{1}{10}\iff x<-\frac{9}{10}}\\ & \text{Syarat (3)},\log (x+1)>1\\ & \log (x+1)>\log {10}^1\\ & (x+1)>10\iff x>9\\ & \text{Jadi},-1<x<-0,9\text{atau}x>9\end{array}\end{array}$.

$\begin{array}{l}\\ 106.& \text{Himpunan penyelesaian pertidaksamaan}\\ & \log 4+\log (x+3)\le \log x^2\\ & \text{a}.\{x|x\ge 6,x\in \mathbb{R}\}\\ & \text{b}.\{x|-3<x\le -2\text{atau}x\ge 6x\in \mathbb{R}\}\\ & \text{c}.\{x|-3<x\le -2\text{atau}0\le x\le 6x\in \mathbb{R}\}\\ & \text{d}.\{x|x\le -2\text{atau}x\ge 6,x\in \mathbb{R}\}\\ & \text{e}.\{x|x\le -4\text{atau}x\ge 4,x\in \mathbb{R}\}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & \log 4+\log (x+3)\le \log x^2\\ & \text{adalah bentuk}{}^a\log f(x)\le {}^a\log g(x)\\ & \text{Syarat penyelesaian ada 2}\\ & \bullet \text{basis}:a=10>0,\ne 1\\ & \bullet \text{numerus}:\{\begin{array}{ll}(1)& x+3>0\Rightarrow x>-3\\ (2)& x^2>0\Rightarrow x\ne 0\end{array}\\ & \text{Proses penyelesaian}\\ & \log 4(x+3)\le \log x^2\\ & 4(x+3)\le x^2\iff x^2\ge 4x+12\\ & \iff x^2-4x-12\ge 0\\ & \iff (x+2)(x-6)\ge 0\\ & x\le -2\text{atau}x\ge 6\\ & \text{Jadi},\text{HP}=\{x|-3<x\le -2\text{atau}x\ge 6x\in \mathbb{R}\}\end{array}\end{array}$.

$.\begin{array}{rl}& \text{Proses penyelesaian}\\ & {}^6\log (x^2-x-6)>1\\ & \iff {}^6\log (x^2-x-6)>1.{}^6\log 6\\ & \iff {}^6\log (x^2-x-6)>{}^6\log 6\\ & \iff {}^a\log f(x)>{}^a\log p,\\ & \text{Karena}\text{basis}=a=6,\text{maka tanda}\\ & \text{pertidaksamaan tetap. Selanjutnya}\\ & f(x)>p)\\ & \iff x^2-x-6>6\\ & \iff x^2-x-12>0\\ & \iff (x+3)(x-4)<0\\ & \iff x<-3\text{atau}x>4\\ & \text{Karena}-2<x\text{atau}x>3,\text{maka}\\ & \text{HP}=\left\{x<-3\text{atau}x>4\right\}\end{array}$.

$\begin{array}{ll}108.& \text{Himpunan penyelesaian dari}\\ & \text{pertidaksamaan bentuk}\\ & \log x^2<\log (x+3)+2\log 2\\ & \text{adalah}....\\ & \begin{array}{ll}\text{a}.& \{-3<x<0\text{atau}0<x<6\}\\ \text{b}.& \{-2<x<0\text{atau}0<x<6\}\\ \text{c}.& \{-1<x<0\text{atau}0<x<6\}\\ \text{d}.& \{-2<x<0\text{atau}0<x<7\}\\ \text{e}.& \{-1<x<0\text{atau}0<x<8\}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{|cc|}\hline \text{Syarat Numerus}& \text{Syarat Numerus}\\ \begin{array}{rl}f(x)>0& \\ x^2>0& \\ x\ne 0& \end{array}& \begin{array}{rl}g(x)& >0\\ x+3& >0\\ x& >-3\end{array}\\ \text{Kita pilih}& \{\begin{array}{ll}& -3<x<0\\ & \text{atau}\\ & x>0\end{array}\\ \hline\end{array}\\ & \begin{array}{rl}& \log x^2<\log (x+3)+2\log 2\\ & \iff \log x^2<\log (x+3)+\log 2^2\\ & \iff \log x^2<\log (x+3).2^2\\ & \iff {}^a\log f(x)<{}^a\log g(x),\\ & \text{Karena}\text{basis}=a=10,\text{maka tanda}\\ & \text{pertidaksamaan tetap. Selanjutnya}\\ & f(x)<g(x)\\ & \iff x^2<(x+3){.2}^2\\ & \iff x^2<(x+3).4\\ & \iff x^2<4x+12\\ & \iff x^2-4x-12<0\\ & \iff (x+2)(x-6)<0\\ & \iff -2<x<6\\ & \text{Karena}-3<x<0\text{atau}x>0,\text{maka}\\ & \text{HP}=\{-2<x<0\text{atau}0<x<6\}\end{array}\end{array}$.

$\begin{array}{ll}109.& \text{Suatu larutan memiliki konsentrasi}\\ & \text{ion}{\text{H}}^{+}\text{sebesar}2\times {10}^{-6}.\\ & \text{PH dari larutan tersebutadalah}....\\ & (\log 2=0,3010)\\ & \begin{array}{llllll}\text{a}.& 4.3& & & \text{d}.& 5,7\\ \text{b}.& 4,7& \text{c}.& 5,3& \text{e}.& 6,3\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{pH}& =-\log \left[{\text{ H}}^{+}\right]\\ & =-\log (2\times {10}^{-6})\\ & =-\log 2-\log {10}^{-6}\\ & =-0,3010-(-6)\log 10\\ & =-0,3010+6.1\\ & =-0,3010+6\\ & =6-0,3010\\ & =5,699\text{dibulatkan}\\ & =5,7\end{array}\end{array}$

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

 $\begin{array}{l}\\ 93.& \text{Persamaan}{}^x\log 2+{}^x\log (3x-4)=2\\ & \text{mempunyai akar}{x}_1\text{dan}{x}_2,\text{maka}\\ & \text{nilai}{x}_1+x_2\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 3\\ \text{c}.& 4\\ \text{d}.& 6\\ \text{e}.& 8\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Alternatif 1}\\ & {}^x\log 2+{}^x\log (3x-4)=2\\ & \iff {}^x\log 2(3x-4)=2\\ & \iff {}^x\log 6x-8=2\\ & \iff 6x-8=x^2\\ & \iff x^2-6x+8=0,\text{dengan}\{\begin{array}{ll}a& =1\\ b& =-6\\ c& =8\end{array}\\ & \iff x_1+x_2=-{\displaystyle \frac{b}{a}}\\ & \iff x_1+x_2=-{\displaystyle \frac{-6}{1}=6}\\ & \text{Alternatif 2}\\ & \iff x^2-6x+8=0\\ & \iff (x-2)(x-4)\\ & \iff x_1=2\text{atau}{x}_2=4\\ & \iff x_1+x_2=2+4=6\end{array}\end{array}$

$\begin{array}{l}\\ 94.& \text{Jika}{x}_1\text{dan}{x}_2\text{memenuhi}\\ & (\log x)(2\log x-3)=\log 100\\ & \text{maka}{x}_1\times x_2\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 100\\ \text{b}.& 10\sqrt{10}\\ \text{c}.& \sqrt{10}\\ \text{d}.& -\sqrt{10}\\ \text{e}.& -10\sqrt{10}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& (\log x)(2\log x-3)=\log 100\\ & \iff (\log x)(2\log x-3)=2\\ & \iff 2{\log }^{2}x-3\log x-2=0\{\begin{array}{ll}a& =2\\ b& =-3\\ c& =-2\end{array}\\ & \iff \log x_1+\log x_2=-{\displaystyle \frac{-3}{2}=\frac{3}{2}}\\ & \iff \log (x_1\times x_2)=1{\displaystyle \frac{1}{2}}\\ & \iff (x_1\times x_2)={10}^{1\frac{1}{2}}=10\sqrt{10}\end{array}\end{array}$

$\begin{array}{l}\\ 95.& \text{Persamaan}\\ & {10}^{{}^{{}^2}\log x^2}-7\left({10}^{{}^{{}^2}\log x}\right)+10=0\\ & \text{mempunyai dua akar yaitu}{x}_1\text{dan}{x}_2\\ & \text{Nilai}{x}_1\times x_2=....\\ & \begin{array}{l}\\ \text{a}.& -2\\ \text{b}.& -5\\ \text{c}.& 2\\ \text{d}.& 5\\ \text{e}.& 10\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {10}^{{}^{{}^2}\log x^2}-7\left({10}^{{}^{{}^2}\log x}\right)+10=0\\ & {10}^{2^{{}^2}\log x}-7\left({10}^{{}^{{}^2}\log x}\right)+10=0\\ & \text{adalah persamaan kuadrat dalam}{10}^{{}^{{}^2}\log x}\\ & \text{Misalkan}p={10}^{{}^{{}^2}\log x},\text{maka persamaan}\\ & \text{menjadi}{p}^2-7p+10=0\{\begin{array}{ll}a& =1\\ b& =-7\\ c& =10\end{array}\\ & \text{Karena nilai}{p}_1\times p_2={\displaystyle \frac{c}{a}\text{maka}}\\ & {10}^{{}^{{}^2}\log x_1}\times {10}^{{}^{{}^2}\log x_2}={\displaystyle \frac{10}{1}=10}\\ & {10}^{{}^{{}^2}\log x_1+{}^2\log x_2}=10\\ & {10}^{{}^{{}^2}\log x_1+{}^2\log x_2}={10}^1\\ & \iff {}^2\log x_1+{}^2\log x_2=1\\ & \iff {}^2\log x_1\times x_2=1\\ & \iff x_1\times x_2=2^1=2\end{array}\end{array}$

