Lanjutan 1 Contoh Soal dan Pembahasan Persiapan PHB Gasal Materi Fungsi Eksponensial (Kelas X)

$\begin{array}{ll}\\ 6.&\textrm{Bentuk sederhana dari}\: \: \: \\ &\left (\displaystyle \frac{1}{2}  \right )^{-\left ( \frac{1}{2} \right )^{-1}}+\left (\displaystyle \frac{1}{3}  \right )^{-\left ( \frac{1}{3} \right )^{-1}}+\left (\displaystyle \frac{1}{4}  \right )^{-\left ( \frac{1}{4} \right )^{-1}}+\left (\displaystyle \frac{1}{5}  \right )^{-\left ( \frac{1}{5} \right )^{-1}}\\ &\textrm{adalah ... .}\\ &\begin{array}{llll}\\ \textrm{a}.&3142\\ \textrm{b}.&287\\ \color{red}\textrm{c}.&3412\\ \textrm{d}.&4116\\ \textrm{e}.&4096 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&\left (\displaystyle \frac{1}{2}  \right )^{-\left ( \frac{1}{2} \right )^{-1}}+\left (\displaystyle \frac{1}{3}  \right )^{-\left ( \frac{1}{3} \right )^{-1}}+\left (\displaystyle \frac{1}{4}  \right )^{-\left ( \frac{1}{4} \right )^{-1}}+\left (\displaystyle \frac{1}{5}  \right )^{-\left ( \frac{1}{5} \right )^{-1}}\\ &=\left (\displaystyle \frac{1}{2}  \right )^{-2}+\left (\displaystyle \frac{1}{3}  \right )^{-3}+\left (\displaystyle \frac{1}{4}  \right )^{-4}+\left (\displaystyle \frac{1}{5}  \right )^{-5}\\ &=2^{2}+3^{3}+4^{4}+5^{5}\\ &=4+27+256+3125\\ &=\color{red}3412 \end{aligned}\\  \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Nilai dari}\\ &\underset{89}{\underbrace{(-7)(-7)(-7)\cdots (-7)}}-(-7)^{89}\\ &\textrm{adalah ... .}\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \textrm{b}.&7\\ \textrm{c}.&4\\ \color{red}\textrm{d}.&0\\ \textrm{e}.&9 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\underset{89}{\underbrace{(-7)(-7)(-7)\cdots (-7)}}-(-7)^{89}\\ &=(-7)^{89}-(-7)^{89}\\ &=\color{red}0 \end{aligned}\\  \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Nilai dari}\: \:  \left ( \sqrt{3\times \sqrt[4]{27\times \sqrt[3]{81}}} \right )^{\displaystyle \frac{24}{25}}\\  &\textrm{adalah ... .}\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \textrm{b}.&2\\ \color{red}\textrm{c}.&3\\ \textrm{d}.&4\\ \textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}& \left ( \sqrt{3\times \sqrt[4]{27\times \sqrt[3]{81}}} \right )^{\displaystyle \frac{24}{25}}\\ &=\left ( 3^{\frac{1}{2}}.(3^{3})^{\frac{1}{2.4}}.(3^{4})^{\frac{1}{2.4.3}} \right )^{\displaystyle \frac{24}{25}}\\ &=\left ( 3^{\frac{1}{2}+\frac{3}{8}+\frac{4}{24}} \right )^{\displaystyle \frac{24}{25}}\\ &=3^{\frac{12}{25}+\frac{9}{25}+\frac{4}{25}}\\ &=3^{\frac{25}{25}}\\ &=\color{red}3 \end{aligned}\\  \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Jika}\: \: \sqrt{27^{2x+3}}=\displaystyle \frac{1}{3^{x-2}.9^{3x}}\: ,\: \: \textrm{maka nilai}\\ &8x+2\: \: \textrm{adalah ... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&0\\ \textrm{b}.&2\\ \textrm{c}.&4\\ \textrm{d}.&8\\ \textrm{e}.