Latihan Soal 7 Persiapan PAS Gasal Matematika Peminatan Kelas X (Fungsi Eksponen dan Fungsi Logaritma)

 $\begin{array}{l}\\ 61.& \text{Nilai dari}\\ & {}^2\log {\displaystyle \frac{4}{3}+2.{}^2\log \sqrt{12}\text{adalah}...}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 6}\\ \text{b}.& {\displaystyle 4}\\ \text{c}.& 3{\displaystyle \frac{1}{2}}\\ \text{d}.& 2{\displaystyle \frac{1}{2}}\\ \text{e}.& {\displaystyle 2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& ={}^2\log {\displaystyle \frac{4}{3}+2.{}^2\log \sqrt{12}}\\ & ={}^2\log \frac{4}{3}+{}^2\log {\left(\sqrt{12}\right)}^2\\ & ={}^2\log \frac{4}{3}+{}^2\log 12\\ & ={}^2\log \frac{4}{3}\times 12\\ & ={}^2\log 16\\ & ={}^2\log 2^4\\ & =4\end{array}\end{array}$.

$\begin{array}{l}\\ 62.& \text{Nilai dari}\\ & {\displaystyle \frac{1}{6}.{}^2\log 25-\frac{1}{3}.{}^2\log 10\text{adalah}...}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -3}\\ \text{b}.& {\displaystyle -2}\\ \text{c}.& -2{\displaystyle \frac{1}{2}}\\ \text{d}.& -{\displaystyle \frac{1}{2}}\\ \text{e}.& -{\displaystyle \frac{1}{3}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& ={\displaystyle \frac{1}{6}.{}^2\log 25-\frac{1}{3}.{}^2\log 10}\\ & ={\displaystyle \frac{1}{3}.\frac{1}{2}.{}^2\log 25-\frac{1}{3}.{}^2\log 10}\\ & =\frac{1}{3}({}^2\log {25}^{\frac{1}{2}}-{}^2\log 10)\\ & =\frac{1}{3}({}^2\log 5-{}^2\log 10)\\ & =\frac{1}{3}({}^2\log \frac{5}{10})\\ & =\frac{1}{3}({}^2\log \frac{1}{2})\\ & =\frac{1}{3}({}^2\log 2^{-1})\\ & =-\frac{1}{3}\end{array}\end{array}$

$\begin{array}{l}\\ 63.& \text{Nilai dari}\\ & {}^2\log ({}^2\log \sqrt{\sqrt{2}})\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -4}\\ \text{b}.& {\displaystyle -2}\\ \text{c}.& -1{\displaystyle \frac{1}{2}}\\ \text{d}.& -{\displaystyle \frac{1}{2}}\\ \text{e}.& -{\displaystyle \frac{1}{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& ={}^2\log ({}^2\log \sqrt{\sqrt{2}})\\ & ={}^2\log ({}^2\log 2^{\frac{1}{4}})\\ & ={}^2\log \frac{1}{4}\\ & ={}^2\log {\left(2\right)}^{-2}\\ & =-2\end{array}\end{array}$

$\begin{array}{l}\\ 64.& \mathbf{\text{(UMPTN '99)}}\\ & \text{Diketahui}\log 2=0,3010\text{dan}\log 3=0,4771\\ & \text{maka}\log (\sqrt[3]{2}\times \sqrt{3})\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 0,1505}\\ \text{b}.& {\displaystyle 0,1590}\\ \text{c}.& {\displaystyle 0,2007}\\ \text{d}.& {\displaystyle 0,3389}\\ \text{e}.& {\displaystyle 0,3891}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \log (\sqrt[3]{2}\times \sqrt{3})\\ & =\log \sqrt[3]{2}+\log \sqrt{3}\\ & =\log 2^{\frac{1}{3}}+\log 3^{\frac{1}{2}}\\ & ={\displaystyle \frac{1}{3}\log 2+{\displaystyle \frac{1}{2}\log 3}}\\ & ={\displaystyle \frac{1}{3}(0,3010)+{\displaystyle \frac{1}{2}(0,4771)}}\\ & =0,1003+0,2386\\ & =0,3389\end{array}\end{array}$