$\begin{array}{l}\\ 96.& \text{Nilai}x\text{yang memenuhi}\\ & {}^x\log (x+12)-3.{}^x\log 4+1=0\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{2}}\\ \text{b}.& 2\\ \text{c}.& 4\\ \text{d}.& 8\\ \text{e}.& 16\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {}^x\log (x+12)-3.{}^x\log 4+1=0\\ & {}^x\log (x+12)-{}^x\log 4^3=-1\\ & {}^x\log {\displaystyle \frac{x+12}{64}=-1}\\ & {\displaystyle \frac{x+12}{64}=x^{-1}=\frac{1}{x}}\\ & x+12={\displaystyle \frac{64}{x}}\\ & x^2+12x-64=0\\ & (x+16)(x-4)=0\\ & x=-16\text{atau}x=4\end{array}\end{array}$

$\begin{array}{l}\\ 97.& \text{Nilai}x\text{yang memenuhi}\\ & {\displaystyle \frac{{}^2\log (2x-3)}{{}^2\log x}-{}^x\log (x+6)+{\displaystyle \frac{1}{{}^{(x+2)}\log x}=1}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -1\text{dan}6\\ \text{b}.& -2\text{dan}6\\ \text{c}.& -1\\ \text{d}.& -2\\ \text{e}.& 6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& {\displaystyle \frac{{}^2\log (2x-3)}{{}^2\log x}-{}^x\log (x+6)+{\displaystyle \frac{1}{{}^{(x+2)}\log x}=1}}\\ & {}^x\log (2x-3)-{}^x\log (x+6)+{}^x\log (x+2)=1\\ & {}^x\log (2x-3)(x+2)=1+\log (x+6)\\ & {}^x\log {\displaystyle \frac{(2x^2+x-6)}{(x+6)}=1}\\ & {\displaystyle \frac{(2x^2+x-6)}{(x+6)}=x^1}\\ & (2x^2+x-6)=x^2+6x\\ & x^2-5x-6=0\\ & (x+1)(x-6)=0\\ & x=-1\text{atau}x=6\end{array}\end{array}$.

$\begin{array}{l}\\ 98.& \text{Himpunan penyelesaian dari persamaan}\\ & {}^3\log (x^2-5x+7)=0\text{adalah}....\\ & \begin{array}{lllll}\text{a}.& \{2,3\}& & \text{d}.& \{3,4\}\\ \text{b}.& \{2,4\}& & \text{e}.& \{3,5\}\\ \text{c}.& \{2,5\}& & & \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}:\\ & {}^3\log (x^2-5x+7)=0\\ & \iff {}^3\log (x^2-5x+7)={}^3\log 3^0\\ & \iff {}^3\log (x^2-5x+7)={}^3\log 1\\ & \text{bersesuaian dengan rumus}{}^a\log f(x)={}^a\log p\\ & \underset{―}{\text{Syarat numerus}}\\ & f(x)>0\iff x^2-5x+7>0\\ & \text{adalah definit positif}\\ & \text{sehingga semua nilai}x\text{memenuhi}\\ & \underset{―}{\text{Langkah berikutnya}}\\ & f(x)=p\\ & \iff x^2-5x+7=1\\ & \iff x^2-5x+6=0\\ & \iff (x-2)(x-3)=0\\ & \iff x=2\text{atau}x=3\\ & \text{Jadi, HP}=\{2,3\}\end{array}\end{array}$.

$\begin{array}{l}\\ 99.& \text{Himpunan penyelesaian dari}\\ & \log x^2=\log 4+\log (x+3)\\ & \text{adalah}....\\ & \begin{array}{llllll}\text{a}.& \{1,4\}& & & \text{d}.& \{2,6\}\\ \text{b}.& \{1,6\}& \text{c}.& {\displaystyle \{2,4\}}& \text{e}.& \{4,6\}\end{array}\\ \\ & \text{Jawab}:\text{tidak ada}\\ & \text{ingat formula}:{}^a\log f(x)={}^a\log g(x)\\ & \begin{array}{|cc|}\hline \mathbf{\text{Syarat Numerus}}& \mathbf{\text{Syarat Numerus}}\\ f(x)& g(x)\\ \begin{array}{rl}{x}^2>0& \\ x<0\text{atau}x& >0\end{array}& \begin{array}{rl}x+3& >0\\ x& >-3\end{array}\\ \begin{array}{rl}& \text{Yang digunakan}\\ & \text{adalah yang }\\ & \text{memenuhi}\\ & \text{keduanya}\\ & \text{yaitu}:\end{array}& \begin{array}{rl}& -3<x<0\\ & \text{atau}\\ & x>0\end{array}\\ \hline\end{array}\\ & \begin{array}{rl}& \text{Syarat Penyelesaian}\\ & \log x^2=\log 4+\log (x+3)\\ & \iff \log x^2=\log 4(x+3)\\ & \text{maka},f(x)=g(x)\\ & x^2=4(x+3)\\ & \iff x^2=4x+12\\ & \iff x^2-4x-12=0\\ & \iff (x+2)(x-6)=0\\ & \iff x+2=0\text{atau}x-6=0\\ & \iff x=-2\text{atau}x=6\\ & \text{karena syaratnya},-3<x<0\text{atau}x>0,\\ & \text{maka keduanya memenuhi}\\ & \text{HP}=\{-2,6\}\end{array}\end{array}$

$\begin{array}{l}\\ 100.& \text{Nilai}x\text{yang memenuhi persamaan}\\ & \log x^2=\log x\text{adalah}....\\ & \begin{array}{lllll}\text{a}.& 1& & \text{d}.& 4\\ \text{b}.& 2& & \text{e}.& 5\\ \text{c}.& 3& & & \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Diketahui}\log x^2=\log x\\ & \text{bersesuaian rumus}{}^a\log f(x)={}^a\log g(x)\\ & \underset{―}{\text{Syarat numerus}}\\ & \begin{array}{ll}\begin{array}{rl}& f(x)>0\\ & x^2>0\\ & x>0\text{atau}x>0\end{array}& \begin{array}{rl}& g(x)>0\\ & x>0\\ & \end{array}\end{array}\\ & \text{Sehingga syarat numerusnya}x>0\\ & \underset{―}{\text{Syarat berikutnya}}\\ & \begin{array}{rl}f(x)& =g(x)\\ x^2& =x\\ x^2-x& =0\\ x(x-1)& =0\\ x=0\text{atau}x& =1\end{array}\\ & \text{Jadi, HP}=\left\{1\right\}\end{array}\end{array}$.

$\begin{array}{l}\\ 101.& \text{Salah satu nilai}x\text{yang memenuhi persamaan}\\ & 2{\log }^{2}x-9\log x+4=0\text{adalah}....\\ & \begin{array}{lllll}\text{a}.& {\displaystyle \sqrt{10}}& & \text{d}.& 100\\ \text{b}.& {\displaystyle 1}& & \text{e}.& 1000\\ \text{c}.& 10& & & \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Diketahui}2{\log }^{2}x-9\log x+4=0\\ & \text{bersesuaian rumus}:\\ & A{({}^a\log f(x))}^2+B({}^a\log f(x))+C=0\\ & \underset{―}{\text{Langkah pengerjaan}}\\ & \begin{array}{rl}& 2{\log }^{2}x-9\log x+4=0\\ & \iff (2\log x-1)(\log x-4)=0\\ & \iff 2\log x-1=0\text{atau}\log x-4=0\\ & \iff \log x={\displaystyle \frac{1}{2}\text{atau}\log x=4}\\ & \iff x={10}^{{.}^{\frac{1}{2}}}\text{atau}x={10}^4\\ & \iff x=\sqrt{10}\text{atau}x=10000\end{array}\\ & \text{Jadi, HP}=\{\sqrt{10},10000\}\end{array}\end{array}$.

