$\begin{array}{ll}\\ 61.&\textrm{Nilai dari}\\ & ^{2}\log \displaystyle \frac{4}{3}+\: 2.\, ^{2}\log \sqrt{12}\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 6\\ \color{red}\textrm{b}.&\displaystyle 4\\ \textrm{c}.&3 \displaystyle \frac{1}{2}\\ \textrm{d}.& 2\displaystyle \frac{1}{2}\\ \textrm{e}.& \displaystyle 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&=\: ^{2}\log \displaystyle \frac{4}{3}+\: 2.\, ^{2}\log \sqrt{12}\\ &=\: ^{2}\log \frac{4}{3}+\: ^{2}\log \left ( \sqrt{12} \right )^{2}\\ &=\: ^{2}\log \frac{4}{3}+\: ^{2}\log 12\\ &=\: ^{2}\log \frac{4}{3}\times 12\\ &=\: ^{2}\log 16\\ &=\: ^{2}\log 2^{4}\\ &=\color{red}4 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 62.&\textrm{Nilai dari}\\ & \displaystyle \frac{1}{6}.\, ^{2}\log 25-\: \frac{1}{3}.\, ^{2}\log 10\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -3\\ \textrm{b}.&\displaystyle -2\\ \textrm{c}.&-2 \displaystyle \frac{1}{2}\\ \textrm{d}.& -\displaystyle \frac{1}{2}\\ \color{red}\textrm{e}.&- \displaystyle \frac{1}{3} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&=\: \displaystyle \frac{1}{6}.\, ^{2}\log 25-\: \frac{1}{3}.\, ^{2}\log 10\\ &=\displaystyle \frac{1}{3}.\frac{1}{2}.\, ^{2}\log 25-\: \frac{1}{3}.\, ^{2}\log 10\\ &=\frac{1}{3}\left ( ^{2}\log 25^{\frac{1}{2}} -\: ^{2}\log 10\right )\\ &=\frac{1}{3}\left ( ^{2}\log 5-\: ^{2}\log 10 \right )\\ &=\frac{1}{3}\left ( ^{2}\log \frac{5}{10} \right )\\ &=\frac{1}{3}\left ( ^{2}\log \frac{1}{2} \right )\\ &=\frac{1}{3}\left ( ^{2}\log 2^{-1} \right )\\ &=\color{red}-\frac{1}{3} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 63.&\textrm{Nilai dari}\\ & ^{2}\log \left ( ^{2}\log \sqrt{\sqrt{2}} \right )\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -4\\ \color{red}\textrm{b}.&\displaystyle -2\\ \textrm{c}.&-1 \displaystyle \frac{1}{2}\\ \textrm{d}.& -\displaystyle \frac{1}{2}\\ \textrm{e}.&- \displaystyle \frac{1}{4} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&=\: ^{2}\log \left ( ^{2}\log \sqrt{\sqrt{2}} \right )\\ &=\: ^{2}\log \left ( ^{2}\log 2^{\frac{1}{4}} \right )\\ &=\: ^{2}\log \frac{1}{4}\\ &=\: ^{2}\log \left (2 \right )^{-2}\\ &=\color{red}-2 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 64.&\textbf{(UMPTN '99)}\\ &\textrm{Diketahui}\: \: \log 2=0,3010\: \: \textrm{dan}\: \: \log 3=0,4771\\ &\textrm{maka}\: \: \log \left ( \sqrt[3]{2}\times \sqrt{3} \right )\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 0,1505\\ \textrm{b}.&\displaystyle 0,1590\\ \textrm{c}.&\displaystyle 0,2007\\ \color{red}\textrm{d}.&\displaystyle 0,3389\\ \textrm{e}.&\displaystyle 0,3891 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\log \left ( \sqrt[3]{2}\times \sqrt{3} \right )\\ &=\log \sqrt[3]{2}+\log \sqrt{3}\\ &=\log 2^{\frac{1}{3}}+\log 3^{\frac{1}{2}}\\ &=\displaystyle \frac{1}{3}\log 2+\displaystyle \frac{1}{2}\log 3\\ &=\displaystyle \frac{1}{3}(0,3010)+\displaystyle \frac{1}{2}(0,4771)\\ &=0,1003+0,2386\\ &=\color{red}0,3389 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 65.&\textbf{(UMPTN '98)}\\ &\textrm{Nilai}\: \: ^{a}\log \displaystyle \frac{1}{b}\times \: ^{b}\log \displaystyle \frac{1}{c^{2}}\times \: ^{c}\log \displaystyle \frac{1}{a^{3}}\: =\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle -6\\ \textrm{b}.