Latihan Soal 1 Persiapan PAS Gasal Matematika Peminatan Kelas X (Fungsi Eksponen dan Fungsi Logaritma)

 $\begin{array}{ll}\\ 1.&\textrm{Hasil dari}\: \: \displaystyle \frac{3^{4}.5^{3}.7^{2}}{27.125.49} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \textrm{b}.&3\\ \textrm{c}.&4\\ \textrm{d}.&9\\ \color{red}\textrm{e}.&16 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}\displaystyle \frac{3^{4}.5^{3}.7^{2}}{27.125.49}&=\displaystyle \frac{3^{4}.5^{3}.7^{2}}{3^{3}.5^{3}.7^{2}}\\ &=3^{4-3}.5^{3-3}.7^{2-2}\\ &=3^{1}.5^{0}.7^{0}\\ &=3.1.1\\ &=\color{red}3 \end{aligned} \end{array}$

$\begin{array}{l}\\ 2.&\textrm{Bentuk sederhana dari}\: \: \displaystyle \frac{5a^{4}b^{2}}{a^{2}b^{-3}} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle ab^{2}\\\\ \textrm{b}.&\displaystyle a^{2}b^{2}\\\\ \color{red}\textrm{c}.&\displaystyle 5a^{2}b^{5}\\\\ \textrm{d}.&5a^{4}b^{5}\\\\ \textrm{e}.&5a^{5}a^{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\displaystyle \frac{5a^{4}b^{2}}{a^{2}b^{-3}}&=5a^{4-2}b^{2-(-3)}\\ &=\color{red}5a^{2}b^{5} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Bentuk sederhana dari}\: \: \left ( \displaystyle \frac{ab^{2}}{a^{2}b^{3}} \right )^{4} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{b}\\\\ \textrm{b}.&\displaystyle \frac{1}{ab}\\\\ \textrm{c}.&\displaystyle \frac{1}{a^{2}b^{2}}\\\\ \textrm{d}.&\displaystyle \frac{1}{ab^{2}}\\\\ \color{red}\textrm{e}.&\displaystyle \frac{1}{a^{4}b^{4}} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}\left ( \displaystyle \frac{ab^{2}}{a^{2}b^{3}} \right )^{4}&=\left ( a^{1-2}b^{2-3} \right )^{4}\\ &=\left ( a^{-1}b^{-1} \right )^{4}\\ &=\left ( \displaystyle \frac{1}{ab} \right )^{4}\\ &=\color{red}\displaystyle \frac{1}{a^{4}b^{4}} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Bentuk sederhana dari}\: \: \displaystyle \frac{3^{\frac{5}{6}}\times 12^{\frac{7}{12}}}{2^{-\frac{1}{4}}\times 6^{\frac{2}{3}} }\: \: \textrm{adalah} ...\: .\\ &\begin{array}{lll}\\ \textrm{a}.\quad 6^{\frac{1}{4}}&&\textrm{d}.\quad \left ( \displaystyle \frac{2}{3} \right )^{\frac{3}{4}}\\ \color{red}\textrm{b}.\quad 6^{\frac{3}{4}}&\textrm{c}.\quad 6^{\frac{2}{3}}&\textrm{e}.\quad \left ( \displaystyle \frac{3}{2} \right )^{\frac{3}{4}}\end{array}\\\\ &\textrm{Jawab}:\qquad \color{red}\textbf{b}\\ &\begin{aligned}\displaystyle \frac{3^{\frac{5}{6}}\times 12^{\frac{7}{12}}}{6^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}&=\displaystyle \frac{3^{\frac{5}{6}}\times (3\times 4)^{\frac{7}{12}}}{(2\times 3)^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}\\ &=\displaystyle \frac{3^{\frac{5}{6}}\times 3^{\frac{7}{12}}\times 4^{\frac{7}{12}}}{2^{\frac{2}{3}}\times 3^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}\\ &=\displaystyle \frac{3^{\frac{5}{6}}\times 3^{\frac{7}{12}}\times \left ( 2^{2} \right )^{\frac{7}{12}}}{2^{\frac{2}{3}}\times 3^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}\\ &=\displaystyle \frac{3^{\frac{5}{6}}\times 3^{\frac{7}{12}}\times 2^{\frac{14}{12}}}{2^{\frac{2}{3}}\times 3^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}\\ &=2^{(\frac{14}{12}-\frac{2}{3}+\frac{1}{4})}\times 3^{(\frac{5}{6}+\frac{7}{12}-\frac{2}{3})}\\ &=2^{(\frac{14-8+3}{12})}\times 3^{(\frac{10+7-8}{12})}\\ &=2^{\frac{9}{12}}\times 3^{\frac{9}{12}}\\ &=(2\times 3)^{\frac{9}{12}}\\ &=6^{\frac{9}{12}}\\ &=\color{red}6^{\frac{3}{4}} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Bentuk sederhana dari}\: \: \left ( \displaystyle \frac{a^{\frac{1}{2}}b^{-3}}{a^{-1}b^{-\frac{3}{2}}} \right )^{\frac{3}{2}}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{a}{b}\\\\ \textrm{b}.&\displaystyle \frac{b}{a}\\\\ \textrm{c}.&\displaystyle ab\\\\ \textrm{d}.&\sqrt[a]{b}\\\\ \color{red}\textrm{e}.