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

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

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

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

$\begin{array}{l}\\ 88.& \text{Fungsi invers dari}f(x)=5^x\\ & \begin{array}{lll}\text{a}.& f^{-1}(x)=5^{-x}& \\ \text{b}.& f^{-1}(x)={\left({\displaystyle \frac{1}{5}}\right)}^x\\ \text{c}.& f^{-1}(x)={\left({\displaystyle \frac{1}{5}}\right)}^{-x}& \\ \text{d}.& f^{-1}(x)={}^5\log x\\ \text{e}.& f^{-1}(x)={}^x\log 5\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}:f(x)=5^x,\text{maka inversnya}\\ & \text{adalah}:\\ & \text{Langkah mula-mula dilogkan}\\ & \text{masing-masing ruas untuk mencari nilai}\\ & x,\text{yaitu}:\\ & \log f(x)=\log 5^x\iff \log f(x)=x\log 5\\ & \iff x={\displaystyle \frac{\log f(x)}{\log 5}={}^5\log f(x)}\\ & \text{Selanjutnya kita ganti}x\text{dengan}{f}^{-1}(x),\text{dan}\\ & f(x)\text{dengan}x,\text{sehingga menjadi bentuk}\\ & \iff f^{-1}(x)={}^5\log x\end{array}\end{array}$.

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

$\begin{array}{ll}90.& \text{Diketahui}f(x)={}^a\log (x+1).\text{Nilai}f(7)\\ & \text{jika nilai}f(1)=1\text{adalah}....\\ & \begin{array}{llllll}\text{a}.& {\displaystyle \frac{1}{4}}& & & \text{d}.& 3\\ \\ \text{b}.& {\displaystyle \frac{1}{2}}& \text{c}.& 2& \text{e}.& 4\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}& f(x)={}^a\log (x+1)\\ & f(1)={}^a\log (1+1)=1\\ & \iff {}^a\log 2=1\\ & \iff 2=a^1\\ & \iff 2=a\end{array}\end{array}$.

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

$.\begin{array}{rl}& \text{Selanjutnya}\\ & =5\left({\displaystyle \frac{A+2B}{A+B}}\right)+3\left({\displaystyle \frac{2A-3B}{A-B}}\right)\\ & -2\left({\displaystyle \frac{A+2B}{A+B}}\right)\left({\displaystyle \frac{2A-3B}{A-B}}\right)\\ & =\left({\displaystyle \frac{5A+10B}{A+B}}\right)+\left({\displaystyle \frac{6A-9B}{A-B}}\right)\\ & -\left({\displaystyle \frac{2A+4B}{A+B}}\right)\left({\displaystyle \frac{2A-3B}{A-B}}\right)\\ & ={\displaystyle \frac{(5A+10B)(A-B)+(6A-9B)(A+B)}{A^2-B^2}}\\ & -\left({\displaystyle \frac{4A^2-6AB+8AB-12B^2}{A^2-B^2}}\right)\\ & ={\displaystyle \frac{5A^2-5AB+10AB-10B^2}{A^2-B^2}}\\ & +{\displaystyle \frac{6A^2+6AB-9AB-9B^2}{A^2-B^2}}\\ & -\left({\displaystyle \frac{4A^2-6AB+8AB-12B^2}{A^2-B^2}}\right)\\ & ={\displaystyle \frac{7A^2-7B^2}{A^2-B^2}}\\ & ={\displaystyle \frac{7(A^2-B^2)}{A^2-B^2}}\\ & =7\end{array}$

$\begin{array}{l}\\ 92.& \text{Diberikan}A={}^6\log 16\text{dan}B={}^{12}\log 27\\ & \text{Terdapat bilangan-bilangan bulat positif}\\ & a,b,\text{dan}c\text{sehingga}(A+a)(B+b)=c\\ & \text{Nilai dari}a+b+c\text{adalah}....\\ \\ & \text{KOMPETISI HARDIKNAS ONLINE}\\ & \text{POSI}(\text{Pelatihan Olimpiade Sain Indonesia})\\ & \text{Bidang Matematika 2020}\\ & \begin{array}{l}\\ \text{a}.& 23\\ \text{b}.& 24\\ \text{c}.& 27\\ \text{d}.& 30\\ \text{e}.& 34\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{....}}\\ & \begin{array}{rl}& \text{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}}}\\ & \iff \log 2+\log 3={\displaystyle \frac{4\log 2}{A}...........(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}}\\ & \iff 2\log 2+\log 3={\displaystyle \frac{3\log 3}{B}.........(2)}\\ & \text{ELIMINASI}\\ & \text{Dari persamaan (1) dan (2) diperoleh}:\\ & \bullet \log 2={\displaystyle \frac{3\log 3}{B}-{\displaystyle \frac{4\log 2}{A}}}\\ & \iff \log 2={\displaystyle \frac{3A\log 3-4B\log 2}{AB}}\\ & \iff AB\log 2=3A\log 3-4B\log 2\\ & \iff AB\log 2+4B\log 2=3A\log 3\\ & \iff (AB+4B)\log 2=3A\log 3\\ & \iff {\displaystyle \frac{\log 2}{\log 3}={\displaystyle \frac{3A}{AB+4B}..........(3)}}\\ & \bullet \log 3={\displaystyle \frac{8\log 2}{A}-{\displaystyle \frac{3\log 3}{B}}}\\ & \iff \log 3={\displaystyle \frac{8B\log 2-3A\log 3}{AB}}\\ & \iff AB\log 3=8B\log 2-3A\log 3\\ & \iff AB\log 3+3A\log 3=8B\log 2\\ & \iff (AB+3A)\log 3=8B\log 2\\ & \iff \frac{\log 2}{\log 3}={\displaystyle \frac{AB+3A}{8B}...........(4)}\\ & \text{KESAMAAN}\\ & \frac{\log 2}{\log 3}=\frac{\log 2}{\log 3}\\ & {\displaystyle \frac{AB+3A}{8B}={\displaystyle \frac{3A}{AB+4B}}}\\ & \iff (AB+3A)(AB+4B)=(8B).(3A)\\ & \iff (B+3)(A+4)=24\\ & \iff (A+4)(B+3)=24\\ & \text{KESIMPULAN}\\ & a=4,b=3,\text{dan}c=24,\\ & \text{maka}a+b+c=4+3+24=31\end{array}\end{array}$.

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

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

$\begin{array}{l}\\ 72.& \mathbf{\text{(SPMB '04)}}\\ & \text{Jika}a>1,\text{maka penyelesaian untuk}\\ & ({}^a\log (2x+1))({}^3\log \sqrt{a})=1\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 1}\\ \text{b}.& {\displaystyle 2}\\ \text{c}.& {\displaystyle 3}\\ \text{d}.& {\displaystyle 4}\\ \text{e}.& {\displaystyle 5}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}({}^a\log (2x+1))({}^3\log \sqrt{a})& =1\\ ({}^3\log \sqrt{a})({}^a\log (2x+1))& =1\\ ({}^3\log a^{{.}^{{}^{\frac{1}{2}}}})({}^a\log (2x+1))& =1\\ {\displaystyle \frac{1}{2}({}^3\log a)({}^a\log (2x+1))}& =1\\ {}^3\log (2x+1)& =2\\ 2x+1& =3^2\\ 2x& =9-1\\ 2x& =8\\ x& =4\end{array}\end{array}$.

$\begin{array}{l}\\ 73.& \mathbf{\text{(SPMB '04)}}\\ & \text{Nilai}{\displaystyle \frac{{({}^5\log 10)}^2-{({}^5\log 2)}^2}{{}^5\log \sqrt{20}}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{2}}\\ \text{b}.& {\displaystyle 1}\\ \text{c}.& {\displaystyle 2}\\ \text{d}.& {\displaystyle 4}\\ \text{e}.& {\displaystyle 5}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {\displaystyle \frac{{({}^5\log 10)}^2-{({}^5\log 2)}^2}{{}^5\log \sqrt{20}}}\\ & ={\displaystyle \frac{({}^5\log 10+{}^5\log 2)({}^5\log 10-{}^5\log 2)}{{}^5\log (20{)}^{{.}^{\frac{1}{2}}}}}\\ & ={\displaystyle \frac{{}^5\log (10.2){\times }^5\log \left(\frac{10}{2}\right)}{{\displaystyle \frac{1}{2}\times {}^5\log 20}}}\\ & =2\times \left({\displaystyle \frac{{}^5\log 20}{{}^5\log 20}}\right)\times {}^5\log 5\\ & =2.1.1\\ & =2\end{array}\end{array}$.

