Contoh Soal 6 Fungsi Logaritma (Uraian)

$\begin{array}{l}\\ 26.& \text{Diketahui bahwa}\\ & {}^{{}^2}\log 3=p\text{dan}{}^{{}^3}\log 11=q,\\ & \text{maka nilai}{}^{{}^{44}}\log 66=....\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}{}^{{}^{44}}\log 66& ={\displaystyle \frac{{}^{{}^{{}^{...}}}\log 66}{{}^{{}^{{}^{...}}}\log 44}}\\ & ={\displaystyle \frac{{}^{{}^{{}^{...}}}\log (2\times 3\times 11)}{{}^{{}^{{}^{...}}}\log (2^2\times 11)}}\\ & ={\displaystyle \frac{{}^{{}^{{}^3}}\log 2+{}^{{}^{{}^3}}\log 3+{}^{{}^{{}^3}}\log 11}{{}^{{}^{{}^3}}\log 2^2+{}^{{}^{{}^3}}\log 11}}\\ & ={\displaystyle \frac{\frac{1}{{}^{{}^{{}^2}}\log 3}+{}^{{}^{{}^3}}\log 3+{}^{{}^{{}^3}}\log 11}{\frac{2}{{}^{{}^{{}^3}}\log 2^2}+{}^{{}^{{}^3}}\log 11}}\\ & ={\displaystyle \frac{\frac{1}{p}+1+q}{\frac{2}{p}+q}}\\ & ={\displaystyle \frac{\frac{1}{p}+1+q}{\frac{2}{p}+q}\times {\displaystyle \frac{p}{p}}}\\ & =\frac{1+p+pq}{2+pq}\end{array}\end{array}$

$\begin{array}{l}\\ 27.& \mathbf{\text{(AIME 1984)}}\\ & \text{Diketahui bahwa}x\text{dan}y\\ & \text{adalah bilangan real yang memenuhi}\\ \\ & \{\begin{array}{c}{}^{{}^8}\log x+{}^{{}^4}\log y^2=5\\ \\ {}^{{}^8}\log y+{}^{{}^4}\log x^2=7\end{array}\\ \\ & \text{Tentukanlah nilai dari}xy\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {}^{{}^{{}^8}}\log x+{}^{{}^4}\log y^2=5\\ & \iff {}^{{}^{{}^{2^3}}}\log x+{}^{{}^{{}^{2^2}}}\log y^2=5....(1)\\ & {}^{{}^{{}^8}}\log y+{}^{{}^4}\log x^2=7\\ & \iff {}^{{}^{{}^{2^3}}}\log y+{}^{{}^{{}^{2^2}}}\log x^2=7....(2)\\ & \text{selanjutnya},& \\ & \frac{1}{3}{.}^{{}^{{}^2}}\log x+{}^{{}^{{}^2}}\log y=5|\times \frac{1}{3}|\\ & \Rightarrow \frac{1}{9}{.}^{{}^{{}^2}}\log x+\frac{1}{3}.{}^{{}^{{}^2}}\log y=\frac{5}{3}....(3)\\ & {}^{{}^{{}^2}}\log x+\frac{1}{3}{.}^{{}^{{}^2}}\log y=7|\times 1|\\ & {\Rightarrow }^{{}^{{}^2}}\log x+\frac{1}{3}{.}^{{}^{{}^2}}\log y=7....(4)\\ & \text{saat persamaan}(3)-(4)\\ & =-\frac{8}{9}.{}^{{}^{{}^2}}\log x=\frac{5}{3}-7=-\frac{16}{3}\\ & \text{maka}\\ & {}^{{}^{{}^2}}\log x=(-\frac{16}{3})(-\frac{9}{8})\\ & {}^{{}^{{}^2}}\log x=6\iff x=2^6\iff x=64\\ & \text{Pada persamaan 1 selanjutnya}\\ & \frac{1}{3}.{}^{{}^{{}^2}}\log x+{}^{{}^{{}^2}}\log y=5\\ & \iff \frac{1}{3}.{}^{{}^{{}^2}}\log 2^6+{}^{{}^{{}^2}}\log y=5\\ & \iff \frac{1}{3}.6+{}^{{}^{{}^2}}\log y=5\\ & \iff 2+{}^{{}^{{}^2}}\log y=5\\ & \iff {}^{{}^{{}^2}}\log y=5-2=3\iff y=2^3=8\\ & \text{Jadi},x.y=64.8=512\end{array}\end{array}$

