Contoh Soal 8 Fungsi Logaritma (Persamaan Logaritma)

$\begin{array}{l}\\ 36.& \text{Persamaan}{}^x\log 2+{}^x\log (3x-4)=2\\ & \text{mempunyai akar}{x}_1\text{dan}{x}_2,\text{maka}\\ & \text{nilai}{x}_1+x_2\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 3\\ \text{c}.& 4\\ \text{d}.& 6\\ \text{e}.& 8\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Alternatif 1}\\ & {}^x\log 2+{}^x\log (3x-4)=2\\ & \iff {}^x\log 2(3x-4)=2\\ & \iff {}^x\log 6x-8=2\\ & \iff 6x-8=x^2\\ & \iff x^2-6x+8=0,\text{dengan}\{\begin{array}{ll}a& =1\\ b& =-6\\ c& =8\end{array}\\ & \iff x_1+x_2=-{\displaystyle \frac{b}{a}}\\ & \iff x_1+x_2=-{\displaystyle \frac{-6}{1}=6}\\ & \text{Alternatif 2}\\ & \iff x^2-6x+8=0\\ & \iff (x-2)(x-4)\\ & \iff x_1=2\text{atau}{x}_2=4\\ & \iff x_1+x_2=2+4=6\end{array}\end{array}$

$\begin{array}{l}\\ 37.& \text{Jika}{x}_1\text{dan}{x}_2\text{memenuhi}\\ & (\log x)(2\log x-3)=\log 100\\ & \text{maka}{x}_1\times x_2\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 100\\ \text{b}.& 10\sqrt{10}\\ \text{c}.& \sqrt{10}\\ \text{d}.& -\sqrt{10}\\ \text{e}.& -10\sqrt{10}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& (\log x)(2\log x-3)=\log 100\\ & \iff (\log x)(2\log x-3)=2\\ & \iff 2{\log }^{2}x-3\log x-2=0\{\begin{array}{ll}a& =2\\ b& =-3\\ c& =-2\end{array}\\ & \iff \log x_1+\log x_2=-{\displaystyle \frac{-3}{2}=\frac{3}{2}}\\ & \iff \log (x_1\times x_2)=1{\displaystyle \frac{1}{2}}\\ & \iff (x_1\times x_2)={10}^{1\frac{1}{2}}=10\sqrt{10}\end{array}\end{array}$

$\begin{array}{l}\\ 38.& \text{Persamaan}\\ & {10}^{{}^{{}^2}\log x^2}-7\left({10}^{{}^{{}^2}\log x}\right)+10=0\\ & \text{mempunyai dua akar yaitu}{x}_1\text{dan}{x}_2\\ & \text{Nilai}{x}_1\times x_2=....\\ & \begin{array}{l}\\ \text{a}.& -2\\ \text{b}.& -5\\ \text{c}.& 2\\ \text{d}.& 5\\ \text{e}.& 10\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {10}^{{}^{{}^2}\log x^2}-7\left({10}^{{}^{{}^2}\log x}\right)+10=0\\ & {10}^{2^{{}^2}\log x}-7\left({10}^{{}^{{}^2}\log x}\right)+10=0\\ & \text{adalah persamaan kuadrat dalam}{10}^{{}^{{}^2}\log x}\\ & \text{Misalkan}p={10}^{{}^{{}^2}\log x},\text{maka persamaan}\\ & \text{menjadi}{p}^2-7p+10=0\{\begin{array}{ll}a& =1\\ b& =-7\\ c& =10\end{array}\\ & \text{Karena nilai}{p}_1\times p_2={\displaystyle \frac{c}{a}\text{maka}}\\ & {10}^{{}^{{}^2}\log x_1}\times {10}^{{}^{{}^2}\log x_2}={\displaystyle \frac{10}{1}=10}\\ & {10}^{{}^{{}^2}\log x_1+{}^2\log x_2}=10\\ & {10}^{{}^{{}^2}\log x_1+{}^2\log x_2}={10}^1\\ & \iff {}^2\log x_1+{}^2\log x_2=1\\ & \iff {}^2\log x_1\times x_2=1\\ & \iff x_1\times x_2=2^1=2\end{array}\end{array}$

$\begin{array}{l}\\ 39.& \text{Nilai}x\text{yang memenuhi}\\ & {}^x\log (x+12)-3.{}^x\log 4+1=0\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{2}}\\ \text{b}.& 2\\ \text{c}.& 4\\ \text{d}.& 8\\ \text{e}.& 16\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {}^x\log (x+12)-3.{}^x\log 4+1=0\\ & {}^x\log (x+12)-{}^x\log 4^3=-1\\ & {}^x\log {\displaystyle \frac{x+12}{64}=-1}\\ & {\displaystyle \frac{x+12}{64}=x^{-1}=\frac{1}{x}}\\ & x+12={\displaystyle \frac{64}{x}}\\ & x^2+12x-64=0\\ & (x+16)(x-4)=0\\ & x=-16\text{atau}x=4\end{array}\end{array}$

