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

 $\begin{array}{l}\\ 21.& \text{Jika}f(x)=b^x,\text{di mana konstan positif},\\ \\ & {\displaystyle \frac{f(x^2+x)}{f(x+1)}=....}\\ & \begin{array}{l}\\ \text{a}.& f\left(x^2\right)& & & \\ \text{b}.& f(x+1)f(x-1)\\ \text{c}.& f(x+1)+f(x-1)\\ \text{d}.& f(x+1)-f(x-1)\\ \text{e}.& f(x^2-1)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{\displaystyle \frac{f(x^2+x)}{f(x+1)}}& =\frac{b^{x^2+x}}{b^{x+1}}\\ & =b^{x^2+x-(x+1)}\\ & =b^{x^2-1}\\ & =f(x^2-1)\end{array}\end{array}$.

$\begin{array}{l}\\ 22.& \text{Daerah hasil dari fungsi eksponen}\\ & y=x^{-\frac{2}{3}}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& y<0\\ \text{b}.& y>0\\ \text{c}.& y\ge 0\\ \text{d}.& y\le 0\\ \text{e}.& \text{Semua bilangan real}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \text{Perhatikanlah gambar berikut}\end{array}$

$.\begin{array}{rl}\text{diketahui}& \\ y& =x^{-{\displaystyle \frac{2}{3}}}\\ y^3& =x^{-2}\\ y^3& ={\displaystyle \frac{1}{x^2},\text{atau}}\\ y^3\times x^2& =1,\\ \text{sehingga}& y\text{tidak mungkin berharga}0\end{array}$.

$\begin{array}{l}\\ 23.& \text{Nilai}p-q^{p-q}\text{untuk}p=2\text{dan}q=-2\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -18\\ \text{b}.& -14\\ \text{c}.& 1\\ \text{d}.& 18\\ \text{e}.& 256\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}p-q^{p-q}& =(2-(-2){)}^{2-(-2)}\\ & =4^4\\ & =256\end{array}\end{array}$

$\begin{array}{l}\\ 24.& \text{Bentuk sederhana dari}\\ & {\displaystyle \frac{3^{x+8}{(7^{3x+1}\times 3^{4x-1})}^2}{(7^{2x}\times 3^{3x+2}{)}^3}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& 49\\ \text{b}.& 9\\ \text{c}.& 7\\ \text{d}.& 7^{2x+2}\\ \text{e}.& 3^{2x-1}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {\displaystyle \frac{3^{x+8}{(7^{3x+1}\times 3^{4x-1})}^2}{(7^{2x}\times 3^{3x+2}{)}^3}}\\ & ={\displaystyle \frac{3^{x+8+2(4x-1)}\times 7^{2(3x+1)}}{7^{3.2x}\times 3^{3(3x+2)}}}\\ & ={\displaystyle \frac{3^{x+8x+8-2}\times 7^{6x+2}}{3^{9x+6}\times 7^{6x}}}\\ & =3^0\times 7^2\\ & =1\times 49\\ & =49\end{array}\end{array}$

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

$\begin{array}{l}\\ 26.& (\mathbf{\text{UM UGM 05}})\text{Hasil dari}\\ & \sqrt{0,3+\sqrt{0,08}}=\sqrt{a}+\sqrt{b},\text{maka}{\displaystyle \frac{1}{a}+\frac{1}{b}=....}\\ & \begin{array}{l}\\ \text{a}.& 25\\ \text{b}.& 20\\ \text{c}.& 15\\ \text{d}.& 10\\ \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\sqrt{0,3+\sqrt{0,08}}& =\sqrt{0,2+0,1+\sqrt{4\times 0,2\times 0,1}}\\ & =\sqrt{0,2+0,1+2\sqrt{\times 0,2\times 0,1}}\\ & =\sqrt{0,2}+\sqrt{0,1}\\ \text{maka},a=0,2& ,b=0,1\\ \text{sehingga}{\displaystyle \frac{1}{a}+\frac{1}{b}}& ={\displaystyle \frac{1}{0,2}+\frac{1}{0,1}=5+10=15}\end{array}\end{array}$