$\begin{array}{l}\\ 74.& \mathbf{\text{(SPMB '03)}}\\ & \text{Jika diketahui bahwa}\\ & {}^4{\log }^4\log x-{}^4{\log }^4{\log }^4\log 16=2\\ & \text{maka}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle {}^2\log x=8}\\ \text{b}.& {\displaystyle {}^2\log x=4}\\ \text{c}.& {\displaystyle {}^4\log x=8}\\ \text{d}.& {\displaystyle {}^4\log x=16}\\ \text{e}.& {\displaystyle {}^{16}\log x=8}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {}^4\log {}^4\log x-{}^4\log {}^4\log {}^4\log 16=2\\ & \iff {}^4\log {}^4\log x-{}^4\log {}^4\log {}^4\log 4^2=2\\ & \iff {}^4\log {}^4\log x-{}^4\log {}^4\log 2=2\\ & \iff {}^4\log {}^4\log x-{}^4\log {}^{2^2}\log 2^1=2\\ & \iff {}^4\log {}^4\log x-{}^4\log \left({\displaystyle \frac{1}{2}}\right)=2\\ & \iff {}^4\log {}^4\log x-{}^{2^2}\log 2^{-1}=2\\ & \iff {}^4\log {}^4\log x-(-{\displaystyle \frac{1}{2}})=2\\ & \iff {}^4\log {}^4\log x+\frac{1}{2}=2\\ & \iff {}^4\log {}^4\log x=2-{\displaystyle \frac{1}{2}=\frac{3}{2}}\\ & \iff {}^4\log x=4^{.\frac{3}{2}}\\ & \iff {}^4\log x={\left(2^2\right)}^{{.}^{\frac{3}{2}}}\\ & \iff {}^4\log x=2^3\\ & \iff {}^4\log x=8\end{array}\end{array}$.

$\begin{array}{l}\\ 75.& \mathbf{\text{(UMPTN '92)}}\\ & \text{Jika}x\text{memenuhi persamaan}\\ & {}^4{\log }^4\log x-{}^4{\log }^4{\log }^4\log 16=2\\ & \text{maka nilai}{}^{16}\log x=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 4}\\ \text{b}.& {\displaystyle 2}\\ \text{c}.& {\displaystyle 1}\\ \text{d}.& {\displaystyle -2}\\ \text{e}.& {\displaystyle -4}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {}^4\log {}^4\log x-{}^4\log {}^4\log {}^4\log 16=2\\ & \text{menyebabkan}\\ & {}^4\log x=8\Rightarrow x=4^8\\ & (\text{lihat pembahasan no.23})\\ & \text{maka},\\ & {}^{16}\log x={}^{16}\log 4^8={}^{4^2}\log 4^8\\ & ={\displaystyle \frac{8}{2}}\\ & =4\end{array}\end{array}$.

$\begin{array}{l}\\ 78.& \mathbf{\text{(UMPTN '94)}}\\ & \text{Hasil kali akar-akar persamaan}\\ & {}^3\log x^{.\left({}^{2+{}^3\log x}\right)}=15\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{9}}\\ \text{b}.& {\displaystyle \frac{1}{3}}\\ \text{c}.& {\displaystyle 1}\\ \text{d}.& {\displaystyle 3}\\ \text{e}.& {\displaystyle 9}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {}^3\log x^{.\left({}^{2+{}^3\log x}\right)}=15\\ & \iff {\left(2+{}^3\log x\right)}^3\log x-15=0\\ & \iff 2{}^3\log x+{({}^3\log x)}^2-15=0\\ & \iff {({}^3\log x)}^2+2{}^3\log x-15=0\\ & \iff ({}^3\log x_1+5)({}^3\log x_2-3)=0\\ & \iff {}^3\log x_1+5=0\text{atau}{}^3\log x_1-3=0\\ & \iff {}^3\log x_1=-5\text{atau}{}^3\log x_2=3\\ & \iff x_1=3^{-5}\text{atau}{x}_2=3^3\\ & \text{maka}\\ & \iff x_1\times x_2=3^{-5}\times 3^3=3^{-5+3}=3^{-2}\\ & \iff ={\displaystyle \frac{1}{3^2}}\\ & \iff =\frac{1}{9}\end{array}\end{array}$.

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

$\begin{array}{l}\\ 80.& \mathbf{\text{(AIME 1984)}}\\ & \text{Diketahui bahwa}x\text{dan}y\\ & \text{adalah bilangan real yang memenuhi}\\ \\ & \{\begin{array}{c}{}^{{}^8}\log x+{}^{{}^4}\log y^2=5\\ \\ {}^{{}^8}\log y+{}^{{}^4}\log x^2=7\end{array}\\ \\ & \text{Tentukanlah nilai dari}xy\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {}^{{}^{{}^8}}\log x+{}^{{}^4}\log y^2=5\\ & \iff {}^{{}^{{}^{2^3}}}\log x+{}^{{}^{{}^{2^2}}}\log y^2=5....(1)\\ & {}^{{}^{{}^8}}\log y+{}^{{}^4}\log x^2=7\\ & \iff {}^{{}^{{}^{2^3}}}\log y+{}^{{}^{{}^{2^2}}}\log x^2=7....(2)\\ & \text{selanjutnya},& \\ & \frac{1}{3}{.}^{{}^{{}^2}}\log x+{}^{{}^{{}^2}}\log y=5|\times \frac{1}{3}|\\ & \Rightarrow \frac{1}{9}{.}^{{}^{{}^2}}\log x+\frac{1}{3}.{}^{{}^{{}^2}}\log y=\frac{5}{3}....(3)\\ & {}^{{}^{{}^2}}\log x+\frac{1}{3}{.}^{{}^{{}^2}}\log y=7|\times 1|\\ & {\Rightarrow }^{{}^{{}^2}}\log x+\frac{1}{3}{.}^{{}^{{}^2}}\log y=7....(4)\\ & \text{saat persamaan}(3)-(4)\\ & =-\frac{8}{9}.{}^{{}^{{}^2}}\log x=\frac{5}{3}-7=-\frac{16}{3}\\ & \text{maka}\\ & {}^{{}^{{}^2}}\log x=(-\frac{16}{3})(-\frac{9}{8})\\ & {}^{{}^{{}^2}}\log x=6\iff x=2^6\iff x=64\\ & \text{Pada persamaan 1 selanjutnya}\\ & \frac{1}{3}.{}^{{}^{{}^2}}\log x+{}^{{}^{{}^2}}\log y=5\\ & \iff \frac{1}{3}.{}^{{}^{{}^2}}\log 2^6+{}^{{}^{{}^2}}\log y=5\\ & \iff \frac{1}{3}.6+{}^{{}^{{}^2}}\log y=5\\ & \iff 2+{}^{{}^{{}^2}}\log y=5\\ & \iff {}^{{}^{{}^2}}\log y=5-2=3\iff y=2^3=8\\ & \text{Jadi},x.y=64.8=512\end{array}\end{array}$

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

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

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

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

$.\begin{array}{rl}& \text{dari persamaan}(4)\text{dan}(5)\\ & \log (2y)={\displaystyle \frac{\log x\times \log (2z)}{\log 2}}\\ & \log \left({\displaystyle \frac{x}{z}}\right)={\displaystyle \frac{\log x\times \log (2z)}{\log 2}}\\ & \log 2(\log x-\log z)=\log x\times \log (2z)\\ & \log 2\times \log x-\log 2\times \log z=\log x\times (\log 2+\log z)\\ & \log 2\times \log x-\log 2\times \log z=\log x\times \log 2+\log x\times \log z\\ & -\log 2\times \log z=\log x\times \log z\\ & \log 2^{-1}=\log x\\ & {\displaystyle \frac{1}{2}=x.....(6)}\end{array}$

$.\begin{array}{|cc|}\hline \text{persamaan}(2)& \text{Menentukan nilai}z\\ \begin{array}{rl}\log 2y\times \log (4x)& =\log x\times \log (8yz)\\ \log 2y\times \log (4(2yz))& =\log x\times \log (8yz)\\ \log 2y\times \log (8yz)& =\log x\times \log (8yz)\\ \log (2y)& =\log x\\ 2y& =x\\ y& ={\displaystyle \frac{1}{2}x}\\ & ={\displaystyle \frac{1}{2}\times \frac{1}{2}}\\ & ={\displaystyle \frac{1}{4}.....(7)}\\ & \end{array}& \begin{array}{rl}& \\ x& =2yz\\ {\displaystyle \frac{1}{2}}& =2\left({\displaystyle \frac{1}{4}}\right)z\\ 1& =z\\ & \\ & \\ & \\ & \\ & \\ & \\ & \end{array}\\ \hline\end{array}$

$.\begin{array}{rl}& \text{maka nilai untuk}x{y}^{5}z\text{adalah}\\ & x{y}^{5}z=\left({\displaystyle \frac{1}{2}}\right).{\left({\displaystyle \frac{1}{4}}\right)}^5.1\\ & ={\displaystyle \frac{1}{2\times 4^5}}\\ & ={\displaystyle \frac{1}{2\times {\left(2^2\right)}^5}={\displaystyle \frac{1}{2^{1+10}}}}\\ & ={\displaystyle \frac{1}{2^{11}}={\displaystyle \frac{1}{2^{{}^{\frac{11}{1}}}}={\displaystyle \frac{1}{2^{{}^{\frac{p}{q}}}}}}}\\ & \{\begin{array}{ll}p& =11\\ q& =1\end{array}\text{dan jelas bahwa}p\text{dan}q\text{saling prima}\\ & \text{Jadi},\\ & p+q=11+1=12\end{array}$

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)

 $\begin{array}{l}\\ 61.& \text{Nilai dari}\\ & {}^2\log {\displaystyle \frac{4}{3}+2.{}^2\log \sqrt{12}\text{adalah}...}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 6}\\ \text{b}.& {\displaystyle 4}\\ \text{c}.& 3{\displaystyle \frac{1}{2}}\\ \text{d}.& 2{\displaystyle \frac{1}{2}}\\ \text{e}.& {\displaystyle 2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& ={}^2\log {\displaystyle \frac{4}{3}+2.{}^2\log \sqrt{12}}\\ & ={}^2\log \frac{4}{3}+{}^2\log {\left(\sqrt{12}\right)}^2\\ & ={}^2\log \frac{4}{3}+{}^2\log 12\\ & ={}^2\log \frac{4}{3}\times 12\\ & ={}^2\log 16\\ & ={}^2\log 2^4\\ & =4\end{array}\end{array}$.

