Latihan Soal 3 Persiapan PAS Gasal Matematika Peminatan Kelas X (Fungsi Eksponen dan Fungsi Logaritma)

 $\begin{array}{ll}\\ 21.&\textrm{Jika}\: \: f(x)=b^{x},\: \: \textrm{di mana konstan positif},\\\\ &\displaystyle \frac{f\left ( x^{2}+x \right )}{f(x+1)}= ....\\ &\begin{array}{lllllllll}\\ \textrm{a}.&f\left ( x^{2} \right )&&&\\ \textrm{b}.&f(x+1)f(x-1)\\ \textrm{c}.&f(x+1)+f(x-1)\\ \textrm{d}.&f(x+1)-f(x-1)\\ \color{red}\textrm{e}.&f\left ( x^{2}-1 \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}\displaystyle \frac{f\left ( x^{2}+x \right )}{f(x+1)}&=\frac{b^{x^{2}+x}}{b^{x+1}}\\ &=b^{x^{2}+x-(x+1)}\\ &=b^{x^{2}-1}\\ &=\color{red}f\left ( x^{2}-1 \right ) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 22.&\textrm{Daerah hasil dari fungsi eksponen}\\ &y\: =x^{- \frac{2}{3}}\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\\ \textrm{a}.&y< 0\\ \color{red}\textrm{b}.&y> 0\\ \textrm{c}.&y\geq 0\\ \textrm{d}.&y\leq 0\\ \textrm{e}.&\textrm{Semua bilangan real} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\textrm{Perhatikanlah gambar berikut} \end{array}$

$.\quad\: \, \begin{aligned}\textrm{diketahui}&\\ y\: &=x^{-\displaystyle \frac{2}{3}}\\ y^{3}\: &=x^{-2}\\ y^{3}\: &=\displaystyle \frac{1}{x^{2}},\: \textrm{atau}\\ y^{3}\times x^{2}\: &=1,\\ \textrm{sehingga}&\: \: y\: \: \textrm{tidak mungkin berharga}\: \: \color{red}0 \end{aligned}$.

$\begin{array}{ll}\\ 23.&\textrm{Nilai}\: \: p-q^{p-q}\: \: \textrm{untuk}\: \:p=2\: \: \textrm{dan}\: \: q=-2\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-18\\ \textrm{b}.&-14\\ \textrm{c}.&1\\ \textrm{d}.&18\\ \color{red}\textrm{e}.&256 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}p-q^{p-q}&=(2-(-2))^{2-(-2)}\\ &=4^{4}\\ &=\color{red}256 \end{aligned} \end{array}$

$\begin{array}{l}\\ 24.&\textrm{Bentuk sederhana dari}\\ &\displaystyle \frac{3^{x+8}\left (7^{3x+1}\times 3^{4x-1} \right )^{2}}{(7^{2x}\times 3^{3x+2})^{3}}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \color{red}\textrm{a}.&49\\ \textrm{b}.&9\\ \textrm{c}.&7\\ \textrm{d}.&7^{2x+2}\\ \textrm{e}.&3^{2x-1} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\displaystyle \frac{3^{x+8}\left (7^{3x+1}\times 3^{4x-1} \right )^{2}}{(7^{2x}\times 3^{3x+2})^{3}}\\ &=\displaystyle \frac{3^{x+8+2(4x-1)}\times 7^{2(3x+1)}}{7^{3.2x}\times 3^{3(3x+2)}}\\ &=\displaystyle \frac{3^{x+8x+8-2}\times 7^{6x+2}}{3^{9x+6}\times 7^{6x}}\\ &=3^{0}\times 7^{2}\\ &=1\times 49\\ &=\color{red}49 \end{aligned} \end{array}$