$\begin{array}{l}\\ 27.& (\mathbf{\text{SPMB 06}})\text{Jika bilangan bulat}a\text{dan}b\text{memenuhi}\\ & {\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}=a+b\sqrt{30},\text{maka}ab=....}\\ & \begin{array}{l}\\ \text{a}.& -22\\ \text{b}.& -11\\ \text{c}.& -9\\ \text{d}.& 2\\ \text{e}.& 13\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}}& ={\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}\times {\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}-\sqrt{6}}}}\\ & ={\displaystyle \frac{5-2\sqrt{30}+6}{5-6}}\\ & ={\displaystyle \frac{11-2\sqrt{30}}{-1}}\\ & =-11+2\sqrt{30}\\ \text{sehingga}& a=-11,b=2,\text{maka}\\ ab& =(-11)\times 2\\ & =-22\end{array}\end{array}$

$\begin{array}{l}\\ 28.& (\mathbf{\text{OSK 2013}})\text{Misal}a\text{dan}b\text{bilangan asli}\\ & \text{dengan}a>b.\text{Jika}\sqrt{94+2\sqrt{2013}}=\sqrt{a}+\sqrt{b}\\ & \text{maka nilai}a-b\text{adalah... .}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\sqrt{94+2\sqrt{2013}}& =\sqrt{61+33+2\sqrt{61\times 33}}\\ & =\sqrt{61}+\sqrt{33}\\ & =\sqrt{a}+\sqrt{b}\\ \text{Sehingga}a& =61,b=33,\text{maka}\\ a-b& =61-33\\ & =28\end{array}\end{array}$.

$\begin{array}{l}\\ 29.& \text{Nilai dari}\\ & \sqrt{54+14\sqrt{5}}+\sqrt{12-2\sqrt{35}}+\sqrt{32-10\sqrt{7}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 10\\ \text{b}.& 11\\ \text{c}.& 12\\ \text{d}.& 5\sqrt{6}\\ \text{e}.& 6\sqrt{6}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{misal diketah}& \text{ui}\\ x& =\sqrt{54+14\sqrt{5}}+\sqrt{12-2\sqrt{35}}+\sqrt{32-10\sqrt{7}}\\ \text{untuk}& \\ \sqrt{54+14\sqrt{5}}& =\sqrt{49+5+2.7\sqrt{5}}=7+\sqrt{5}\\ \sqrt{12-2\sqrt{35}}& =\sqrt{7+5-2\sqrt{7.5}}=\sqrt{7}-\sqrt{5}\\ \sqrt{32-10\sqrt{7}}& =\sqrt{25+7-2.5\sqrt{7}}=5-\sqrt{7}+\\ & ---------------\\ & =7+5\\ & =12\end{array}\end{array}$.

$\begin{array}{l}\\ 30.& \text{Nilai eksak dari}\\ & {\displaystyle \frac{1}{{10}^{-2020}+1}+\frac{1}{{10}^{-2019}+1}+...+{\displaystyle \frac{1}{{10}^{2019}+1}+\frac{1}{{10}^{2020}+1}}}\\ & \begin{array}{l}\\ \text{a}.& 2020& & & \\ \text{b}.& 2020,5\\ \text{c}.& 2021\\ \text{d}.& 2021,5\\ \text{e}.& 2022\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Misal},x& =\frac{1}{{10}^{-2020}+1}+\frac{1}{{10}^{-2019}+1}+...+\frac{1}{{10}^{2019}+1}+\frac{1}{{10}^{2020}+1}\\ \text{maka},x& =\frac{1}{\frac{1}{{10}^{2020}}+1}+\frac{1}{\frac{1}{{10}^{2019}}+1}+...+\frac{1}{{10}^{2019}+1}+\frac{1}{{10}^{2020}+1}\\ & =\frac{{10}^{2020}}{1+{10}^{2020}}+\frac{{10}^{2019}}{1+{10}^{2019}}+...+\frac{1}{{10}^{2019}+1}+\frac{1}{{10}^{2020}+1}\\ & =\frac{{10}^{2020}}{1+{10}^{2020}}+\frac{1}{{10}^{2020}+1}+\frac{{10}^{2019}}{1+{10}^{2019}}+\frac{1}{{10}^{2019}+1}+...+\frac{1}{1^0+1}\\ & =\frac{{10}^{2020}+1}{{10}^{2020}+1}+\frac{{10}^{2019}+1}{{10}^{2019}+1}+\frac{{10}^{2018}+1}{{10}^{2018}+1}+...+\frac{{10}^1+1}{{10}^1+1}+\frac{1}{1^0+1}\\ & =\underset{\text{sebanyak}2020}{\underbrace{1+1+1+1+1+...+1}}+\frac{1}{2}\\ & =2020,5\end{array}\end{array}$