Contoh Soal 4 Fungsi Logaritma

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

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

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

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

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