$\begin{array}{l}\\ 40.& \text{Nilai}x\text{yang memenuhi}\\ & {\displaystyle \frac{{}^2\log (2x-3)}{{}^2\log x}-{}^x\log (x+6)+{\displaystyle \frac{1}{{}^{(x+2)}\log x}=1}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -1\text{dan}6\\ \text{b}.& -2\text{dan}6\\ \text{c}.& -1\\ \text{d}.& -2\\ \text{e}.& 6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& {\displaystyle \frac{{}^2\log (2x-3)}{{}^2\log x}-{}^x\log (x+6)+{\displaystyle \frac{1}{{}^{(x+2)}\log x}=1}}\\ & {}^x\log (2x-3)-{}^x\log (x+6)+{}^x\log (x+2)=1\\ & {}^x\log (2x-3)(x+2)=1+\log (x+6)\\ & {}^x\log {\displaystyle \frac{(2x^2+x-6)}{(x+6)}=1}\\ & {\displaystyle \frac{(2x^2+x-6)}{(x+6)}=x^1}\\ & (2x^2+x-6)=x^2+6x\\ & x^2-5x-6=0\\ & (x+1)(x-6)=0\\ & x=-1\text{atau}x=6\end{array}\end{array}$

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.

Contoh Soal 5 Fungsi Logaritma

$\begin{array}{l}\\ 21.& \mathbf{\text{(SPMB '04)}}\\ & \text{Jika}a>1,\text{maka penyelesaian untuk}\\ & ({}^a\log (2x+1))({}^3\log \sqrt{a})=1\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 1}\\ \text{b}.& {\displaystyle 2}\\ \text{c}.& {\displaystyle 3}\\ \text{d}.& {\displaystyle 4}\\ \text{e}.& {\displaystyle 5}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}({}^a\log (2x+1))({}^3\log \sqrt{a})& =1\\ ({}^3\log \sqrt{a})({}^a\log (2x+1))& =1\\ ({}^3\log a^{{.}^{{}^{\frac{1}{2}}}})({}^a\log (2x+1))& =1\\ {\displaystyle \frac{1}{2}({}^3\log a)({}^a\log (2x+1))}& =1\\ {}^3\log (2x+1)& =2\\ 2x+1& =3^2\\ 2x& =9-1\\ 2x& =8\\ x& =4\end{array}\end{array}$

$\begin{array}{l}\\ 22.& \mathbf{\text{(SPMB '04)}}\\ & \text{Nilai}{\displaystyle \frac{{({}^5\log 10)}^2-{({}^5\log 2)}^2}{{}^5\log \sqrt{20}}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{2}}\\ \text{b}.& {\displaystyle 1}\\ \text{c}.& {\displaystyle 2}\\ \text{d}.& {\displaystyle 4}\\ \text{e}.& {\displaystyle 5}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {\displaystyle \frac{{({}^5\log 10)}^2-{({}^5\log 2)}^2}{{}^5\log \sqrt{20}}}\\ & ={\displaystyle \frac{({}^5\log 10+{}^5\log 2)({}^5\log 10-{}^5\log 2)}{{}^5\log (20{)}^{{.}^{\frac{1}{2}}}}}\\ & ={\displaystyle \frac{{}^5\log (10.2){\times }^5\log \left(\frac{10}{2}\right)}{{\displaystyle \frac{1}{2}\times {}^5\log 20}}}\\ & =2\times \left({\displaystyle \frac{{}^5\log 20}{{}^5\log 20}}\right)\times {}^5\log 5\\ & =2.1.1\\ & =2\end{array}\end{array}$

$\begin{array}{l}\\ 23.& \mathbf{\text{(SPMB '03)}}\\ & \text{Jika diketahui bahwa}\\ & {}^4{\log }^4\log x-{}^4{\log }^4{\log }^4\log 16=2\\ & \text{maka}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle {}^2\log x=8}\\ \text{b}.& {\displaystyle {}^2\log x=4}\\ \text{c}.& {\displaystyle {}^4\log x=8}\\ \text{d}.& {\displaystyle {}^4\log x=16}\\ \text{e}.& {\displaystyle {}^{16}\log x=8}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {}^4\log {}^4\log x-{}^4\log {}^4\log {}^4\log 16=2\\ & \iff {}^4\log {}^4\log x-{}^4\log {}^4\log {}^4\log 4^2=2\\ & \iff {}^4\log {}^4\log x-{}^4\log {}^4\log 2=2\\ & \iff {}^4\log {}^4\log x-{}^4\log {}^{2^2}\log 2^1=2\\ & \iff {}^4\log {}^4\log x-{}^4\log \left({\displaystyle \frac{1}{2}}\right)=2\\ & \iff {}^4\log {}^4\log x-{}^{2^2}\log 2^{-1}=2\\ & \iff {}^4\log {}^4\log x-(-{\displaystyle \frac{1}{2}})=2\\ & \iff {}^4\log {}^4\log x+\frac{1}{2}=2\\ & \iff {}^4\log {}^4\log x=2-{\displaystyle \frac{1}{2}=\frac{3}{2}}\\ & \iff {}^4\log x=4^{.\frac{3}{2}}\\ & \iff {}^4\log x={\left(2^2\right)}^{{.}^{\frac{3}{2}}}\\ & \iff {}^4\log x=2^3\\ & \iff {}^4\log x=8\end{array}\end{array}$