&\displaystyle 6\\ \textrm{c}.&\displaystyle \frac{b}{a^{2}c}\\ \textrm{d}.&\displaystyle \frac{a^{2}c}{b}\\ \textrm{e}.&\displaystyle -\frac{1}{6} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&^{a}\log \displaystyle \frac{1}{b}\times \: ^{b}\log \displaystyle \frac{1}{c^{2}}\times \: ^{c}\log \displaystyle \frac{1}{a^{3}}\\ &=\: ^{a}\log \displaystyle b^{-1}\times \: ^{b}\log \displaystyle c^{-2}\times \: ^{c}\log \displaystyle a^{-3}\\ &=(-1).(-2).(-3)\times \: ^{a}\log \displaystyle a\times \: ^{b}\log \displaystyle c\times \: ^{c}\log \displaystyle a\\ &=-6\times \: ^{a}\log a\\ &=-6\times 1\\ &=\color{red}-6 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 66.&\textbf{(UMPTN '01)}\\ &\textrm{Jika}\: \: \displaystyle \frac{^{2}\log a}{^{3}\log b}=m\: \: \textrm{dan}\: \: \displaystyle \frac{^{3}\log a}{^{2}\log b}=n\\ &\textrm{dengan}\: \: a> 1,\: b> 1,\: \textrm{maka}\: \: \displaystyle \frac{m}{n}=....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle ^{2}\log 3\\ \textrm{b}.&\displaystyle ^{3}\log 2\\ \textrm{c}.&\displaystyle ^{4}\log 9\\ \textrm{d}.&\displaystyle \left ( ^{3}\log 2 \right )^{2}\\ \color{red}\textrm{e}.&\displaystyle \left ( ^{2}\log 3 \right )^{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}\displaystyle \frac{m}{n}&=\displaystyle \frac{\displaystyle \frac{^{2}\log a}{^{3}\log b}}{\displaystyle \frac{^{3}\log a}{^{2}\log b}}\\ &=\displaystyle \frac{^{2}\log a\times \: ^{2}\log b}{^{3}\log b\times \: ^{3}\log a}\\ &=\displaystyle \frac{^{2}\log a\times \: \displaystyle \frac{1}{^{b}\log 2}}{^{3}\log b\times \: \displaystyle \frac{1}{^{a}\log 3}}\\ &=\displaystyle \frac{^{2}\log a\times \: ^{a}\log 3}{^{3}\log b\times \: ^{b}\log 2}\\ &=\displaystyle \frac{^{2}\log 3}{^{3}\log 2}=\displaystyle \frac{^{2}\log 3}{\displaystyle \frac{1}{^{2}\log 3}}\\ &=\color{red}\left ( ^{2}\log 3 \right )^{2} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 67.&\textbf{(UMPTN '01)}\\ &\textrm{Jika}\: \: ^{10}\log x=b\: ,\: \textrm{maka}\: \: ^{10x}\log 100=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{b+1}\\ \color{red}\textrm{b}.&\displaystyle \frac{2}{b+1}\\ \textrm{c}.&\displaystyle \frac{1}{b}\\ \textrm{d}.&\displaystyle \frac{2}{b}\\ \textrm{e}.&\displaystyle \frac{2}{10b} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&^{10x}\log 100\\ &=\displaystyle \frac{\log 100}{\log 10x}\\ &=\displaystyle \frac{^{10}\log 100}{^{10}\log 10x},\quad \color{magenta}\textrm{pilih basis 10}\\ &\color{purple}\textrm{alasannya: supaya sama dengan soal}\\ &=\displaystyle \frac{^{10}\log 10^{2}}{^{10}\log 10+\: ^{10}\log x}\\ &=\displaystyle \frac{2}{1+b}\: \: \textrm{atau}\\ &=\color{red}\frac{2}{b+1} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 68.&\textbf{(UM UGM '03)}\\ &\textrm{Jika}\: \: ^{4}\log 6=m+1\: ,\: \textrm{maka}\: \: ^{9}\log 8=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{3}{4m-2}\\ \color{red}\textrm{b}.&\displaystyle \frac{3}{4m+2}\\ \textrm{c}.&\displaystyle \frac{3}{2m+4}\\ \textrm{d}.&\displaystyle \frac{3}{2m-4}\\ \textrm{e}.