$\begin{array}{l}\\ 28.& \text{Tentukanlah nilai dari}\\ & \text{a}.\left(2^{{}^{{}^{{}^2}}\log 6}\right)\left(3^{{}^{{}^{{}^9}}\log 5}\right)\left(5^{{}^{{}^{{}^{\frac{1}{5}}}}\log 2}\right)\\ & \text{b}.({}^{27}\log 125)\left({}^{25}\log {\displaystyle \frac{1}{64}}\right)({}^{64}\log \frac{1}{9})\\ & \text{c}.({}^{625}\log 19)\left({}^7\log {\displaystyle \frac{1}{25}}\right)({}^{19}\log 7)\\ \\ & \text{Jawab}:\\ & \begin{array}{|c|}\hline \begin{array}{rl}\text{a}.& \left(2^{{}^{{}^{{}^2}}\log 6}\right)\left(3^{{}^{{}^{{}^9}}\log 5}\right)\left(5^{{}^{{}^{{}^{\frac{1}{5}}}}\log 2}\right)\\ & =\left(2^{{}^{{}^{{}^2}}\log 6}\right)\left(3^{{}^{{}^{{}^{3^2}}}\log 5}\right)\left(5^{{}^{{}^{{}^{5^{-1}}}}\log 2}\right)\\ & =\left(2^{{}^{{}^{{}^2}}\log 6}\right)\left(3^{{}^{{}^{{}^3}}\log 5^{{}^{\frac{1}{2}}}}\right)\left(5^{{}^{{}^{{}^5}}\log 2^{-1}}\right)\\ & =\left(2^{{}^{{}^2}\log 6}\right)\left(3^{{}^{{}^3}\log \sqrt{5}}\right)\left(5^{{}^{{}^5}\log \frac{1}{2}}\right)\\ & =6\times \sqrt{5}\times \frac{1}{2}\\ & =3\sqrt{5}\end{array}\\ \hline\end{array}\\ & \begin{array}{|c|}\hline \begin{array}{rl}\text{b}.& ({}^{27}\log 125)\left({}^{25}\log {\displaystyle \frac{1}{64}}\right)({}^{64}\log \frac{1}{9})\\ & =({}^{{}^{3^3}}\log 5^3)({}^{{}^{5^2}}\log 4^{-3})({}^{{}^{4^3}}\log 3^{-2})\\ & ={\displaystyle \frac{3}{3}.(-\frac{3}{2}).(-\frac{2}{3}).{}^{{}^3}\log 5.{}^{{}^5}\log 4.{}^{{}^4}\log 3}\\ & =1\end{array}\\ \hline\end{array}\\ & \text{Pembahasan diserahkan kepada}\\ & \text{Pembaca yang budiman untuk poin c}\end{array}$

$\begin{array}{l}\\ 29.& \text{Tentukanlah nilai}a+b\text{dimana}a\text{dan}b\\ & \text{adalah bilangan riil positif}.\\ & {}^7\log (1+a^2)-{}^7\log 25={}^7\log (2ab-15)-{}^7\log (25+b^2)\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {}^7\log {\displaystyle \frac{(1+a^2)}{25}={}^7\log {\displaystyle \frac{(2ab-15)}{(25+b^2)}}}\\ & \text{diambil}\text{persamaannya, maka}\\ & {\displaystyle \frac{(1+a^2)}{25}={\displaystyle \frac{(2ab-15)}{(25+b^2)}}}\\ & {\displaystyle (1+a^2)(25+b^2)=25(2ab-15)}\\ & \{\begin{array}{ll}(1+a^2)& \text{ faktor dari}25,a>0,a\in \mathbb{R}\\ & \text{atau}\\ (25+b^2)& \text{ faktor dari }25,b>0,b\in \mathbb{R}\text{juga}\end{array}\end{array}\\ \\ & \begin{array}{|clclccc|}\hline \text{No}& a& (1+a^2)& b& (25+b^2)& \text{Keterangan}& a+b\\ 1& 2& 5& 10& 125& \text{Memenuhi}& 12\\ 2& 7& 50& \cdots & \cdots & \text{tidak}& \cdots \\ 3& \cdots & \cdots & 5& 50& \text{tidak}& \cdots \\ \hline\end{array}\end{array}$

$\begin{array}{l}\\ 30.& \text{Jika}{60}^a=3\text{dan}{60}^b=5\\ & \text{maka hasil dari}{12}^{{}^{\left({\displaystyle \frac{1-x-y}{2(1-b)}}\right)}}\\ \\ & \text{Jawab}:\\ & \text{Perhatikanlah bahwa}\\ & \begin{array}{rl}& \{\begin{array}{ll}{60}^a=3& \Rightarrow {}^{60}\log 3=a\\ {60}^b=5& \Rightarrow {}^{60}\log 5=b\end{array}\end{array}\\ & \text{Selanjutnya}\\ & \text{Untuk}(1-a-b),\text{maka}\\ & \begin{array}{rl}1-a-b& =1-{}^{60}\log 3-{}^{60}\log 5\\ & ={}^{60}\log 60-{}^{60}\log 3-{}^{60}\log 5\\ & ={}^{60}\log {\displaystyle \frac{60}{3\times 5}}\\ & ={}^{60}\log 4\\ & ={}^{60}\log 2^2\end{array}\\ & \text{Untuk}2(1-b),\text{maka}\\ & \begin{array}{rl}2(1-b)& =2(1-{}^{60}\log 5)\\ & =2({}^{60}\log 60-{}^{60}\log 5)\\ & =2\left({}^{60}\log {\displaystyle \frac{60}{5}}\right)\\ & =2({}^{60}\log 12)\\ & ={}^{60}\log {12}^2\end{array}\\ & \text{Untuk}\left({\displaystyle \frac{1-x-y}{2(1-b)}}\right),\text{maka}\\ & \begin{array}{rl}\left({\displaystyle \frac{1-x-y}{2(1-b)}}\right)& ={\displaystyle \frac{{}^{60}\log 2^2}{{}^{60}\log {12}^2}}\\ & ={\displaystyle \frac{2{}^{60}\log 2}{2{}^{60}\log 12}}\\ & ={\displaystyle \frac{{}^{60}\log 2}{{}^{60}\log 12}}\\ & ={}^{12}\log 2\end{array}\\ & \text{Jadi},{12}^{{}^{\left({\displaystyle \frac{1-x-y}{2(1-b)}}\right)}}={12}^{{}^{{}^{12}\log 2}}=2\end{array}$

$\begin{array}{l}\\ 31.& \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.