$\begin{array}{l}\\ 24.& \mathbf{\text{(UMPTN '92)}}\\ & \text{Jika}x\text{memenuhi persamaan}\\ & {}^4{\log }^4\log x-{}^4{\log }^4{\log }^4\log 16=2\\ & \text{maka nilai}{}^{16}\log x=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 4}\\ \text{b}.& {\displaystyle 2}\\ \text{c}.& {\displaystyle 1}\\ \text{d}.& {\displaystyle -2}\\ \text{e}.& {\displaystyle -4}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {}^4\log {}^4\log x-{}^4\log {}^4\log {}^4\log 16=2\\ & \text{menyebabkan}\\ & {}^4\log x=8\Rightarrow x=4^8\\ & (\text{lihat pembahasan no.23})\\ & \text{maka},\\ & {}^{16}\log x={}^{16}\log 4^8={}^{4^2}\log 4^8\\ & ={\displaystyle \frac{8}{2}}\\ & =4\end{array}\end{array}$

$\begin{array}{l}\\ 25.& \mathbf{\text{(UMPTN '94)}}\\ & \text{Hasil kali akar-akar persamaan}\\ & {}^3\log x^{.\left({}^{2+{}^3\log x}\right)}=15\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{9}}\\ \text{b}.& {\displaystyle \frac{1}{3}}\\ \text{c}.& {\displaystyle 1}\\ \text{d}.& {\displaystyle 3}\\ \text{e}.& {\displaystyle 9}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {}^3\log x^{.\left({}^{2+{}^3\log x}\right)}=15\\ & \iff {\left(2+{}^3\log x\right)}^3\log x-15=0\\ & \iff 2{}^3\log x+{({}^3\log x)}^2-15=0\\ & \iff {({}^3\log x)}^2+2{}^3\log x-15=0\\ & \iff ({}^3\log x_1+5)({}^3\log x_2-3)=0\\ & \iff {}^3\log x_1+5=0\text{atau}{}^3\log x_1-3=0\\ & \iff {}^3\log x_1=-5\text{atau}{}^3\log x_2=3\\ & \iff x_1=3^{-5}\text{atau}{x}_2=3^3\\ & \text{maka}\\ & \iff x_1\times x_2=3^{-5}\times 3^3=3^{-5+3}=3^{-2}\\ & \iff ={\displaystyle \frac{1}{3^2}}\\ & \iff =\frac{1}{9}\end{array}\end{array}$

Contoh Soal 4 Fungsi Logaritma

$\begin{array}{l}\\ 16.& \mathbf{\text{(UMPTN '01)}}\\ & \text{Jika}{}^{10}\log x=b,\text{maka}{}^{10x}\log 100=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{b+1}}\\ \text{b}.& {\displaystyle \frac{2}{b+1}}\\ \text{c}.& {\displaystyle \frac{1}{b}}\\ \text{d}.& {\displaystyle \frac{2}{b}}\\ \text{e}.& {\displaystyle \frac{2}{10b}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& {}^{10x}\log 100\\ & ={\displaystyle \frac{\log 100}{\log 10x}}\\ & ={\displaystyle \frac{{}^{10}\log 100}{{}^{10}\log 10x},\text{pilih basis 10}}\\ & \text{alasannya: supaya sama dengan soal}\\ & ={\displaystyle \frac{{}^{10}\log {10}^2}{{}^{10}\log 10+{}^{10}\log x}}\\ & ={\displaystyle \frac{2}{1+b}\text{atau}}\\ & =\frac{2}{b+1}\end{array}\end{array}$