$\begin{array}{l}\\ 65.& \mathbf{\text{(UMPTN '98)}}\\ & \text{Nilai}{}^a\log {\displaystyle \frac{1}{b}\times {}^b\log {\displaystyle \frac{1}{c^2}\times {}^c\log {\displaystyle \frac{1}{a^3}=....}}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -6}\\ \text{b}.& {\displaystyle 6}\\ \text{c}.& {\displaystyle \frac{b}{a^{2}c}}\\ \text{d}.& {\displaystyle \frac{a^{2}c}{b}}\\ \text{e}.& {\displaystyle -\frac{1}{6}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {}^a\log {\displaystyle \frac{1}{b}\times {}^b\log {\displaystyle \frac{1}{c^2}\times {}^c\log {\displaystyle \frac{1}{a^3}}}}\\ & ={}^a\log {\displaystyle b^{-1}\times {}^b\log {\displaystyle c^{-2}\times {}^c\log {\displaystyle a^{-3}}}}\\ & =(-1).(-2).(-3)\times {}^a\log {\displaystyle a\times {}^b\log {\displaystyle c\times {}^c\log {\displaystyle a}}}\\ & =-6\times {}^a\log a\\ & =-6\times 1\\ & =-6\end{array}\end{array}$

$\begin{array}{l}\\ 66.& \mathbf{\text{(UMPTN '01)}}\\ & \text{Jika}{\displaystyle \frac{{}^2\log a}{{}^3\log b}=m\text{dan}{\displaystyle \frac{{}^3\log a}{{}^2\log b}=n}}\\ & \text{dengan}a>1,b>1,\text{maka}{\displaystyle \frac{m}{n}=....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle {}^2\log 3}\\ \text{b}.& {\displaystyle {}^3\log 2}\\ \text{c}.& {\displaystyle {}^4\log 9}\\ \text{d}.& {\displaystyle {({}^3\log 2)}^2}\\ \text{e}.& {\displaystyle {({}^2\log 3)}^2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{\displaystyle \frac{m}{n}}& ={\displaystyle \frac{{\displaystyle \frac{{}^2\log a}{{}^3\log b}}}{{\displaystyle \frac{{}^3\log a}{{}^2\log b}}}}\\ & ={\displaystyle \frac{{}^2\log a\times {}^2\log b}{{}^3\log b\times {}^3\log a}}\\ & ={\displaystyle \frac{{}^2\log a\times {\displaystyle \frac{1}{{}^b\log 2}}}{{}^3\log b\times {\displaystyle \frac{1}{{}^a\log 3}}}}\\ & ={\displaystyle \frac{{}^2\log a\times {}^a\log 3}{{}^3\log b\times {}^b\log 2}}\\ & ={\displaystyle \frac{{}^2\log 3}{{}^3\log 2}={\displaystyle \frac{{}^2\log 3}{{\displaystyle \frac{1}{{}^2\log 3}}}}}\\ & ={({}^2\log 3)}^2\end{array}\end{array}$.

$\begin{array}{l}\\ 67.& \mathbf{\text{(UMPTN '01)}}\\ & \text{Jika}{}^{10}\log x=b,\text{maka}{}^{10x}\log 100=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{b+1}}\\ \text{b}.& {\displaystyle \frac{2}{b+1}}\\ \text{c}.& {\displaystyle \frac{1}{b}}\\ \text{d}.& {\displaystyle \frac{2}{b}}\\ \text{e}.& {\displaystyle \frac{2}{10b}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& {}^{10x}\log 100\\ & ={\displaystyle \frac{\log 100}{\log 10x}}\\ & ={\displaystyle \frac{{}^{10}\log 100}{{}^{10}\log 10x},\text{pilih basis 10}}\\ & \text{alasannya: supaya sama dengan soal}\\ & ={\displaystyle \frac{{}^{10}\log {10}^2}{{}^{10}\log 10+{}^{10}\log x}}\\ & ={\displaystyle \frac{2}{1+b}\text{atau}}\\ & =\frac{2}{b+1}\end{array}\end{array}$.

