Contoh Soal 5 Fungsi Eksponen (Matematika Peminatan Kelas X)

 $\begin{array}{l}\\ 21.& \text{Nilai}p-q^{p-q}\text{untuk}p=2\text{dan}q=-2\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -18\\ \text{b}.& -14\\ \text{c}.& 1\\ \text{d}.& 18\\ \text{e}.& 256\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}p-q^{p-q}& =(2-(-2){)}^{2-(-2)}\\ & =4^4\\ & =256\end{array}\end{array}$

$\begin{array}{l}\\ 22.& \text{Bentuk sederhana dari}{\displaystyle \frac{3^{x+8}{(7^{3x+1}\times 3^{4x-1})}^2}{(7^{2x}\times 3^{3x+2}{)}^3}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& 49\\ \text{b}.& 9\\ \text{c}.& 7\\ \text{d}.& 7^{2x+2}\\ \text{e}.& 3^{2x-1}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\displaystyle \frac{3^{x+8}{(7^{3x+1}\times 3^{4x-1})}^2}{(7^{2x}\times 3^{3x+2}{)}^3}}& ={\displaystyle \frac{3^{x+8+2(4x-1)}\times 7^{2(3x+1)}}{7^{3.2x}\times 3^{3(3x+2)}}}\\ & ={\displaystyle \frac{3^{x+8x+8-2}\times 7^{6x+2}}{3^{9x+6}\times 7^{6x}}}\\ & =3^0\times 7^2\\ & =1\times 49\\ & =49\end{array}\end{array}$

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

$\begin{array}{l}\\ 24.& \text{Nilai eksak dari}\\ & {\displaystyle \frac{1}{{10}^{-2020}+1}+\frac{1}{{10}^{-2019}+1}+...+{\displaystyle \frac{1}{{10}^{2019}+1}+\frac{1}{{10}^{2020}+1}}}\\ & \begin{array}{l}\\ \text{a}.& 2020& & & \\ \text{b}.& 2020,5\\ \text{c}.& 2021\\ \text{d}.& 2021,5\\ \text{e}.& 2022\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Misal},x& =\frac{1}{{10}^{-2020}+1}+\frac{1}{{10}^{-2019}+1}+...+\frac{1}{{10}^{2019}+1}+\frac{1}{{10}^{2020}+1}\\ \text{maka},x& =\frac{1}{\frac{1}{{10}^{2020}}+1}+\frac{1}{\frac{1}{{10}^{2019}}+1}+...+\frac{1}{{10}^{2019}+1}+\frac{1}{{10}^{2020}+1}\\ & =\frac{{10}^{2020}}{1+{10}^{2020}}+\frac{{10}^{2019}}{1+{10}^{2019}}+...+\frac{1}{{10}^{2019}+1}+\frac{1}{{10}^{2020}+1}\\ & =\frac{{10}^{2020}}{1+{10}^{2020}}+\frac{1}{{10}^{2020}+1}+\frac{{10}^{2019}}{1+{10}^{2019}}+\frac{1}{{10}^{2019}+1}+...+\frac{1}{1^0+1}\\ & =\frac{{10}^{2020}+1}{{10}^{2020}+1}+\frac{{10}^{2019}+1}{{10}^{2019}+1}+\frac{{10}^{2018}+1}{{10}^{2018}+1}+...+\frac{{10}^1+1}{{10}^1+1}+\frac{1}{1^0+1}\\ & =\underset{\text{sebanyak}2020}{\underbrace{1+1+1+1+1+...+1}}+\frac{1}{2}\\ & =2020,5\end{array}\end{array}$

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