$\begin{array}{l}\\ 17.& \mathbf{\text{(UM UGM '03)}}\\ & \text{Jika}{}^4\log 6=m+1,\text{maka}{}^9\log 8=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{3}{4m-2}}\\ \text{b}.& {\displaystyle \frac{3}{4m+2}}\\ \text{c}.& {\displaystyle \frac{3}{2m+4}}\\ \text{d}.& {\displaystyle \frac{3}{2m-4}}\\ \text{e}.& {\displaystyle \frac{3}{2m+2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Sebelumnya perhatikanlah}\\ & {}^4\log 6=m+1\\ & \iff {}^{2^2}\log (2.3{)}^1=m+1\\ & \iff {\displaystyle \frac{1}{2}\times {}^2\log (2.3)=m+1}\\ & \iff {\displaystyle \frac{1}{2}\times ({}^2\log 2+{}^2\log 3)=m+1}\\ & \iff {\displaystyle \frac{1}{2}\times (1+{}^2\log 3)=m+1}\\ & \iff 1+{}^2\log 3=2m+2\\ & \iff {}^2\log 3=2m+1\\ & \text{Selanjutnya adalah}:\\ & {}^9\log 8={\displaystyle \frac{1}{{}^8\log 9}}\\ & ={\displaystyle \frac{1}{{}^{2^3}\log 3^2}}\\ & ={\displaystyle \frac{1}{{\displaystyle \frac{2}{3}{}^2\log 3}}}\\ & ={\displaystyle \frac{3}{2{}^2\log 3}}\\ & ={\displaystyle \frac{3}{2(2m+1)}}\\ & ={\displaystyle \frac{3}{4m+2}}\end{array}\end{array}$

$\begin{array}{l}\\ 18.& \mathbf{\text{(UMPTN '00)}}\\ & \text{Jika}{}^3\log 5=p\text{dan}{}^3\log 4=q,\\ & \text{maka}{}^4\log 15=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{pq}{1+p}}\\ \text{b}.& {\displaystyle \frac{p+q}{pq}}\\ \text{c}.& {\displaystyle \frac{p+1}{pq}}\\ \text{d}.& {\displaystyle \frac{p+1}{q+1}}\\ \text{e}.& {\displaystyle \frac{pq}{1-p}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {}^4\log 15\\ & ={\displaystyle \frac{{}^{...}\log 15}{{}^{...}\log 4},\text{pilih basis 5}}\\ & \text{mengapa tidak pilih basis selain 5}\\ & \text{lihat penyebut, di sana terdapat numerus 4}\\ & \text{pada soal, pasangan numerus 4 adalah 5},\\ & \text{makanya basis 5 dipilih, bukan yang lain}\\ & ={\displaystyle \frac{{}^5\log 15}{{}^5\log 4}={\displaystyle \frac{{}^5\log (3.5)}{{}^5\log 4}}}\\ & ={\displaystyle \frac{{}^5\log 3+{}^5\log 5}{{}^5\log 4}={\displaystyle \frac{{\displaystyle \frac{1}{{}^3\log 5}+{}^5\log 5}}{{}^5\log 4}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{p}+1}}{q}={\displaystyle \frac{1+p}{pq},\text{atau}}}\\ & ={\displaystyle \frac{p+1}{pq}}\end{array}\end{array}$

$\begin{array}{l}\\ 19.& \mathbf{\text{(UMPTN '94)}}\\ & \text{Jika}{}^6\log 5=a\text{dan}{}^5\log 4=b,\\ & \text{maka}{}^4\log 0,24=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{a+2}{ab}}\\ \text{b}.& {\displaystyle \frac{2a+1}{ab}}\\ \text{c}.& {\displaystyle \frac{a-2}{ab}}\\ \text{d}.& {\displaystyle \frac{2a+1}{2ab}}\\ \text{e}.& {\displaystyle \frac{1-2a}{ab}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& {}^4\log 0,24\\ & ={\displaystyle \frac{{}^{...}\log 0,24}{{}^{...}\log 4}=\frac{{}^{...}\log {\displaystyle \frac{6}{25}}}{{}^{...}\log 4},\text{pilih basis 5}}\\ & \text{mengapa tidak pilih basis selain 5}\\ & \text{lihat penyebut, di sana terdapat numerus 4}\\ & \text{pada soal, pasangan numerus 4 adalah 5},\\ & \text{makanya basis 5 dipilih, bukan yang lain}\\ & =\frac{{}^5\log {\displaystyle \frac{6}{25}}}{{}^5\log 4}={\displaystyle \frac{{}^5\log 6-{}^5\log 25}{{}^5\log 4}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{{}^6\log 5}-{}^5\log 5^2}}{{}^5\log 4}={\displaystyle \frac{{\displaystyle \frac{1}{a}-2}}{b}=\frac{1-2a}{ab}}}\end{array}\end{array}$

