Contoh Soal 7 Fungsi Eksponen (Matematika Peminatan Kelas X)

$\begin{array}{ll}\\ 31.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ & \sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}=3 \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&3\\ \color{red}\textrm{b}.&6\\ \textrm{c}.&7\\ \textrm{d}.&8\\ \textrm{e}.&9 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\textrm{Misalkan}\quad A&=\sqrt{+\sqrt{x+\sqrt{+\cdots }}}\\ \sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}&=3\\ \textrm{dikuadratkan}&\\ x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}&=9\\ x+3&=9\\ x&=9-3\\ x&=6 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 32.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &x=\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&7\sqrt[7]{7}\\ \color{red}\textrm{b}.&7\\ \textrm{c}.&14\\ \textrm{d}.&49\\ \textrm{e}.&\sqrt[3]{81} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}x&=\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\\ x^{3}&=49\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\\ x^{3}&=49x\\ x^{2}&=49\\ x&=\sqrt{49}\\ &=7 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 33.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &x^{x^{x^{x^{x^{\cdots }}}}}=2020 \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\sqrt{2020}\\ \color{red}\textrm{b}.&\sqrt[2020]{2020}\\ \textrm{c}.&2020^{\sqrt{2020}}\\ \textrm{d}.&\sqrt{2020}^{\sqrt{2020}}\\ \textrm{e}.&\sqrt{2020\sqrt{2020}} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}x^{x^{x^{x^{x^{\cdots }}}}}&=2020\\ x^{2020}&=2020\\ x&=\sqrt[2020]{2020} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 34.&\textrm{Nilai dari}\\ &\displaystyle \frac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}}\\ &\textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \textrm{b}.&\sqrt[3]{2}+1\\ \textrm{c}.&\sqrt[3]{2}-1\\ \textrm{d}.&\sqrt[3]{4}+1\\ \color{red}\textrm{e}.&\sqrt[3]{4}-1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&\displaystyle \frac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}}\times \frac{\sqrt[3]{2}-1}{\sqrt[3]{2}-1}\\ &=\displaystyle \frac{\left ( \sqrt[3]{2} \right )^{2}-1}{\sqrt[3]{2}+\sqrt[3]{4}+\sqrt[3]{8}-1-\sqrt[3]{2}-\sqrt[3]{4}}\\ &=\displaystyle \frac{\sqrt[3]{4}-1}{\sqrt[3]{8}-1}\\ &=\displaystyle \frac{\sqrt[3]{4}-1}{2-1}\\ &=\sqrt[3]{4}-1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 35.&\textrm{Nilai dari}\\ &\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\\ &\textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&1\\ \textrm{b}.&2\sqrt{2}-1\\ \textrm{c}.&\displaystyle \frac{1}{2}\sqrt{2}\\ \textrm{d}.&\sqrt{\displaystyle \frac{5}{3}}\\ \textrm{e}.&\sqrt{\displaystyle \frac{2}{5}} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\\ &=\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\times \frac{\sqrt{\sqrt{5}+1}}{\sqrt{\sqrt{5}+1}} -\sqrt{3-2\sqrt{2}}\\ &=\displaystyle \frac{\sqrt{7+3\sqrt{5}}+\sqrt{3-\sqrt{5}}}{\sqrt{5}+1}-\left ( \sqrt{2}-1 \right )\\ &=\displaystyle \frac{\left ( \displaystyle \frac{3+\sqrt{5}}{\sqrt{2}} \right )+\left ( \displaystyle \frac{\sqrt{5}-1}{\sqrt{2}} \right )}{\sqrt{5}+1}+1-\sqrt{2}\\ &=\displaystyle \frac{\displaystyle \frac{2+2\sqrt{5}}{\sqrt{2}}}{1+\sqrt{5}}+1-\sqrt{2}\\ &=\displaystyle \frac{2}{\sqrt{2}}+1-\sqrt{2}\\ &=\sqrt{2}+1-\sqrt{2}\\ &=1 \end{aligned} \end{array}$