Contoh Soal 5 Fungsi Eksponen (Matematika Peminatan Kelas X)

 $\begin{array}{ll}\\ 21.&\textrm{Nilai}\: \: p-q^{p-q}\: \: \textrm{untuk}\: \:p=2\: \: \textrm{dan}\: \: q=-2\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-18\\ \textrm{b}.&-14\\ \textrm{c}.&1\\ \textrm{d}.&18\\ \color{red}\textrm{e}.&256 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}p-q^{p-q}&=(2-(-2))^{2-(-2)}\\ &=4^{4}\\ &=256 \end{aligned} \end{array}$

$\begin{array}{l}\\ 22.&\textrm{Bentuk sederhana dari}\: \: \displaystyle \frac{3^{x+8}\left (7^{3x+1}\times 3^{4x-1} \right )^{2}}{(7^{2x}\times 3^{3x+2})^{3}}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \color{red}\textrm{a}.&49\\ \textrm{b}.&9\\ \textrm{c}.&7\\ \textrm{d}.&7^{2x+2}\\ \textrm{e}.&3^{2x-1} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\displaystyle \frac{3^{x+8}\left (7^{3x+1}\times 3^{4x-1} \right )^{2}}{(7^{2x}\times 3^{3x+2})^{3}}&=\displaystyle \frac{3^{x+8+2(4x-1)}\times 7^{2(3x+1)}}{7^{3.2x}\times 3^{3(3x+2)}}\\ &=\displaystyle \frac{3^{x+8x+8-2}\times 7^{6x+2}}{3^{9x+6}\times 7^{6x}}\\ &=3^{0}\times 7^{2}\\ &=1\times 49\\ &=49 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 23.&\textrm{Jika}\: \: \displaystyle \frac{(0,24)^{3}(0,243)^{5}}{(3,6)^{7}}=\frac{2^{p}3^{q}}{5^{r}},\\ & \textrm{maka nilai}\: \: p+q+r\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&7\\ \textrm{b}.&8\\ \color{red}\textrm{c}.&9\\ \textrm{d}.&10\\ \textrm{e}.&11 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\displaystyle \frac{(0,24)^{3}(0,243)^{5}}{(3,6)^{7}}&=\displaystyle \frac{\left (\frac{24}{100} \right )^{3}\times \left ( \frac{243}{1000} \right )^{5}}{\left (\frac{36}{10} \right )^{7}}\\ &=\displaystyle \frac{(8\times 3)^{3}\times \left (3^{5} \right )^{5}}{(2^{2}\times 3^{2})^{7}}\times \displaystyle \frac{10^{7}}{100^{3}\times 1000^{5}}\\ &=\displaystyle \frac{(2^{3}\times 3)^{3}\times 3^{25}}{(2^{14}\times 3^{14})}\times \displaystyle \frac{10^{7}}{\left ( 10^{2} \right )^{3}\times \left ( 10^{3} \right )^{5}}\\ &=2^{9-14}.3^{3+25-14}.10^{7-6-15}\\ &=2^{-5}.3^{14}.10^{-14}=2^{-5}.3^{14}.(2.5)^{-14}\\ &=2^{-5-14}.3^{21}.5^{-14}\\ &=\displaystyle \frac{2^{-19}.3^{14}}{5^{14}}\\ \textrm{Sehingga}\: \: &p+q+r=-19+14+14=9 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 24.&\textrm{Nilai eksak dari}\\ &\displaystyle \frac{1}{10^{-2020}+1}+\frac{1}{10^{-2019}+1}+...+\displaystyle \frac{1}{10^{2019}+1}+\frac{1}{10^{2020}+1} \\ &\begin{array}{lllllllll}\\ \textrm{a}.&2020&&&\\ \color{red}\textrm{b}.&2020,5\\ \textrm{c}.&2021\\ \textrm{d}.&2021,5\\ \textrm{e}.&2022 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\textrm{Misal},\: \: x&=\frac{1}{10^{-2020}+1}+\frac{1}{10^{-2019}+1}+...+\frac{1}{10^{2019}+1}+\frac{1}{10^{2020}+1}\\ \textrm{maka},\, \: x&=\frac{1}{\frac{1}{10^{2020}}+1}+\frac{1}{\frac{1}{10^{2019}}+1}+...+\frac{1}{10^{2019}+1}+\frac{1}{10^{2020}+1}\\ &=\frac{10^{2020}}{1+10^{2020}}+\frac{10^{2019}}{1+10^{2019}}+...+\frac{1}{10^{2019}+1}+\frac{1}{10^{2020}+1}\\ &=\frac{10^{2020}}{1+10^{2020}}+\frac{1}{10^{2020}+1}+\frac{10^{2019}}{1+10^{2019}}+\frac{1}{10^{2019}+1}+...+\frac{1}{1^{0}+1}\\ &=\frac{10^{2020}+1}{10^{2020}+1}+\frac{10^{2019}+1}{10^{2019}+1}+\frac{10^{2018}+1}{10^{2018}+1}+...+\frac{10^{1}+1}{10^{1}+1}+\frac{1}{1^{0}+1}\\ &=\underset{\textrm{sebanyak}\: 2020}{\underbrace{1+1+1+1+1+...+1}}+\frac{1}{2}\\ &=2020,5 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 25.&\textrm{Nilai dari}\\ &\sqrt{54+14\sqrt{5}}+\sqrt{12-2\sqrt{35}}+\sqrt{32-10\sqrt{7}}\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&10\\ \textrm{b}.&11\\ \color{red}\textrm{c}.&12\\ \textrm{d}.&5\sqrt{6}\\ \textrm{e}.&6\sqrt{6} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\textrm{misal diketah}&\textrm{ui}\\ x&=\sqrt{54+14\sqrt{5}}+\sqrt{12-2\sqrt{35}}+\sqrt{32-10\sqrt{7}}\\ \textrm{untuk}&\\ \sqrt{54+14\sqrt{5}}&=\sqrt{49+5+2.7\sqrt{5}}=7+\sqrt{5}\\ \sqrt{12-2\sqrt{35}}&=\sqrt{7+5-2\sqrt{7.5}}=\sqrt{7}-\sqrt{5}\\ \sqrt{32-10\sqrt{7}}&=\sqrt{25+7-2.5\sqrt{7}}=5-\sqrt{7}\qquad +\\ &---------------\\ &\qquad\qquad\qquad\quad\qquad=7+5\\ &\qquad\qquad\qquad\quad\qquad=12 \end{aligned} \end{array}$