Contoh Soal 8 Fungsi Logaritma (Persamaan Logaritma)

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

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

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

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

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