$\begin{array}{l}\\ 62.& \text{Nilai dari}\\ & {\displaystyle \frac{1}{6}.{}^2\log 25-\frac{1}{3}.{}^2\log 10\text{adalah}...}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -3}\\ \text{b}.& {\displaystyle -2}\\ \text{c}.& -2{\displaystyle \frac{1}{2}}\\ \text{d}.& -{\displaystyle \frac{1}{2}}\\ \text{e}.& -{\displaystyle \frac{1}{3}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& ={\displaystyle \frac{1}{6}.{}^2\log 25-\frac{1}{3}.{}^2\log 10}\\ & ={\displaystyle \frac{1}{3}.\frac{1}{2}.{}^2\log 25-\frac{1}{3}.{}^2\log 10}\\ & =\frac{1}{3}({}^2\log {25}^{\frac{1}{2}}-{}^2\log 10)\\ & =\frac{1}{3}({}^2\log 5-{}^2\log 10)\\ & =\frac{1}{3}({}^2\log \frac{5}{10})\\ & =\frac{1}{3}({}^2\log \frac{1}{2})\\ & =\frac{1}{3}({}^2\log 2^{-1})\\ & =-\frac{1}{3}\end{array}\end{array}$

$\begin{array}{l}\\ 63.& \text{Nilai dari}\\ & {}^2\log ({}^2\log \sqrt{\sqrt{2}})\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -4}\\ \text{b}.& {\displaystyle -2}\\ \text{c}.& -1{\displaystyle \frac{1}{2}}\\ \text{d}.& -{\displaystyle \frac{1}{2}}\\ \text{e}.& -{\displaystyle \frac{1}{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& ={}^2\log ({}^2\log \sqrt{\sqrt{2}})\\ & ={}^2\log ({}^2\log 2^{\frac{1}{4}})\\ & ={}^2\log \frac{1}{4}\\ & ={}^2\log {\left(2\right)}^{-2}\\ & =-2\end{array}\end{array}$

$\begin{array}{l}\\ 64.& \mathbf{\text{(UMPTN '99)}}\\ & \text{Diketahui}\log 2=0,3010\text{dan}\log 3=0,4771\\ & \text{maka}\log (\sqrt[3]{2}\times \sqrt{3})\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 0,1505}\\ \text{b}.& {\displaystyle 0,1590}\\ \text{c}.& {\displaystyle 0,2007}\\ \text{d}.& {\displaystyle 0,3389}\\ \text{e}.& {\displaystyle 0,3891}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \log (\sqrt[3]{2}\times \sqrt{3})\\ & =\log \sqrt[3]{2}+\log \sqrt{3}\\ & =\log 2^{\frac{1}{3}}+\log 3^{\frac{1}{2}}\\ & ={\displaystyle \frac{1}{3}\log 2+{\displaystyle \frac{1}{2}\log 3}}\\ & ={\displaystyle \frac{1}{3}(0,3010)+{\displaystyle \frac{1}{2}(0,4771)}}\\ & =0,1003+0,2386\\ & =0,3389\end{array}\end{array}$

$\begin{array}{l}\\ 65.& \mathbf{\text{(UMPTN '98)}}\\ & \text{Nilai}{}^a\log {\displaystyle \frac{1}{b}\times {}^b\log {\displaystyle \frac{1}{c^2}\times {}^c\log {\displaystyle \frac{1}{a^3}=....}}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -6}\\ \text{b}.& {\displaystyle 6}\\ \text{c}.& {\displaystyle \frac{b}{a^{2}c}}\\ \text{d}.& {\displaystyle \frac{a^{2}c}{b}}\\ \text{e}.& {\displaystyle -\frac{1}{6}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {}^a\log {\displaystyle \frac{1}{b}\times {}^b\log {\displaystyle \frac{1}{c^2}\times {}^c\log {\displaystyle \frac{1}{a^3}}}}\\ & ={}^a\log {\displaystyle b^{-1}\times {}^b\log {\displaystyle c^{-2}\times {}^c\log {\displaystyle a^{-3}}}}\\ & =(-1).(-2).(-3)\times {}^a\log {\displaystyle a\times {}^b\log {\displaystyle c\times {}^c\log {\displaystyle a}}}\\ & =-6\times {}^a\log a\\ & =-6\times 1\\ & =-6\end{array}\end{array}$

$\begin{array}{l}\\ 66.& \mathbf{\text{(UMPTN '01)}}\\ & \text{Jika}{\displaystyle \frac{{}^2\log a}{{}^3\log b}=m\text{dan}{\displaystyle \frac{{}^3\log a}{{}^2\log b}=n}}\\ & \text{dengan}a>1,b>1,\text{maka}{\displaystyle \frac{m}{n}=....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle {}^2\log 3}\\ \text{b}.& {\displaystyle {}^3\log 2}\\ \text{c}.& {\displaystyle {}^4\log 9}\\ \text{d}.& {\displaystyle {({}^3\log 2)}^2}\\ \text{e}.& {\displaystyle {({}^2\log 3)}^2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{\displaystyle \frac{m}{n}}& ={\displaystyle \frac{{\displaystyle \frac{{}^2\log a}{{}^3\log b}}}{{\displaystyle \frac{{}^3\log a}{{}^2\log b}}}}\\ & ={\displaystyle \frac{{}^2\log a\times {}^2\log b}{{}^3\log b\times {}^3\log a}}\\ & ={\displaystyle \frac{{}^2\log a\times {\displaystyle \frac{1}{{}^b\log 2}}}{{}^3\log b\times {\displaystyle \frac{1}{{}^a\log 3}}}}\\ & ={\displaystyle \frac{{}^2\log a\times {}^a\log 3}{{}^3\log b\times {}^b\log 2}}\\ & ={\displaystyle \frac{{}^2\log 3}{{}^3\log 2}={\displaystyle \frac{{}^2\log 3}{{\displaystyle \frac{1}{{}^2\log 3}}}}}\\ & ={({}^2\log 3)}^2\end{array}\end{array}$.

$\begin{array}{l}\\ 67.& \mathbf{\text{(UMPTN '01)}}\\ & \text{Jika}{}^{10}\log x=b,\text{maka}{}^{10x}\log 100=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{b+1}}\\ \text{b}.& {\displaystyle \frac{2}{b+1}}\\ \text{c}.& {\displaystyle \frac{1}{b}}\\ \text{d}.& {\displaystyle \frac{2}{b}}\\ \text{e}.& {\displaystyle \frac{2}{10b}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& {}^{10x}\log 100\\ & ={\displaystyle \frac{\log 100}{\log 10x}}\\ & ={\displaystyle \frac{{}^{10}\log 100}{{}^{10}\log 10x},\text{pilih basis 10}}\\ & \text{alasannya: supaya sama dengan soal}\\ & ={\displaystyle \frac{{}^{10}\log {10}^2}{{}^{10}\log 10+{}^{10}\log x}}\\ & ={\displaystyle \frac{2}{1+b}\text{atau}}\\ & =\frac{2}{b+1}\end{array}\end{array}$.

$\begin{array}{l}\\ 68.& \mathbf{\text{(UM UGM '03)}}\\ & \text{Jika}{}^4\log 6=m+1,\text{maka}{}^9\log 8=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{3}{4m-2}}\\ \text{b}.& {\displaystyle \frac{3}{4m+2}}\\ \text{c}.& {\displaystyle \frac{3}{2m+4}}\\ \text{d}.& {\displaystyle \frac{3}{2m-4}}\\ \text{e}.& {\displaystyle \frac{3}{2m+2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Sebelumnya perhatikanlah}\\ & {}^4\log 6=m+1\\ & \iff {}^{2^2}\log (2.3{)}^1=m+1\\ & \iff {\displaystyle \frac{1}{2}\times {}^2\log (2.3)=m+1}\\ & \iff {\displaystyle \frac{1}{2}\times ({}^2\log 2+{}^2\log 3)=m+1}\\ & \iff {\displaystyle \frac{1}{2}\times (1+{}^2\log 3)=m+1}\\ & \iff 1+{}^2\log 3=2m+2\\ & \iff {}^2\log 3=2m+1\\ & \text{Selanjutnya adalah}:\\ & {}^9\log 8={\displaystyle \frac{1}{{}^8\log 9}}\\ & ={\displaystyle \frac{1}{{}^{2^3}\log 3^2}}\\ & ={\displaystyle \frac{1}{{\displaystyle \frac{2}{3}{}^2\log 3}}}\\ & ={\displaystyle \frac{3}{2{}^2\log 3}}\\ & ={\displaystyle \frac{3}{2(2m+1)}}\\ & ={\displaystyle \frac{3}{4m+2}}\end{array}\end{array}$.