$\begin{array}{l}\\ 20.& \mathbf{\text{(SPMB '05)}}\\ & \text{Jika}{}^3\log 2=p\text{dan}{}^2\log 7=q,\\ & \text{maka}{}^{14}\log 54=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{p+3}{p+q}}\\ \text{b}.& {\displaystyle \frac{2p}{p+q}}\\ \text{c}.& {\displaystyle \frac{p+3}{p(q+1)}}\\ \text{d}.& {\displaystyle \frac{p+q}{p(q+1)}}\\ \text{e}.& {\displaystyle \frac{p(q+1)}{p+q}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {}^{14}\log 54\\ & ={\displaystyle \frac{{}^{...}\log 54}{{}^{...}\log 14}=\frac{{}^{...}\log (2.27)}{{}^{...}\log (2.7)},\text{pilih basis 2}}\\ & \text{mengapa tidak pilih basis selain 2}\\ & \text{lihat penyebut, di sana terdapat numerus 7}\\ & \text{pada soal, pasangan numerus 7 adalah 2},\\ & \text{makanya basis 2 dipilih, bukan yang lain}\\ & =\frac{{}^2\log (2.27)}{{}^2\log (2.7)}={\displaystyle \frac{{}^2\log 2+{}^2\log 27}{{}^2\log 2+{}^2\log 7}}\\ & ={\displaystyle \frac{{}^2\log 2+{}^2\log 3^3}{{}^2\log 2+{}^2\log 7}={\displaystyle \frac{{}^2\log 2+(3\times {\displaystyle \frac{1}{{}^3\log 2}})}{{}^2\log 2+{}^2\log 7}}}\\ & ={\displaystyle \frac{1+{\displaystyle \frac{3}{p}}}{1+q}}\\ & ={\displaystyle \frac{p+3}{p(q+1)}}\end{array}\end{array}$

Contoh Soal 3 Fungsi Logaritma

$\begin{array}{l}\\ 11.& \text{Nilai dari}\\ & {\displaystyle \frac{1}{6}.{}^2\log 25-\frac{1}{3}.{}^2\log 10\text{adalah}...}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -3}\\ \text{b}.& {\displaystyle -2}\\ \text{c}.& -2{\displaystyle \frac{1}{2}}\\ \text{d}.& -{\displaystyle \frac{1}{2}}\\ \text{e}.& -{\displaystyle \frac{1}{3}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& ={\displaystyle \frac{1}{6}.{}^2\log 25-\frac{1}{3}.{}^2\log 10}\\ & ={\displaystyle \frac{1}{3}.\frac{1}{2}.{}^2\log 25-\frac{1}{3}.{}^2\log 10}\\ & =\frac{1}{3}({}^2\log {25}^{\frac{1}{2}}-{}^2\log 10)\\ & =\frac{1}{3}({}^2\log 5-{}^2\log 10)\\ & =\frac{1}{3}({}^2\log \frac{5}{10})\\ & =\frac{1}{3}({}^2\log \frac{1}{2})\\ & =\frac{1}{3}({}^2\log 2^{-1})\\ & =-\frac{1}{3}\end{array}\end{array}$

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

$\begin{array}{l}\\ 13.& \mathbf{\text{(UMPTN '99)}}\\ & \text{Diketahui}\log 2=0,3010\text{dan}\log 3=0,4771\\ & \text{maka}\log (\sqrt[3]{2}\times \sqrt{3})\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 0,1505}\\ \text{b}.& {\displaystyle 0,1590}\\ \text{c}.& {\displaystyle 0,2007}\\ \text{d}.& {\displaystyle 0,3389}\\ \text{e}.& {\displaystyle 0,3891}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \log (\sqrt[3]{2}\times \sqrt{3})\\ & =\log \sqrt[3]{2}+\log \sqrt{3}\\ & =\log 2^{\frac{1}{3}}+\log 3^{\frac{1}{2}}\\ & ={\displaystyle \frac{1}{3}\log 2+{\displaystyle \frac{1}{2}\log 3}}\\ & ={\displaystyle \frac{1}{3}(0,3010)+{\displaystyle \frac{1}{2}(0,4771)}}\\ & =0,1003+0,2386\\ & =0,3389\end{array}\end{array}$

$\begin{array}{l}\\ 14.& \mathbf{\text{(UMPTN '98)}}\\ & \text{Nilai}{}^a\log {\displaystyle \frac{1}{b}\times {}^b\log {\displaystyle \frac{1}{c^2}\times {}^c\log {\displaystyle \frac{1}{a^3}=....}}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -6}\\ \text{b}.& {\displaystyle 6}\\ \text{c}.& {\displaystyle \frac{b}{a^{2}c}}\\ \text{d}.& {\displaystyle \frac{a^{2}c}{b}}\\ \text{e}.& {\displaystyle -\frac{1}{6}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {}^a\log {\displaystyle \frac{1}{b}\times {}^b\log {\displaystyle \frac{1}{c^2}\times {}^c\log {\displaystyle \frac{1}{a^3}}}}\\ & ={}^a\log {\displaystyle b^{-1}\times {}^b\log {\displaystyle c^{-2}\times {}^c\log {\displaystyle a^{-3}}}}\\ & =(-1).(-2).(-3)\times {}^a\log {\displaystyle a\times {}^b\log {\displaystyle c\times {}^c\log {\displaystyle a}}}\\ & =-6\times {}^a\log a\\ & =-6\times 1\\ & =-6\end{array}\end{array}$