&\sqrt[4]{\left ( \displaystyle \frac{a}{b} \right )^{9}} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}\left ( \displaystyle \frac{a^{\frac{1}{2}}b^{-3}}{a^{-1}b^{-\frac{3}{2}}} \right )^{\frac{3}{2}}&=\displaystyle \frac{a^{\frac{3}{4}}b^{\frac{-9}{2}}}{a^{-\frac{3}{2}}b^{-\frac{9}{4}}}\\ &=a^{\frac{3}{4}+\frac{3}{2}}b^{\frac{-9}{2}+\frac{9}{4}}\\ &=a^{\frac{3+6}{4}}b^{\frac{-18+9}{4}}\\ &=a^{\frac{9}{4}}b^{-\frac{9}{4}}\\ &=\left ( \displaystyle \frac{a}{b} \right )^{\frac{9}{4}}\\ &=\color{red}\sqrt[4]{\left ( \displaystyle \frac{a}{b} \right )^{9}} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\displaystyle 5^{2}\left ( \left ( \displaystyle \frac{1}{25} \right )^{2x+6} \right )^{\frac{1}{6}}=\displaystyle \frac{1}{25} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&-5\\ \textrm{b}.&-4\\ \textrm{c}.&-3\\ \textrm{d}.&1\\ \textrm{e}.&3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\displaystyle 5^{2}\left ( \left ( \displaystyle \frac{1}{25} \right )^{2x+6} \right )^{\frac{1}{6}}&=\displaystyle \frac{1}{25}\\ 25.\displaystyle 5^{2}\left ( \left ( \displaystyle \frac{1}{25} \right )^{2x+6} \right )^{\frac{1}{6}}&=1\\ 5^{2}.5^{2}.5^{-\frac{4x+12}{6}}&=5^{0}\\ 5^{2+2-\frac{2}{3}x-2}&=5^{0}\\ \displaystyle 2+2-\frac{2}{3}x-2&=0\\ 2-\displaystyle \frac{2}{3}x&=0\\ -\displaystyle \frac{2}{3}x&=-2\\ x&=\color{red}3 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&(\textbf{UM-UGM 03})\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\left ( \displaystyle \frac{1}{25} \right )^{x-\frac{5}{2}}=\sqrt{\displaystyle \frac{625}{5^{2-x}}} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{3}{5}\\ \color{red}\textrm{b}.&\displaystyle \frac{8}{5}\\ \textrm{c}.&2\\ \textrm{d}.&3\\ \textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\left ( \displaystyle \frac{1}{25} \right )^{x-\frac{5}{2}}&=\sqrt{\displaystyle \frac{625}{5^{2-x}}}\\ 5^{-2x+5}&=5^{\frac{1}{2}(4-(2-x))}\\ -2x+5&=\displaystyle \frac{1}{2}(4-2+x)\\ -4x+10&=2+x\\ -5x&=2-10\\ x&=\displaystyle \frac{-8}{-5}\\ &=\color{red}\frac{8}{5} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &2^{\frac{x}{3}-1}=16 \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&5\\ \textrm{b}.&10\\ \color{red}\textrm{c}.&15\\ \textrm{d}.&20\\ \textrm{e}.&25 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}2^{\frac{x}{3}-1}&=16\\ 2^{\frac{x}{3}-1}&=2^{4}\\ \displaystyle \frac{x}{3}-1&=4\\ \displaystyle \frac{x}{3}&=5\\ x&=\color{red}15 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 9.&(\textbf{SPMB 05})\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\displaystyle \frac{\sqrt[3]{(0,08)^{7-2x}}}{(0,2)^{-4x+5}}=1 \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-3\\ \textrm{b}.&-2\\ \color{red}\textrm{c}.&-1\\ \textrm{d}.&0\\ \textrm{e}.&1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\displaystyle \frac{\sqrt[3]{(0,08)^{7-2x}}}{(0,2)^{-4x+5}}&=1\\ \sqrt[3]{(0,08)^{7-2x}}&=(0,2)^{-4x+5}\\ (0,2)^{\frac{3(7-2x)}{3}}&=(0,2)^{-4x+5}\\ \displaystyle 7-2x&=-4x+5\\ 4x-2x&=5-7\\ 2x&=-2\\ x&=\color{red}-1 \end{aligned} \end{array}$

$\begin{array}{l}\\ 10.&(\textbf{SPMB 05})\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\displaystyle \frac{\sqrt[3]{\displaystyle \frac{1}{9^{2-x}}}}{27}=3^{x+1} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&-16\\ \textrm{b}.&-7\\ \textrm{c}.&4\\ \textrm{d}.&5\\ \textrm{e}.&6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\displaystyle \frac{\sqrt[3]{\displaystyle \frac{1}{9^{2-x}}}}{27}&=3^{x+1}\\ \sqrt[3]{\displaystyle \frac{1}{9^{2-x}}}&=27\times 3^{x+1}\\ \sqrt[3]{3^{-2(2-x)}}&=3^{3}.3^{x+1}\\ 3^{\frac{-4+2x}{3}}&=3^{3+(x+1)}\\ \displaystyle \frac{-4+2x}{3}&=4+x\\ -4+2x&=12+3x\\ 2x-3x&=12+4\\ -x&=16\\ x&=\color{red}-16 \end{aligned} \end{array}$