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

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

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

 $\begin{array}{l}\\ 52.& \text{Nilai dari}{}^2\log {\displaystyle \frac{1}{32}\text{adalah}...}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -7}\\ \text{b}.& {\displaystyle -5}\\ \text{c}.& {\displaystyle -3}\\ \text{d}.& {\displaystyle -2}\\ \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& ={}^2\log {\displaystyle \frac{1}{32}}\\ & ={}^{2^1}\log 2^{-5}\\ & ={\displaystyle \frac{-5}{1}\times {}^2\log 2}\\ & =-5\\ & \\ & \end{array}\end{array}$.

$\begin{array}{l}\\ 53.& \text{Nilai dari}{}^{0,333...}\log 0,111....\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{3}}\\ \text{b}.& {\displaystyle \frac{1}{2}}\\ \text{c}.& 2\\ \text{d}.& 3\\ \text{e}.& 6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& ={}^{0,333...}\log 0,111...\\ & ={}^{\frac{1}{3}}\log \frac{1}{9}\\ & ={}^{\frac{1}{3}}\log {\left(\frac{1}{3}\right)}^2\\ & =2\end{array}\end{array}$.

$\begin{array}{l}\\ 54.& \text{Nilai dari}{}^5\log 25\sqrt{5}\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{5}{2}}\\ \text{b}.& {\displaystyle \frac{3}{2}}\\ \text{c}.& {\displaystyle \frac{1}{2}}\\ \text{d}.& {\displaystyle 2}\\ \text{e}.& 3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& ={}^5\log 25\sqrt{5}\\ & ={}^{5^1}\log 5^2{.5}^{\frac{1}{2}}\\ & ={}^{5^1}\log 5^{\frac{5}{2}}\\ & ={\displaystyle \frac{\frac{5}{2}}{1}\times {}^5\log 5}\\ & ={\displaystyle \frac{5}{2}}\end{array}\end{array}$.

$\begin{array}{l}\\ 55.& \text{Nilai dari}{}^{\sqrt{3}}\log 81\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 12}\\ \text{b}.& {\displaystyle 10}\\ \text{c}.& {\displaystyle 9}\\ \text{d}.& {\displaystyle 8}\\ \text{e}.& 6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& ={}^{\sqrt{3}}\log 81\\ & ={}^{{\displaystyle 3^{\frac{1}{2}}}}\log 3^4\\ & ={\displaystyle \frac{4}{\frac{1}{2}}\times {}^3\log 3}\\ & =8\end{array}\end{array}$

$\begin{array}{l}\\ 56.& \text{Nilai dari}{}^{\frac{1}{3}}\log {\displaystyle \frac{1}{243}\text{adalah}...}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 6}\\ \text{b}.& {\displaystyle 5}\\ \text{c}.& {\displaystyle 4}\\ \text{d}.& {\displaystyle 3}\\ \text{e}.& {\displaystyle 2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& ={}^{\frac{1}{3}}\log {\displaystyle \frac{1}{243}}\\ & ={}^{{\left(\frac{1}{3}\right)}^1}\log {\displaystyle {\left(\frac{1}{3}\right)}^5}\\ & ={\displaystyle \frac{5}{1}\times {}^{\frac{1}{3}}\log \frac{1}{3}}\\ & =5\end{array}\end{array}$

$\begin{array}{l}\\ 57.& \text{Nilai dari}{}^{\sqrt{2}}\log 16\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 10}\\ \text{b}.& {\displaystyle 9}\\ \text{c}.& {\displaystyle 8}\\ \text{d}.& {\displaystyle 6}\\ \text{e}.& {\displaystyle 4}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& ={}^{\sqrt{2}}\log 16\\ & ={}^{{\displaystyle 2^{\frac{1}{2}}}}\log 2^4\\ & ={\displaystyle \frac{4}{\frac{1}{2}}\times {}^2\log 2}\\ & =8\end{array}\end{array}$

$\begin{array}{l}\\ 58.& \text{Nilai dari}{}^{\sqrt{5}}\log \sqrt{125}\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -3}\\ \text{b}.& {\displaystyle -2}\\ \text{c}.& {\displaystyle 2}\\ \text{d}.& {\displaystyle 3}\\ \text{e}.& {\displaystyle 5}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& ={}^{\sqrt{5}}\log \sqrt{125}\\ & ={}^{{\sqrt{5}}^1}\log {\left(\sqrt{5}\right)}^3\\ & ={\displaystyle \frac{3}{1}\times {}^{\sqrt{5}}\log \sqrt{5}}\\ & =3\end{array}\end{array}$

$\begin{array}{l}\\ 59.& \text{Nilai dari}{}^{\sqrt{\sqrt{2}}}\log \sqrt{8\sqrt{8}}\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 4}\\ \text{b}.& {\displaystyle 6}\\ \text{c}.& {\displaystyle 8}\\ \text{d}.& {\displaystyle 9}\\ \text{e}.& {\displaystyle 12}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& ={}^{\sqrt{\sqrt{2}}}\log \sqrt{8\sqrt{8}}\\ & ={}^{\sqrt[4]{2}}\log {\left(8{\left(8\right)}^{\frac{1}{2}}\right)}^{\frac{1}{2}}\\ & ={}^{2^{\frac{1}{4}}}\log 8^{(\frac{1}{2}+\frac{1}{4})}\\ & ={}^{2^{\frac{1}{4}}}\log 2^{3\left(\frac{3}{4}\right)}\\ & ={\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}\times {}^2\log 2}\\ & =9\end{array}\end{array}$

$\begin{array}{l}\\ 60.& \text{Nilai dari}\\ & {}^6\log 8+{}^6\log {\displaystyle \frac{9}{2}\text{adalah}...}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 4}\\ \text{b}.& {\displaystyle 3}\\ \text{c}.& 3{\displaystyle \frac{1}{2}}\\ \text{d}.& 2{\displaystyle \frac{1}{2}}\\ \text{e}.& {\displaystyle 2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& ={}^6\log 8+{}^6\log {\displaystyle \frac{9}{2}}\\ & ={}^6\log 8\times \frac{9}{2}\\ & ={}^6\log 36\\ & ={}^6\log 6^2\\ & =2\\ & \\ & \\ & \end{array}\end{array}$

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

 $\begin{array}{l}\\ 41.& \text{Penyelesaian pertidaksamaan eksponen}\\ & {\left({\displaystyle \frac{1}{3}}\right)}^{2x+1}>\sqrt{{\displaystyle \frac{27}{3^{x-1}}}}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& x>{\displaystyle \frac{6}{5}}\\ \text{b}.& x<-{\displaystyle \frac{6}{5}}\\ \text{c}.& x>{\displaystyle \frac{5}{6}}\\ \text{d}.& x<-2\\ \text{e}.& x<2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\left({\displaystyle \frac{1}{3}}\right)}^{2x+1}& >\sqrt{{\displaystyle \frac{27}{3^{x-1}}}}\\ 3^{-(2x+1)}& >3^{\frac{1}{2}(3-(x-1))}\\ -(2x+1)& >{\displaystyle \frac{1}{2}(3-(x-1))}\\ -4x-2& >4-x\\ -4x+x& >4+2\\ -3x& >6\\ 3x& <-6\\ x& <-2\end{array}\end{array}$

$\begin{array}{l}\\ 42.& (\mathbf{\text{UMPTN 01}})\text{Nilai}x\text{yang memenuhi}\\ & 4^{x^2-x-2}{.2}^{x^2+3x-10}<{\displaystyle \frac{1}{16}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& x<-5\text{atau}x>-2\\ \text{b}.& x<-2\text{atau}x>{\displaystyle \frac{5}{3}}\\ \text{c}.& -2<x<-1\\ \text{d}.& -2<x<{\displaystyle \frac{5}{3}}\\ \text{e}.& -5<x<2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{4}^{x^2-x-2}{.2}^{x^2+3x-10}& <{\displaystyle \frac{1}{16}}\\ 2^{2(x^2-x-2)+(x^2+3x-10)}& <2^{-4}\\ 2(x^2-x-2)+(x^2+3x-10)& <-4\\ 3x^2-2x+3x-4-10+4& <0\\ 3x^2+x-10+& <0\\ (x+2)(3x-5)& <0\\ \therefore -2<x& <{\displaystyle \frac{5}{3}}\end{array}\end{array}$