$\begin{array}{l}\\ 15.& \mathbf{\text{(UMPTN '01)}}\\ & \text{Jika}{\displaystyle \frac{{}^2\log a}{{}^3\log b}=m\text{dan}{\displaystyle \frac{{}^3\log a}{{}^2\log b}=n}}\\ & \text{dengan}a>1,b>1,\text{maka}{\displaystyle \frac{m}{n}=....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle {}^2\log 3}\\ \text{b}.& {\displaystyle {}^3\log 2}\\ \text{c}.& {\displaystyle {}^4\log 9}\\ \text{d}.& {\displaystyle {({}^3\log 2)}^2}\\ \text{e}.& {\displaystyle {({}^2\log 3)}^2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{\displaystyle \frac{m}{n}}& ={\displaystyle \frac{{\displaystyle \frac{{}^2\log a}{{}^3\log b}}}{{\displaystyle \frac{{}^3\log a}{{}^2\log b}}}}\\ & ={\displaystyle \frac{{}^2\log a\times {}^2\log b}{{}^3\log b\times {}^3\log a}}\\ & ={\displaystyle \frac{{}^2\log a\times {\displaystyle \frac{1}{{}^b\log 2}}}{{}^3\log b\times {\displaystyle \frac{1}{{}^a\log 3}}}}\\ & ={\displaystyle \frac{{}^2\log a\times {}^a\log 3}{{}^3\log b\times {}^b\log 2}}\\ & ={\displaystyle \frac{{}^2\log 3}{{}^3\log 2}={\displaystyle \frac{{}^2\log 3}{{\displaystyle \frac{1}{{}^2\log 3}}}}}\\ & ={({}^2\log 3)}^2\end{array}\end{array}$

Contoh Soal 2 Fungsi Logaritma

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

$\begin{array}{l}\\ 7.& \text{Nilai dari}{}^{\sqrt{5}}\log \sqrt{125}\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -3}\\ \text{b}.& {\displaystyle -2}\\ \text{c}.& {\displaystyle 2}\\ \text{d}.& {\displaystyle 3}\\ \text{e}.& {\displaystyle 5}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& ={}^{\sqrt{5}}\log \sqrt{125}\\ & ={}^{{\sqrt{5}}^1}\log {\left(\sqrt{5}\right)}^3\\ & ={\displaystyle \frac{3}{1}\times {}^{\sqrt{5}}\log \sqrt{5}}\\ & =3\end{array}\end{array}$

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

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

$\begin{array}{l}\\ 10.& \text{Nilai dari}\\ & {}^2\log {\displaystyle \frac{4}{3}+2.{}^2\log \sqrt{12}\text{adalah}...}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 6}\\ \text{b}.& {\displaystyle 4}\\ \text{c}.& 3{\displaystyle \frac{1}{2}}\\ \text{d}.& 2{\displaystyle \frac{1}{2}}\\ \text{e}.& {\displaystyle 2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& ={}^2\log {\displaystyle \frac{4}{3}+2.{}^2\log \sqrt{12}}\\ & ={}^2\log \frac{4}{3}+{}^2\log {\left(\sqrt{12}\right)}^2\\ & ={}^2\log \frac{4}{3}+{}^2\log 12\\ & ={}^2\log \frac{4}{3}\times 12\\ & ={}^2\log 16\\ & ={}^2\log 2^4\\ & =4\end{array}\end{array}$

Contoh Soal 1 Fungsi Logaritma

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

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

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

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

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

Lanjutan Materi Fungsi Logaritma

 $\text{B. Sifat-Sifat Logaritma}$

Jika syarat logaritma memenuhi untuk bilangan yang diposisikan sebagai basis dan numerus, maka akan berlaku sifat-sifat loaritma berikut:

$\begin{array}{rl}(1)& a^{{}^a\text{log b}}=b\\ (2)& {}^a\log (b.c)={}^a\log b+{}^a\log c\\ (3)& {}^a\log \left({\displaystyle \frac{b}{c}}\right)={}^a\log b-{}^a\log c\\ (4)& {}^a\log b={\displaystyle \frac{{}^x\log b}{{}^x\log c}}\\ (5)& {}^a\log b={\displaystyle \frac{1}{{}^b\log a}}\\ (6)& {}^a\log b=n\Rightarrow {}^b\log a={\displaystyle \frac{1}{n}}\\ (7)& {}^{a^m}\log b^n={\displaystyle \frac{n}{m}\times {}^a\log b}\\ (8)& {}^a\log b\times {}^b\log c\times {}^c\log p={}^a\log p\\ (9)& {}^a\log a=1\\ (10)& {}^a\log a^n=n\\ (11)& {}^a\log 1=0\\ (12)& {}^{.}\log b={}^{10}\log b\end{array}$

ada yang tak kalah penting untuk diketahui walaupun kadang sebagian orang menganggap tidak perlu dituliskan, di sini saya tuliskan, yaitu:

$\begin{array}{rl}(\text{a})& \log 2=0,3010\\ (\text{b})& \log 3=0,4771\\ (\text{c})& \log 5=0,6990\\ (\text{d})& \log 7=0,8451\end{array}$

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{l}\\ 1.& {}^2\log 3+{}^2\log 8-{}^2\log 24=....\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}{}^2\log 3+{}^2\log 8-{}^2\log 24& ={}^2\log \left({\displaystyle \frac{3\times 8}{24}}\right)\\ & ={}^2\log 1\\ & ={}^2\log 2^0\\ & =0\end{array}\end{array}$

$\begin{array}{l}\\ 2.& {}^2\log 12+{}^2\log 8-{}^2\log 24=....\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}{}^2\log 12+{}^2\log 8-{}^2\log 24& ={}^2\log \left({\displaystyle \frac{12\times 8}{24}}\right)\\ & ={}^2\log 4\\ & ={}^2\log 2^2\\ & =2\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Diketahui}{}^3\log 7=a,{}^5\log 2=b,\text{dan}{}^2\log 3=c\\ & \text{Nyatakanlah logaritma berikut dalam bentuk}a,b,\text{dan}c,\text{yaitu}:\\ & \text{a}.{}^7\log 3\\ & \text{b}.{}^4\log 5\\ & \text{c}.{}^{21}\log 5\\ & \text{d}.{}^6\log 7\\ \\ & \text{Jawab}:\\ & \begin{array}{|ll|}\hline \begin{array}{rl}& \\ {}^7\log 3& ={\displaystyle \frac{1}{{}^3\log 7}}\\ & ={\displaystyle \frac{1}{a}}\\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}{}^4\log 5& ={\displaystyle \frac{1}{{}^5\log 4}}\\ & ={\displaystyle \frac{1}{{}^5\log 2^2}}\\ & ={\displaystyle \frac{1}{2{}^5\log 2}}\\ & ={\displaystyle \frac{1}{2b}}\end{array}\\ \begin{array}{rl}& \\ {}^{21}\log 5& ={\displaystyle \frac{{}^{...}\log 5}{{}^{...}\log 21}}\\ & ={\displaystyle \frac{{}^2\log 5}{{}^2\log 21}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{{}^5\log 2}}}{{\displaystyle \frac{{}^3\log 21}{{}^3\log 2}}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{{}^5\log 2}}}{{\displaystyle \frac{{}^3\log 3\times 7}{{}^3\log 2}}}}\\ & ={\displaystyle \frac{1}{bc(1+a)}}\end{array}& \begin{array}{rl}& \\ {}^6\log 7& ={\displaystyle \frac{{}^3\log 7}{{}^3\log 6}}\\ & ={\displaystyle \frac{{}^3\log 7}{{}^3\log 2\times 3}}\\ & ={\displaystyle \frac{{}^3\log 7}{{}^3\log 2+{}^3\log 3}}\\ & ={\displaystyle \frac{{}^3\log 7}{{\displaystyle \frac{1}{{}^2\log 3}+{}^3\log 3}}}\\ & ={\displaystyle \frac{a}{{\displaystyle \frac{1}{c}+1}}}\\ & ={\displaystyle \frac{ac}{1+c}}\\ & \\ & \end{array}\\ \hline\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Diketahui bahwa}{}^4\log 5=a\\ & \text{a}.\text{Carilah nilai}{}^4\log 10\\ & \text{b}.\text{Tunjukkan bahwa}{}^{0,1}\log 1,25={\displaystyle \frac{2-2a}{2a+1}}\\ \\ & \text{Jawab}:\\ & \begin{array}{|ll|}\hline \begin{array}{rl}& \\ {}^4\log 10& ={}^4\log (2\times 5)\\ & ={}^4\log 2+{}^4\log 5\\ & ={}^{2^2}\log 2+a\\ & ={\displaystyle \frac{1}{2}.{}^2\log 2+a}\\ & ={\displaystyle \frac{1}{2}+a}\\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}& \\ & {}^{0,1}\log 1,25\\ & ={\displaystyle \frac{{}^4\log 1,25}{{}^4\log 0,1}}\\ & ={\displaystyle \frac{{}^4\log {\displaystyle \frac{125}{100}}}{{}^4\log {\displaystyle \frac{1}{10}}}}\\ & ={\displaystyle \frac{{}^4\log 125-{}^4\log 100}{{}^4\log {10}^{-1}}}\\ & ={\displaystyle \frac{{}^4\log 5^3-{}^4\log {10}^2}{-{}^4\log 10}}\\ & ={\displaystyle \frac{3.{}^4\log 5-2.{}^4\log 10}{-{}^4\log 10}}\\ & ={\displaystyle \frac{3a-2\left({\displaystyle \frac{1}{2}+a}\right)}{-\left({\displaystyle \frac{1}{2}+a}\right)}}\\ & ={\displaystyle \frac{a-1}{-a-{\displaystyle \frac{1}{2}}}\times {\displaystyle \frac{-2}{-2}}}\\ & ={\displaystyle \frac{2-2a}{2a+1}◼}\end{array}\\ \hline\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Jika}{}^{2017}\log {\displaystyle \frac{1}{x}={}^x\log {\displaystyle \frac{1}{y}={}^y\log {\displaystyle \frac{1}{2017}}}}\\ & \text{maka hasil dari}(2x-3y)\\ \\ & \text{Jawab}:\\ \\ & \begin{array}{rl}{}^{2017}\log {\displaystyle \frac{1}{x}}& ={}^x\log {\displaystyle \frac{1}{y}={}^y\log {\displaystyle \frac{1}{2017}}}\\ {}^{2017}\log {\displaystyle \frac{1}{x}}& ={}^y\log {\displaystyle \frac{1}{2017}}\\ {}^{2017}\log {\displaystyle x^{-1}}& ={}^y\log {\displaystyle (2017{)}^{-1}}\\ -{}^{2017}\log x& =-{}^y\log 2017\\ {}^{2017}\log x& ={}^y\log 2017\\ & \text{dipenuhi saat}\\ x& =y=2017\end{array}\\ \\ & (2x-3y)=2x-3x=-x=-2017\end{array}$