&12 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\textbf{Alternatif 1}\\ &\color{blue}\begin{aligned}\sqrt{27^{2x+3}}&=\displaystyle \frac{1}{3^{x-2}.9^{3x}}\\ 27^{\displaystyle \frac{2x+3}{2}}&=3^{-(x-2)}.9^{-3x}\\ 3^{3.\left ( \displaystyle \frac{2x+3}{2} \right )}&=3^{2-x}.3^{2(-3x)}\\ 3^{3.\left ( \displaystyle \frac{2x+3}{2} \right )}&=3^{2-x-6x}\\ \color{red}3\color{black}^{3.\left ( \displaystyle \frac{2x+3}{2} \right )}&=\color{red}3\color{black}^{2-7x}\\ \color{black}3\frac{(2x+3)}{2}&=\color{black}2-7x\\ 6x+9&=4-14x\\ 6x+14x&=4-9\\ 20x&=-5\\ &x=-\displaystyle \frac{1}{4}\\ \textrm{maka nilai}&\\ 8x+2&=8\left ( -\displaystyle \frac{1}{4} \right )+2=-2+2=\color{red}0  \end{aligned}\\ &\textbf{Alternatif 2}\\ &\color{blue}\begin{aligned}\sqrt{27^{2x+3}}&=\displaystyle \frac{1}{3^{x-2}.9^{3x}}\\ 27^{\displaystyle \frac{2x+3}{2}}.3^{x-2}.9^{3x}&=1\\ 3^{3\left ( \displaystyle \frac{2x+3}{2} \right )}.3^{x-2}.3^{2(3x)}&=3^{0}\\ 3^{3\left ( \displaystyle \frac{2x+3}{2} \right )+x-2+6x}&=3^{0}\\ \displaystyle \frac{3(2x+3)}{2}+7x-2&=0\\ 6x+9+14x-4&=0\\ 20x+5&=0\\ 20x&=-5\\ x&=-\displaystyle \frac{1}{4}\\ \textrm{Selanjutnya sama}&\\ \textrm{dengan langkah no.1}&\: \: \textrm{di atas} \end{aligned}\\  \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Penyelesaian persamaan}\: \: 3^{2x+1}=81^{x-2}\: \\ &\textrm{adalah ... .}\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&2\\ \textrm{c}.&4\\ \color{red}\textrm{d}.&4\displaystyle \frac{1}{2}\\ \textrm{e}.&16 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}3^{2x+1}&=81^{x-2}\\ 3^{2x+1}&=(3^{4})^{x-2}\\ 3^{2x+1}&=3^{4x-8}\\ \color{black}2x+1&=\color{black}4x-8\\ \color{black}2x-4x&=\color{black}-8-1\\ -2x&=-9\\ x&=\displaystyle \frac{-9}{-2}\\ &=\color{red}4\displaystyle \frac{1}{2} \end{aligned}\\  \end{array}$.

Contoh Soal dan Pembahasan Persiapan PHB Gasal Materi Fungsi Eksponensial (Kelas X)

 $\begin{array}{ll}\\ 1.&\textrm{Bentuk sederhana dari}\: \: \displaystyle \frac{a^{p}.a^{q}}{a^{r}}:a^{2r} \: \: \textrm{adalah ... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&a^{p+q-3r}\\ \textrm{b}.&a^{p+3r-q}\\ \textrm{c}.&a^{p-2q+r}\\ \textrm{d}.&a^{p+q+r}\\ \textrm{e}.&a^{p-3q-r} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\displaystyle \frac{a^{p}.a^{q}}{a^{r}}:a^{2r}&=\displaystyle \frac{a^{p}.a^{q}}{a^{r}.a^{2r}}\\ &=\displaystyle \frac{a^{p+q}}{a^{r+2r}}\\ &=\displaystyle \frac{a^{p+q}}{a^{3r}}\\ &=\color{red}a^{p+q-3r} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Jika}\: \: x=\sqrt[3]{5+\sqrt[3]{8}}\: ,\: \: \textrm{maka nilai}\\ &x^{3}\: \textrm{adalah ... .}\\ &\begin{array}{llll}\\ \textrm{a}.&3\\ \textrm{b}.&4\\ \textrm{c}.&5\\ \textrm{d}.&6\\ \color{red}\textrm{e}.