Contoh Soal 5 Fungsi Logaritma

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

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

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

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

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