&\displaystyle \frac{3}{2m+2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\color{purple}\textrm{Sebelumnya perhatikanlah}\\ &^{4}\log 6=m+1\\ &\Leftrightarrow \: ^{2^{2}}\log (2.3)^{1}=m+1\\ &\Leftrightarrow \: \displaystyle \frac{1}{2}\times \: ^{2}\log (2.3)=m+1\\ &\Leftrightarrow \: \displaystyle \frac{1}{2}\times \: \left (^{2}\log 2 +\: ^{2}\log 3 \right )=m+1\\ &\Leftrightarrow \: \displaystyle \frac{1}{2}\times \: \left (1 +\: ^{2}\log 3 \right )=m+1\\ &\Leftrightarrow \: 1 +\: ^{2}\log 3=2m+2\\ &\Leftrightarrow \: ^{2}\log 3=2m+1\\ &\color{red}\textrm{Selanjutnya adalah}:\\ &^{9}\log 8=\: \displaystyle \frac{1}{^{8}\log 9}\\ &=\: \displaystyle \frac{1}{^{2^{3}}\log 3^{2}}\\ &=\: \displaystyle \frac{1}{\displaystyle \frac{2}{3}\: ^{2}\log 3}\\ &=\: \displaystyle \frac{3}{2\: ^{2}\log 3}\\ &=\: \displaystyle \frac{3}{2(2m+1)}\\ &=\: \color{red}\displaystyle \frac{3}{4m+2} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 69.&\textbf{(UMPTN '00)}\\ &\textrm{Jika}\: \: ^{3}\log 5=p\: \: \textrm{dan}\: \: ^{3}\log 4=q,\\ &\textrm{maka}\: \: ^{4}\log 15=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{pq}{1+p}\\ \textrm{b}.&\displaystyle \frac{p+q}{pq}\\ \color{red}\textrm{c}.&\displaystyle \frac{p+1}{pq}\\ \textrm{d}.&\displaystyle \frac{p+1}{q+1}\\ \textrm{e}.&\displaystyle \frac{pq}{1-p} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&^{4}\log 15\\ &=\displaystyle \frac{^{...}\log 15}{^{...}\log 4},\: \: \color{blue}\textrm{pilih basis 5}\\ &\color{blue}\textrm{mengapa tidak pilih basis selain 5}\\ &\color{blue}\textrm{lihat penyebut, di sana terdapat numerus 4}\\ &\color{blue}\textrm{pada soal, pasangan numerus 4 adalah 5},\\ &\textrm{makanya basis 5 dipilih, bukan yang lain}\\ &=\displaystyle \frac{^{5}\log 15}{^{5}\log 4}=\displaystyle \frac{^{5}\log (3.5)}{^{5}\log 4}\\ &=\displaystyle \frac{^{5}\log 3+\: ^{5}\log 5}{^{5}\log 4}=\displaystyle \frac{\displaystyle \frac{1}{^{3}\log 5}+\: ^{5}\log 5}{^{5}\log 4}\\ &=\displaystyle \frac{\displaystyle \frac{1}{p}+1}{q}=\displaystyle \frac{1+p}{pq},\: \: \textrm{atau}\\ &=\color{red}\displaystyle \frac{p+1}{pq} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 70.&\textbf{(UMPTN '94)}\\ &\textrm{Jika}\: \: ^{6}\log 5=a\: \: \textrm{dan}\: \: ^{5}\log 4=b,\\ &\textrm{maka}\: \: ^{4}\log 0,24=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{a+2}{ab}\\ \textrm{b}.&\displaystyle \frac{2a+1}{ab}\\ \textrm{c}.&\displaystyle \frac{a-2}{ab}\\ \textrm{d}.&\displaystyle \frac{2a+1}{2ab}\\ \color{red}\textrm{e}.&\displaystyle \frac{1-2a}{ab} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&^{4}\log 0,24\\ &=\displaystyle \frac{^{...}\log 0,24}{^{...}\log 4}=\frac{^{...}\log \displaystyle \frac{6}{25}}{^{...}\log 4},\: \: \color{purple}\textrm{pilih basis 5}\\ &\color{red}\textrm{mengapa tidak pilih basis selain 5}\\ &\color{red}\textrm{lihat penyebut, di sana terdapat numerus 4}\\ &\color{red}\textrm{pada soal, pasangan numerus 4 adalah 5},\\ &\textrm{makanya basis 5 dipilih, bukan yang lain}\\ &=\frac{^{5}\log \displaystyle \frac{6}{25}}{^{5}\log 4}=\displaystyle \frac{^{5}\log 6-\: ^{5}\log 25}{^{5}\log 4}\\ &=\displaystyle \frac{\displaystyle \frac{1}{^{6}\log 5}-\: ^{5}\log 5^{2}}{^{5}\log 4}=\displaystyle \frac{\displaystyle \frac{1}{a}-2}{b}=\color{red}\frac{1-2a}{ab} \end{aligned} \end{array}$.