&7 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}x&=\sqrt[3]{5+\sqrt[3]{8}}\quad \color{black}\textrm{dipangkatkan 3}\\ &\qquad \color{black}\textrm{masing-masing ruas}\\ x^{3}&=\left ( \sqrt[3]{5+\sqrt[3]{8}} \right )^{3}\\ &=5+\sqrt[3]{8}\\ &=5+\sqrt[3]{2^{3}}=5+2=\color{red}7  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Jika}\: \: 4^{a}\times 4^{b}=64\: \: \textrm{dan}\: \:  \displaystyle \frac{4^{a}}{4^{b}}=16\\ &\textrm{maka nilai dari}\: \: a:b\: \: \textrm{adalah ... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&5\\ \textrm{b}.&\displaystyle \frac{5}{4}\\ \textrm{c}.&\displaystyle \frac{1}{3}\\ \textrm{d}.&3\\ \textrm{e}.&\displaystyle \frac{3}{4} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui}\\ &\bullet \quad 4^{a}\times 4^{b}=64\color{black}\Leftrightarrow 4^{a+b}=4^{3}\Leftrightarrow a+b=3\: \: ....(1)\\ &\bullet \quad \displaystyle \frac{4^{a}}{4^{b}}=16\color{black}\Leftrightarrow 4^{a-b}=4^{2}\Leftrightarrow a-b=2\: \: ...........(2) \\ &\color{black}\textrm{Dari persamaan}\: \: (1)\: \:  \&\: \:  (2)\: \: \textrm{akan didapatkan}\\ &\begin{array}{|l|l|}\hline \begin{array}{llllll} a+b&=&3\\ a-b&=&2&+\\\hline 2a&=&5\\ \: \: a&=&\displaystyle \frac{5}{2} \end{array}&\begin{array}{rlllll} a+b&=&3\\ a-b&=&2&-\\\hline 2b&=&1\\ \: \: b&=&\displaystyle \frac{1}{2} \end{array}\\\hline  \end{array}\\ &\textrm{Sehingga nilai}\: \: a:b=\displaystyle \frac{a}{b}=\displaystyle \frac{\displaystyle \frac{5}{2}}{\displaystyle \frac{1}{2}}=\color{red}5 \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 4.&(\textbf{SPMB 2003})\\ &\textrm{Jika}\: \: a\neq 0,\: \textrm{maka nilai}\: \:  \displaystyle \frac{(-2a)^{3}(2a)^{-\frac{2}{3}}}{\left ( 16a^{4} \right )^{\frac{1}{3}}}\: \: \textrm{adalah ... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-2^{2}a\\ \color{red}\textrm{b}.&\displaystyle -2a\\ \textrm{c}.&\displaystyle 2a^{2}\\ \textrm{d}.&2^{2}a\\ \textrm{e}.&\displaystyle -2a^{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\displaystyle \frac{(-2a)^{3}(2a)^{-\frac{2}{3}}}{\left ( 16a^{4} \right )^{\frac{1}{3}}}&=\displaystyle \frac{-8a^{3}}{(2a)^{\frac{2}{3}}.\left ( 16a^{4} \right )^{\frac{1}{3}}}\\ &=-\displaystyle \frac{(2a)^{3}}{(2a)^{\frac{2}{3}}.(2a)^{\frac{4}{3}}}\\ &=-\displaystyle \frac{(2a)^{3}}{(2a)^{\frac{2}{3}+\frac{4}{3}}}\\ &=-\displaystyle \frac{(2a)^{3}}{(2a)^{2}}\\ &=-(2a)^{3-2}\\ &=\color{red}-2a \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Bentuk sederhana dari}\: \: \displaystyle \frac{4^{n+3}-4^{n+1}}{4(4^{n-1})}\: \\ &\textrm{adalah ... .}\\ &\begin{array}{llll}\\ \textrm{a}.&64\\ \color{red}\textrm{b}.&60\\ \textrm{c}.&18\\ \textrm{d}.&16\\ \textrm{e}.&15 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned} \displaystyle \frac{4^{n+3}-4^{n+1}}{4(4^{n-1})}&=\displaystyle \frac{4^{n}.4^{3}-4^{n}.4}{4.\displaystyle \frac{4^{n}}{4}}\\ &=\displaystyle \frac{4^{n}(64-4)}{4^{n}}=\color{red}60\\ \end{aligned}\\  \end{array}$.