Contoh Soal 7 Fungsi Eksponen (Matematika Peminatan Kelas X)

$\begin{array}{l}\\ 31.& \text{Nilai}x\text{yang memenuhi}\\ & \sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}=3\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 3\\ \text{b}.& 6\\ \text{c}.& 7\\ \text{d}.& 8\\ \text{e}.& 9\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Misalkan}A& =\sqrt{+\sqrt{x+\sqrt{+\cdots }}}\\ \sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}& =3\\ \text{dikuadratkan}& \\ x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}& =9\\ x+3& =9\\ x& =9-3\\ x& =6\end{array}\end{array}$

$\begin{array}{l}\\ 32.& \text{Nilai}x\text{yang memenuhi}\\ & x=\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 7\sqrt[7]{7}\\ \text{b}.& 7\\ \text{c}.& 14\\ \text{d}.& 49\\ \text{e}.& \sqrt[3]{81}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}x& =\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\\ x^3& =49\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\\ x^3& =49x\\ x^2& =49\\ x& =\sqrt{49}\\ & =7\end{array}\end{array}$

$\begin{array}{l}\\ 33.& \text{Nilai}x\text{yang memenuhi}\\ & x^{x^{x^{x^{x^{\cdots }}}}}=2020\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& \sqrt{2020}\\ \text{b}.& \sqrt[2020]{2020}\\ \text{c}.& {2020}^{\sqrt{2020}}\\ \text{d}.& {\sqrt{2020}}^{\sqrt{2020}}\\ \text{e}.& \sqrt{2020\sqrt{2020}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{x}^{x^{x^{x^{x^{\cdots }}}}}& =2020\\ x^{2020}& =2020\\ x& =\sqrt[2020]{2020}\end{array}\end{array}$

$\begin{array}{l}\\ 34.& \text{Nilai dari}\\ & {\displaystyle \frac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}}}\\ & \text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& \sqrt[3]{2}+1\\ \text{c}.& \sqrt[3]{2}-1\\ \text{d}.& \sqrt[3]{4}+1\\ \text{e}.& \sqrt[3]{4}-1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& {\displaystyle \frac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}}\times \frac{\sqrt[3]{2}-1}{\sqrt[3]{2}-1}}\\ & ={\displaystyle \frac{{\left(\sqrt[3]{2}\right)}^2-1}{\sqrt[3]{2}+\sqrt[3]{4}+\sqrt[3]{8}-1-\sqrt[3]{2}-\sqrt[3]{4}}}\\ & ={\displaystyle \frac{\sqrt[3]{4}-1}{\sqrt[3]{8}-1}}\\ & ={\displaystyle \frac{\sqrt[3]{4}-1}{2-1}}\\ & =\sqrt[3]{4}-1\end{array}\end{array}$

$\begin{array}{l}\\ 35.& \text{Nilai dari}\\ & {\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}}\\ & \text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& 2\sqrt{2}-1\\ \text{c}.& {\displaystyle \frac{1}{2}\sqrt{2}}\\ \text{d}.& \sqrt{{\displaystyle \frac{5}{3}}}\\ \text{e}.& \sqrt{{\displaystyle \frac{2}{5}}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}}\\ & ={\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\times \frac{\sqrt{\sqrt{5}+1}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}}\\ & ={\displaystyle \frac{\sqrt{7+3\sqrt{5}}+\sqrt{3-\sqrt{5}}}{\sqrt{5}+1}-(\sqrt{2}-1)}\\ & ={\displaystyle \frac{\left({\displaystyle \frac{3+\sqrt{5}}{\sqrt{2}}}\right)+\left({\displaystyle \frac{\sqrt{5}-1}{\sqrt{2}}}\right)}{\sqrt{5}+1}+1-\sqrt{2}}\\ & ={\displaystyle \frac{{\displaystyle \frac{2+2\sqrt{5}}{\sqrt{2}}}}{1+\sqrt{5}}+1-\sqrt{2}}\\ & ={\displaystyle \frac{2}{\sqrt{2}}+1-\sqrt{2}}\\ & =\sqrt{2}+1-\sqrt{2}\\ & =1\end{array}\end{array}$