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

$\begin{array}{ll}\\ 26.&(\textbf{UM UGM 05})\textrm{Hasil dari}\\ &\sqrt{0,3+\sqrt{0,08}}=\sqrt{a}+\sqrt{b}\: ,\: \textrm{maka}\: \: \displaystyle \frac{1}{a}+\frac{1}{b}=....\\ &\begin{array}{llll}\\ \textrm{a}.&25\\ \textrm{b}.&20\\ \color{red}\textrm{c}.&15\\ \textrm{d}.&10\\ \textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\sqrt{0,3+\sqrt{0,08}}&=\sqrt{0,2+0,1+\sqrt{4\times 0,2\times 0,1}}\\ &=\sqrt{0,2+0,1+2\sqrt{\times 0,2\times 0,1}}\\ &=\sqrt{0,2}+\sqrt{0,1}\\ \textrm{maka},\: \: a=0,2&,\: \: b=0,1\\ \textrm{sehingga}\: \displaystyle \frac{1}{a}+\frac{1}{b}&=\displaystyle \frac{1}{0,2}+\frac{1}{0,1}=5+10=\color{red}15\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 27.&(\textbf{SPMB 06})\textrm{Jika bilangan bulat}\: \: a\: \: \: \textrm{dan}\: \: b\: \: \textrm{memenuhi}\\ &\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}=a+b\sqrt{30}\: ,\: \textrm{maka}\: \: ab=....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&-22\\ \textrm{b}.&-11\\ \textrm{c}.&-9\\ \textrm{d}.&2\\ \textrm{e}.&13 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}&=\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}\times \displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}-\sqrt{6}}\\ &=\displaystyle \frac{5-2\sqrt{30}+6}{5-6}\\ &=\displaystyle \frac{11-2\sqrt{30}}{-1}\\ &=-11+2\sqrt{30}\\ \textrm{sehingga}&\: \: \: a=-11,\: \: b=2,\: \: \textrm{maka}\\ ab&=(-11)\times 2\\ &=\color{red}-22 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 28.&(\textbf{OSK 2013})\textrm{Misal}\: \: a\: \: \textrm{dan}\: \: b\: \: \textrm{bilangan asli}\\ &\textrm{dengan}\: \: a>b.\: \: \textrm{Jika} \: \: \sqrt{94+2\sqrt{2013}}=\sqrt{a}+\sqrt{b}\\ &\textrm{maka nilai} \: \: a-b\: \: \textrm{adalah... .}\\\\ &\textrm{Jawab}:\\ &\begin{aligned} \sqrt{94+2\sqrt{2013}}&=\sqrt{61+33+2\sqrt{61\times 33}}\\ &=\sqrt{61}+\sqrt{33}\\ &=\sqrt{a}+\sqrt{b}\\ \textrm{Sehingga}\: \: a&=61,\: \: b=33,\: \: \textrm{maka}\\ a-b&=61-33\\ &=\color{red}28 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 29.&\textrm{Nilai dari}\\ &\sqrt{54+14\sqrt{5}}+\sqrt{12-2\sqrt{35}}+\sqrt{32-10\sqrt{7}}\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&10\\ \textrm{b}.&11\\ \color{red}\textrm{c}.&12\\ \textrm{d}.&5\sqrt{6}\\ \textrm{e}.&6\sqrt{6} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\textrm{misal diketah}&\textrm{ui}\\ x&=\sqrt{54+14\sqrt{5}}+\sqrt{12-2\sqrt{35}}+\sqrt{32-10\sqrt{7}}\\ \textrm{untuk}&\\ \sqrt{54+14\sqrt{5}}&=\sqrt{49+5+2.7\sqrt{5}}=7+\sqrt{5}\\ \sqrt{12-2\sqrt{35}}&=\sqrt{7+5-2\sqrt{7.5}}=\sqrt{7}-\sqrt{5}\\ \sqrt{32-10\sqrt{7}}&=\sqrt{25+7-2.5\sqrt{7}}=5-\sqrt{7}\qquad +\\ &---------------\\ &\qquad\qquad\qquad\quad\qquad=7+5\\ &\qquad\qquad\qquad\quad\qquad=\color{red}12 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 30.&\textrm{Nilai eksak dari}\\ &\displaystyle \frac{1}{10^{-2020}+1}+\frac{1}{10^{-2019}+1}+...+\displaystyle \frac{1}{10^{2019}+1}+\frac{1}{10^{2020}+1} \\ &\begin{array}{lllllllll}\\ \textrm{a}.&2020&&&\\ \color{red}\textrm{b}.&2020,5\\ \textrm{c}.&2021\\ \textrm{d}.&2021,5\\ \textrm{e}.&2022 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\textrm{Misal},\: \: x&=\frac{1}{10^{-2020}+1}+\frac{1}{10^{-2019}+1}+...+\frac{1}{10^{2019}+1}+\frac{1}{10^{2020}+1}\\ \textrm{maka},\, \: x&=\frac{1}{\frac{1}{10^{2020}}+1}+\frac{1}{\frac{1}{10^{2019}}+1}+...+\frac{1}{10^{2019}+1}+\frac{1}{10^{2020}+1}\\ &=\frac{10^{2020}}{1+10^{2020}}+\frac{10^{2019}}{1+10^{2019}}+...+\frac{1}{10^{2019}+1}+\frac{1}{10^{2020}+1}\\ &=\frac{10^{2020}}{1+10^{2020}}+\frac{1}{10^{2020}+1}+\frac{10^{2019}}{1+10^{2019}}+\frac{1}{10^{2019}+1}+...+\frac{1}{1^{0}+1}\\ &=\frac{10^{2020}+1}{10^{2020}+1}+\frac{10^{2019}+1}{10^{2019}+1}+\frac{10^{2018}+1}{10^{2018}+1}+...+\frac{10^{1}+1}{10^{1}+1}+\frac{1}{1^{0}+1}\\ &=\underset{\textrm{sebanyak}\: 2020}{\underbrace{1+1+1+1+1+...+1}}+\frac{1}{2}\\ &=\color{red}2020,5 \end{aligned} \end{array}$