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

$\begin{array}{l}\\ 44.& (\mathbf{\text{SBMPTN 2015 Mat IPA}})\\ & \text{Nilai}c\text{yang memenuhi}\\ & (0,12{)}^{4x^2+8x+c}<(0,0144{)}^{x^2+4x+4}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& c>0\\ \text{b}.& c>2\\ \text{c}.& c>4\\ \text{d}.& c>6\\ \text{e}.& c>8\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}(0,12{)}^{4x^2+8x+c}& <(0,0144{)}^{x^2+4x+4}\\ (0,12{)}^{4x^2+8x+c}& <(0,12{)}^{2(x^2+4x+4)}\\ 4x^2+8x+c& >2(x^2+4x+4)\\ 2x^2+c-8& >0\text{haruslah definit positif}\\ \text{Syaratnya}& \{\begin{array}{ll}a& =2>0\\ D& =b^2-4ac<0\end{array}\\ \text{Maka}D& =b^2-4ac<0\\ \mathbf{\text{ambil dari}}& 2x^2-c-8=0\{\begin{array}{ll}a& =2\\ b& =0\\ c& =c-8\end{array}\\ & =0^2-4.2(c-8)<0\\ -8c& +64<0\\ -8c& <-64\\ 8c& >64\\ c& >8\end{array}\end{array}$

$\begin{array}{l}\\ 45.& \text{Nilai}x\text{yang memenuhi}\\ & 9^{2x}-{10.9}^x+9>0,x\in \mathbb{R}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& x<1\text{atau}x>0\\ \text{b}.& x<0\text{atau}x>1\\ \text{c}.& x<-1\text{atau}x>2\\ \text{d}.& x<1\text{atau}x>2\\ \text{e}.& x<-1\text{atau}x>1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{9}^{2x}-{10.9}^x+9& >0\\ {\left(9^x\right)}^2-10.\left(9^x\right)+9& >0\\ (9^x-1)(9^x-9)& >0\\ 9^x<1\text{atau}{9}^x& >9\\ 9^x<9^0\text{atau}{9}^x& >9^1\\ x<0\text{atau}x& >1\end{array}\end{array}$

$\begin{array}{l}\\ 46.& \text{Jumlah akar-akar persamaan}\\ & 5^{x+1}+5^{2-x}-30=0\text{adalah}....\\ \\ & \begin{array}{l}\\ \text{a}.& -2& & & \\ \text{b}.& -1\\ \text{c}.& 0\\ \text{d}.& 1\\ \text{e}.& 2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{5}^{x+1}+5^{2-x}-30& =0\\ \left(5^x\right){.5}^1+{\displaystyle \frac{5^2}{5^x}-30}& =0\\ 5{\left(5^x\right)}^2+25-30\left(5^x\right)& =0\\ \text{Persamaan kuadrat}& \text{dalam}{5}^x,\text{maka}\\ 5(5^x{)}^2-30(5^x)+25& =0\{\begin{array}{ll}a& =5\\ b& =-30\\ c& =25\end{array}\\ (5^{x_1}).\left(5^{x_2}\right)& ={\displaystyle \frac{c}{a}}\\ 5^{x_1+x_2}& ={\displaystyle \frac{25}{5}=5}\\ 5^{x_1+x_2}& =5^1\\ x_1+x_2& =1\end{array}\end{array}$

$\begin{array}{l}\\ 47.& \text{Jumlah akar-akar persamaan}\\ & {2020}^{x^2-7x+7}={2021}^{x^2-7x+7}\text{adalah}....\\ \\ & \begin{array}{l}\\ \text{a}.& -7\\ \text{b}.& -5\\ \text{c}.& -3\\ \text{d}.& 5\\ \text{e}.& 7\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{2020}^{x^2-7x+7}& ={2021}^{x^2-7x+7}\\ \text{Karena basis}& \text{tidak sama},\\ \text{maka harusl}& \text{ah pangkatnya}=0,\\ x^2-7x+7& =0\\ \text{dan jumlah}& \text{akar-akarnya adalah}:\\ x_1+x_2& =-{\displaystyle \frac{b}{a},\text{dari persamaan}}\\ x^2-7x+7& =0\{\begin{array}{ll}a& =1\\ b& =-7\\ c& =7\end{array}\\ \text{maka}{x}_1+x_2& =-{\displaystyle \frac{b}{a}=-\frac{-7}{1}=7}\end{array}\end{array}$

$\begin{array}{l}\\ 48.& \text{Nilai dari}{\displaystyle \frac{2^{2020}+2^{2018}}{2^{2018}+2^{2016}}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 5\\ \text{c}.& 10\\ \text{d}.& 20\\ \text{e}.& 40\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\displaystyle \frac{2^{2020}+2^{2018}}{2^{2018}+2^{2016}}}& ={\displaystyle \frac{2^4{.2}^{2016}+2^2{.2}^{2016}}{2^2{.2}^{2018}+2^{2016}}}\\ & ={\displaystyle \frac{2^{2016}(2^4+2^2)}{2^{2016}(2^2+1)}}\\ & ={\displaystyle \frac{16+4}{4+1}}\\ & ={\displaystyle \frac{20}{5}}\\ & =4\end{array}\end{array}$

$\begin{array}{l}\\ 49.& (\mathbf{\text{UM IPB}})\text{Jika}ab=a^b\text{dan}{\displaystyle \frac{a}{b}=a^{3b}}\\ & \text{maka nilai}a\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 0,5\\ \text{c}.& 1\\ \text{d}.& 0,25\\ \text{e}.& 0,75\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Diketahui}& \\ ab& =a^b\\ b& ={\displaystyle \frac{a^b}{a}=a^{b-1}.....\mathbf{\text{1}}}\\ \text{maka}& \\ {\displaystyle \frac{a}{b}}& =a^{3b}...............\mathbf{\text{2}}\\ \mathbf{\text{1}}& ke\mathbf{\text{2}}\\ {\displaystyle \frac{a}{a^{b-1}}}& =a^{3b}\\ a^{2-b}& =a^{3b}\\ 2-b& =3b\\ -4b& =-2\\ b& ={\displaystyle \frac{1}{2}................\mathbf{\text{3}}}\\ \mathbf{\text{3}}& ke\mathbf{\text{1}}\\ a\left({\displaystyle \frac{1}{2}}\right)& =a^{\frac{1}{2}}\\ {\displaystyle \frac{1}{4}{a}^2}& =a\\ a^2-4a& =0\\ a(a-4)& =0\\ a=0& \text{atau}a=4\end{array}\end{array}$

$\begin{array}{l}\\ 50.& \text{Jika}{\displaystyle 3.x^{{}^{{}^{{}^{\frac{3}{2}}}}}=4,\text{maka}x=....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 1,1}\\ \text{b}.& {\displaystyle 1,2}\\ \text{c}.& 1,3\\ \text{d}.& {\displaystyle 1,4}\\ \text{e}.& 1,5\\ \\ & & & (\mathbf{\text{SAT Test Math Level 2}})\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}{\displaystyle 3.x^{{}^{{}^{{}^{\frac{3}{2}}}}}}& =4\\ {\left(3.x^{{}^{{}^{{}^{\frac{3}{2}}}}}\right)}^2& =4^2\\ 3^2.x^3& =4^2\\ x^3& ={\displaystyle \frac{4^2}{3^2}}\\ x^3& ={\displaystyle \frac{4^2}{3^2}\times \frac{3}{3}}\\ x^3& \le {\displaystyle \frac{4^2}{3^2}\times \frac{4}{3}}\\ x^3& \le \left({\displaystyle \frac{4^3}{3^3}}\right)\\ x^3& \le {\left({\displaystyle \frac{4}{3}}\right)}^3\\ x& \le {\displaystyle \frac{4}{3}}\\ x& \le 1,\overline{333}\\ x& \approx 1,3\end{array}\end{array}$

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

Pembahasan:

$\begin{array}{rl}{x}^8& =2207-{\displaystyle \underset{x^8}{\underbrace{{\displaystyle \frac{1}{2207-\frac{1}{2207-\frac{1}{2207-...}}}}}}}\\ x^8& =2207-{\displaystyle \frac{1}{x^8}}\\ x^8+{\displaystyle \frac{1}{x^8}}& =2207\\ {(x^4+{\displaystyle \frac{1}{x^4}})}^2& =2207+2\\ (x^4+{\displaystyle \frac{1}{x^4}})& =\sqrt{2209}=47\end{array}$

$\begin{array}{rl}{x}^4+{\displaystyle \frac{1}{x^4}}& =47\\ {(x^2+{\displaystyle \frac{1}{x^2}})}^2& =47+2\\ x^2+{\displaystyle \frac{1}{x^2}}& =\sqrt{49}=7\\ {(x+{\displaystyle \frac{1}{x}})}^2& =7+2\\ x+{\displaystyle \frac{1}{x}}& =\sqrt{9}=3\\ x^2-3x+1& =0,\\ & \text{persamaan kuadrat dalam x,}\\ & \mathbf{\text{gunakan rumus abc}}\\ x_{1,2}=& {\displaystyle \frac{3\pm \sqrt{5}}{2}={\displaystyle \frac{3\pm 1\sqrt{5}}{2}={\displaystyle \frac{a\pm b\sqrt{c}}{d}}}}\\ & \mathbf{\text{Sehingga}},\{\begin{array}{ll}& a=3\\ & b=1\\ & c=5\\ & d=2\end{array}\end{array}$