Fungsi Logaritma

$\text{A. Pendahuluan}$

Logaritma merupakan invers(balikan) dari perpangkatan

Secara definisi:

$\boxed{{\displaystyle a^c=b{\Rightarrow }^a\log b=c}}$, tetapi di sini diberikan syarat bahwa bilangan basis/dasar perpangkatannya harus berupa bilangan real positif dan tidak sama dengan satu serta bilangan pangkatnya(ekponen) harus berupa bilangan real positif juga.

Perhatikanlah ringkasannya

${}^a\log b=c\{\begin{array}{ll}a& \text{syaratnya}:a>0,a\ne 1\\ & \text{selanjutnya disebut basis}\\ b& \text{syaratnya}:b>0\\ & \text{selanjutnya disebut}\mathbf{\text{numerus}}\\ c& \text{tidak ada syarat apapun}\\ & \text{selanjutnya disebut hasil logaritma}\end{array}$

Contoh berikut adalah mengubah bentuk perpangkatan ke dalam logaritma yang memenuhi persyaratan

$\begin{array}{rl}(1)& 2^4=16\Rightarrow {}^2\log 16=4\\ (2)& 2^3=8\Rightarrow {}^2\log 8=3\\ (3)& 2^2=4\Rightarrow {}^2\log 4=2\\ (4)& 2^1=2\Rightarrow {}^2\log 2=1\\ (5)& 2^0=1\Rightarrow {}^2\log 1=0\\ (6)& 2^{-1}={\displaystyle \frac{1}{2}=0,5\Rightarrow {}^2\log {\displaystyle \frac{1}{2}=-1}}\\ (7)& 2^{-2}={\displaystyle \frac{1}{4}=0,25\Rightarrow {}^2\log {\displaystyle \frac{1}{4}=-2}}\\ (8)& 2^{-3}={\displaystyle \frac{1}{8}=0,125\Rightarrow {}^2\log {\displaystyle \frac{1}{8}=-3}}\\ (9)& 2^{-4}={\displaystyle \frac{1}{16}=0,0625\Rightarrow {}^2\log {\displaystyle \frac{1}{16}=-4}}\end{array}$

Berikut contoh kebalikan di atas yang tidak memenuhi definisi logaritma yang ada, yaitu:

$\begin{array}{rl}(1)& (-2{)}^4=16\Rightarrow {}^{(-2)}\log 16=\cdots \\ (2)& (-2{)}^3=-8\Rightarrow {}^{(-2)}\log (-8)=\cdots \\ (3)& (-2{)}^2=4\Rightarrow {}^{(-2)}\log 4=\cdots \\ (4)& (-2{)}^1=-2\Rightarrow {}^{(-2)}\log (-2)=\cdots \\ (5)& (-2{)}^0=1\Rightarrow {}^{(-2)}\log 1=\cdots \\ (6)& (-2{)}^{-1}=-{\displaystyle \frac{1}{2}\Rightarrow {}^{(-2)}\log (-{\displaystyle \frac{1}{2}})=\cdots }\\ (7)& (-2{)}^{-2}={\displaystyle \frac{1}{4}\Rightarrow {}^{(-2)}\log {\displaystyle \frac{1}{4}=\cdots }}\\ (8)& (-2{)}^{-3}=-{\displaystyle \frac{1}{8}\Rightarrow {}^{(-2)}\log (-{\displaystyle \frac{1}{8}})=\cdots }\\ (9)& (-2{)}^{-4}={\displaystyle \frac{1}{16}\Rightarrow {}^{(-2)}\log {\displaystyle \frac{1}{16}=\cdots }}\end{array}$