$\begin{array}{l}\\ 68.& \mathbf{\text{(UM UGM '03)}}\\ & \text{Jika}{}^4\log 6=m+1,\text{maka}{}^9\log 8=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{3}{4m-2}}\\ \text{b}.& {\displaystyle \frac{3}{4m+2}}\\ \text{c}.& {\displaystyle \frac{3}{2m+4}}\\ \text{d}.& {\displaystyle \frac{3}{2m-4}}\\ \text{e}.& {\displaystyle \frac{3}{2m+2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Sebelumnya perhatikanlah}\\ & {}^4\log 6=m+1\\ & \iff {}^{2^2}\log (2.3{)}^1=m+1\\ & \iff {\displaystyle \frac{1}{2}\times {}^2\log (2.3)=m+1}\\ & \iff {\displaystyle \frac{1}{2}\times ({}^2\log 2+{}^2\log 3)=m+1}\\ & \iff {\displaystyle \frac{1}{2}\times (1+{}^2\log 3)=m+1}\\ & \iff 1+{}^2\log 3=2m+2\\ & \iff {}^2\log 3=2m+1\\ & \text{Selanjutnya adalah}:\\ & {}^9\log 8={\displaystyle \frac{1}{{}^8\log 9}}\\ & ={\displaystyle \frac{1}{{}^{2^3}\log 3^2}}\\ & ={\displaystyle \frac{1}{{\displaystyle \frac{2}{3}{}^2\log 3}}}\\ & ={\displaystyle \frac{3}{2{}^2\log 3}}\\ & ={\displaystyle \frac{3}{2(2m+1)}}\\ & ={\displaystyle \frac{3}{4m+2}}\end{array}\end{array}$.

$\begin{array}{l}\\ 69.& \mathbf{\text{(UMPTN '00)}}\\ & \text{Jika}{}^3\log 5=p\text{dan}{}^3\log 4=q,\\ & \text{maka}{}^4\log 15=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{pq}{1+p}}\\ \text{b}.& {\displaystyle \frac{p+q}{pq}}\\ \text{c}.& {\displaystyle \frac{p+1}{pq}}\\ \text{d}.& {\displaystyle \frac{p+1}{q+1}}\\ \text{e}.& {\displaystyle \frac{pq}{1-p}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {}^4\log 15\\ & ={\displaystyle \frac{{}^{...}\log 15}{{}^{...}\log 4},\text{pilih basis 5}}\\ & \text{mengapa tidak pilih basis selain 5}\\ & \text{lihat penyebut, di sana terdapat numerus 4}\\ & \text{pada soal, pasangan numerus 4 adalah 5},\\ & \text{makanya basis 5 dipilih, bukan yang lain}\\ & ={\displaystyle \frac{{}^5\log 15}{{}^5\log 4}={\displaystyle \frac{{}^5\log (3.5)}{{}^5\log 4}}}\\ & ={\displaystyle \frac{{}^5\log 3+{}^5\log 5}{{}^5\log 4}={\displaystyle \frac{{\displaystyle \frac{1}{{}^3\log 5}+{}^5\log 5}}{{}^5\log 4}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{p}+1}}{q}={\displaystyle \frac{1+p}{pq},\text{atau}}}\\ & ={\displaystyle \frac{p+1}{pq}}\end{array}\end{array}$.

$\begin{array}{l}\\ 70.& \mathbf{\text{(UMPTN '94)}}\\ & \text{Jika}{}^6\log 5=a\text{dan}{}^5\log 4=b,\\ & \text{maka}{}^4\log 0,24=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{a+2}{ab}}\\ \text{b}.& {\displaystyle \frac{2a+1}{ab}}\\ \text{c}.& {\displaystyle \frac{a-2}{ab}}\\ \text{d}.& {\displaystyle \frac{2a+1}{2ab}}\\ \text{e}.& {\displaystyle \frac{1-2a}{ab}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& {}^4\log 0,24\\ & ={\displaystyle \frac{{}^{...}\log 0,24}{{}^{...}\log 4}=\frac{{}^{...}\log {\displaystyle \frac{6}{25}}}{{}^{...}\log 4},\text{pilih basis 5}}\\ & \text{mengapa tidak pilih basis selain 5}\\ & \text{lihat penyebut, di sana terdapat numerus 4}\\ & \text{pada soal, pasangan numerus 4 adalah 5},\\ & \text{makanya basis 5 dipilih, bukan yang lain}\\ & =\frac{{}^5\log {\displaystyle \frac{6}{25}}}{{}^5\log 4}={\displaystyle \frac{{}^5\log 6-{}^5\log 25}{{}^5\log 4}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{{}^6\log 5}-{}^5\log 5^2}}{{}^5\log 4}={\displaystyle \frac{{\displaystyle \frac{1}{a}-2}}{b}=\frac{1-2a}{ab}}}\end{array}\end{array}$.