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)

$\begin{array}{l}\\ 31.& \text{Jika bentuk}{\displaystyle \frac{a{b}^{-1}}{a^{-1}-b^{-1}}}\\ & \text{dinyatakan dalam pangkat positif}=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{a^2}{a-b}}& & & \\ \\ \text{b}.& {\displaystyle \frac{a^2}{a-1}}\\ \\ \text{c}.& {\displaystyle \frac{b-a}{ab}}\\ \\ \text{d}.& {\displaystyle \frac{a^2}{b-a}}\\ \\ \text{e}.& {\displaystyle \frac{1}{a-b}}\\ \\ & & & & (\mathbf{\text{SAT Test Math Level 2}})\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\displaystyle \frac{a{b}^{-1}}{a^{-1}-b^{-1}}}& ={\displaystyle \frac{a{b}^{-1}}{a^{-1}-b^{-1}}\times \frac{b}{b}}\\ & ={\displaystyle \frac{a}{a^{-1}b-1}}\\ & ={\displaystyle \frac{a}{a^{-1}b-1}\times \frac{a}{a}}\\ & ={\displaystyle \frac{a^2}{b-a}}\end{array}\end{array}$

$\begin{array}{l}\\ 32.& \text{Nilai}x\text{yang memenuhi}\\ & \sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}=3\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 3\\ \text{b}.& 6\\ \text{c}.& 7\\ \text{d}.& 8\\ \text{e}.& 9\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Misalkan}A& =\sqrt{+\sqrt{x+\sqrt{+\cdots }}}\\ \sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}& =3\\ \text{dikuadratkan}& \\ x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}& =9\\ x+3& =9\\ x& =9-3\\ x& =6\end{array}\end{array}$

$\begin{array}{l}\\ 33.& \text{Nilai}x\text{yang memenuhi}\\ & x=\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 7\sqrt[7]{7}\\ \text{b}.& 7\\ \text{c}.& 14\\ \text{d}.& 49\\ \text{e}.& \sqrt[3]{81}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}x& =\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\\ x^3& =49\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\\ x^3& =49x\\ x^2& =49\\ x& =\sqrt{49}\\ & =7\end{array}\end{array}$

$\begin{array}{l}\\ 34.& \text{Nilai}x\text{yang memenuhi}\\ & x^{x^{x^{x^{x^{\cdots }}}}}=2020\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& \sqrt{2020}\\ \text{b}.& \sqrt[2020]{2020}\\ \text{c}.& {2020}^{\sqrt{2020}}\\ \text{d}.& {\sqrt{2020}}^{\sqrt{2020}}\\ \text{e}.& \sqrt{2020\sqrt{2020}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{x}^{x^{x^{x^{x^{\cdots }}}}}& =2020\\ x^{2020}& =2020\\ x& =\sqrt[2020]{2020}\end{array}\end{array}$

$\begin{array}{l}\\ 35.& \text{Nilai dari}\\ & {\displaystyle \frac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}}}\\ & \text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& \sqrt[3]{2}+1\\ \text{c}.& \sqrt[3]{2}-1\\ \text{d}.& \sqrt[3]{4}+1\\ \text{e}.& \sqrt[3]{4}-1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& {\displaystyle \frac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}}\times \frac{\sqrt[3]{2}-1}{\sqrt[3]{2}-1}}\\ & ={\displaystyle \frac{{\left(\sqrt[3]{2}\right)}^2-1}{\sqrt[3]{2}+\sqrt[3]{4}+\sqrt[3]{8}-1-\sqrt[3]{2}-\sqrt[3]{4}}}\\ & ={\displaystyle \frac{\sqrt[3]{4}-1}{\sqrt[3]{8}-1}}\\ & ={\displaystyle \frac{\sqrt[3]{4}-1}{2-1}}\\ & =\sqrt[3]{4}-1\end{array}\end{array}$

$\begin{array}{l}\\ 36.& \text{Nilai dari}\\ & {\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}}\\ & \text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& 2\sqrt{2}-1\\ \text{c}.& {\displaystyle \frac{1}{2}\sqrt{2}}\\ \text{d}.& \sqrt{{\displaystyle \frac{5}{3}}}\\ \text{e}.& \sqrt{{\displaystyle \frac{2}{5}}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}}\\ & ={\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\times \frac{\sqrt{\sqrt{5}+1}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}}\\ & ={\displaystyle \frac{\sqrt{7+3\sqrt{5}}+\sqrt{3-\sqrt{5}}}{\sqrt{5}+1}-(\sqrt{2}-1)}\\ & ={\displaystyle \frac{\left({\displaystyle \frac{3+\sqrt{5}}{\sqrt{2}}}\right)+\left({\displaystyle \frac{\sqrt{5}-1}{\sqrt{2}}}\right)}{\sqrt{5}+1}+1-\sqrt{2}}\\ & ={\displaystyle \frac{{\displaystyle \frac{2+2\sqrt{5}}{\sqrt{2}}}}{1+\sqrt{5}}+1-\sqrt{2}}\\ & ={\displaystyle \frac{2}{\sqrt{2}}+1-\sqrt{2}}\\ & =\sqrt{2}+1-\sqrt{2}\\ & =1\end{array}\end{array}$.

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

$\begin{array}{l}\\ 38.& \text{Jika}{\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}={\displaystyle \frac{a\sqrt{2}+b\sqrt{3}+c\sqrt{5}}{12},}}\\ \\ & \text{maka}a+b+c=....\\ & (\mathbf{\text{UM UGM 2016 Mat Das}})\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 1\\ \text{c}.& 2\\ \text{d}.& 3\\ \text{e}.& 4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& {\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}}\\ & ={\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\times {\displaystyle \frac{(\sqrt{2}+\sqrt{3}-\sqrt{5})}{(\sqrt{2}+\sqrt{3}-\sqrt{5})}}}\\ & ={\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{{(\sqrt{2}+\sqrt{3})}^2-{\left(\sqrt{5}\right)}^2}}\\ & ={\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2+3+2\sqrt{2.3}-5}}\\ & ={\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2\sqrt{6}}\times {\displaystyle \frac{\sqrt{6}}{\sqrt{6}}}}\\ & ={\displaystyle \frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{12}}\\ & ={\displaystyle \frac{2\sqrt{3}+3\sqrt{2}-\sqrt{30}}{12}}\\ & ={\displaystyle \frac{3\sqrt{2}+2\sqrt{3}-\sqrt{30}}{12}}\\ & \{\begin{array}{ll}a& =3\\ b& =2\\ c& =-1\end{array}\\ a+b+c& =3+2+(-1)\\ & =4\end{array}\end{array}$

$\begin{array}{l}\\ 39.& \text{Tunjukkan bahwa}\\ & \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+\cdots }}}}}=3\\ \\ & \text{Bukti}\\ & \begin{array}{rl}{x}^2& =x^2\\ x^2& =1+(x^2-1)\\ & =1+(x-1)(x+1)\\ & =1+(x-1)\sqrt{{(x+1)}^2}\\ & =1+(x-1)\sqrt{1+{(x+1)}^2-1}\\ & =1+(x-1)\sqrt{1+(x+1-1)(x+1+1)}\\ & =1+(x-1)\sqrt{1+x(x+2)}\\ & =1+(x-1)\sqrt{1+x\sqrt{{(x+2)}^2}}\\ & =1+(x-1)\sqrt{1+x\sqrt{1+{(x+2)}^2-1}}\\ & =1+(x-1)\sqrt{1+x\sqrt{1+(x+2-1)(x+2+1)}}\\ & =1+(x-1)\sqrt{1+x\sqrt{1+(x+1)(x+3)}}\\ & =1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{{(x+3)}^2}}}\\ x^2& =1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{...}}}\\ x& =\sqrt{1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{...}}}}\end{array}\end{array}$

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

$\mathbf{\text{bukti}}$

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

$\begin{array}{rl}{3}^{15a}& =5^{5b}={15}^{3c}\{\begin{array}{ll}3=5^{\frac{5b}{15a}}& \\ 3^{\frac{15a}{5b}}=b& (a^b=c^d\to a=c^{\frac{d}{b}}\text{atau}{a}^{\frac{b}{d}}=c)\end{array}\\ 3^{15a}& ={15}^{3c}\\ 3^{15a}& =(3\times 5{)}^{3c}\\ 3^{15a}& =(3\times 3^{\frac{15a}{5b}}{)}^{3c}\\ 3^{15a}& =3^{3c+\frac{9c}{b}}\\ a^{f(x)}& =a^{g(x)}\\ f(x)& =g(x)\\ 15a& =3c+\frac{9ac}{b}\\ 15ab& =3bc+9ac\\ 5ab& =bc+3ac\\ 5ab-bc-3ac& =0◼\end{array}$