$\begin{array}{ll}\\ 31.&\textrm{Nilai dari}\: \: \displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0}}}}-\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2}}}}-\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2^{\displaystyle 2}}}}=....\\ &\begin{array}{llll}\\ \textrm{a}.&4\\ \textrm{b}.&6\\ \textrm{c}.&8\\ \textrm{d}.&10\\ \textrm{e}.&12 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0}}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 1}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 2}}=\displaystyle 2^{\displaystyle 4}=16\\ &\textrm{Sedangkan untuk}\\ &\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2}}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 1}}=\displaystyle 2^{\displaystyle 2}=4\\ &\textrm{Selanjutnya untuk}\\ &\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2^{\displaystyle 2}}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 4}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 0}}=2^{\displaystyle 1}=2\\ &\textrm{Sehingga}\\ &\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0}}}}-\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2}}}}-\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2^{\displaystyle 2}}}}=16-4-2=10 \end{aligned} \end{array}$.
EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 6)
$\begin{array}{ll}\\ 26.&\textrm{Nilai dari}\: \: \displaystyle \sqrt[4]{4}-\sqrt{2}=....\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \textrm{b}.&-1\\ \textrm{c}.&0\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle \sqrt[4]{4}-\sqrt{2}=\displaystyle \sqrt[2]{\sqrt[2]{2^{2}}}-\sqrt{2}=\sqrt{2}-\sqrt{2}=0 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 27.&\textrm{Nilai dari}\: \: \displaystyle \frac{\sqrt[\displaystyle 3]{64}}{\sqrt[\displaystyle 4]{64}}=....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\sqrt{2}\\ \textrm{b}.&\sqrt{2}\\ \textrm{c}.&2\sqrt{2}\\ \textrm{d}.&3\\ \textrm{e}.&4 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle \frac{\sqrt[\displaystyle 3]{64}}{\sqrt[\displaystyle 4]{64}}=\displaystyle \frac{\sqrt[\displaystyle 3]{4^{\displaystyle 3}}}{\sqrt[\displaystyle 4]{8^{\displaystyle 2}}}=\displaystyle \frac{\sqrt[\displaystyle 3]{4^{\displaystyle 3}}}{\sqrt[\displaystyle 4]{\sqrt{8}^{\displaystyle 4}}}=\frac{4}{\sqrt{8}}\\ &=\frac{4}{\sqrt{8}}\times \frac{\sqrt{8}}{\sqrt{8}}=\frac{4}{8}\left( \sqrt{8} \right)\\ &=\frac{1}{2}\left( 2\sqrt{2} \right)=\sqrt{2} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 28.&\textrm{Nilai dari}\: \: \displaystyle \frac{\sqrt{\sqrt{25}}}{\sqrt{5}}=....\\ &\begin{array}{llll}\\ \textrm{a}.&5\\ \textrm{b}.&\sqrt{5}\\ \textrm{c}.&\displaystyle \frac{1}{2}\sqrt{5}\\ \textrm{d}.&1\\ \textrm{e}.&\displaystyle \frac{1}{2} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle \frac{\sqrt{\sqrt{25}}}{\sqrt{5}}=\frac{\sqrt{\sqrt[\displaystyle 2]{5^{\displaystyle 2}}}}{\sqrt{5}}=\frac{\sqrt{5}}{\sqrt{5}}=1 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 29.&\textrm{Nilai dari}\: \: \displaystyle \frac{\sqrt{49\sqrt{7}}}{\sqrt{7\sqrt{49}}}=....\\ &\begin{array}{llll}\\ \textrm{a}.&7\\ \textrm{b}.&7\sqrt{7}\\ \textrm{c}.&\displaystyle \frac{1}{2}\sqrt{7}\\ \textrm{d}.&\sqrt{7}\\ \textrm{e}.&\displaystyle \sqrt[\displaystyle 4]{7} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle \frac{\sqrt{49\sqrt{7}}}{\sqrt{7\sqrt{49}}}=\displaystyle \frac{7.\sqrt[\displaystyle 4]{7}}{\sqrt{7}.\sqrt[\displaystyle 4]{49}}=\displaystyle \frac{7.\sqrt[\displaystyle 4]{7}}{\sqrt{7}.\sqrt[\displaystyle 4]{7^{\displaystyle 2}}}\\ &=\displaystyle \frac{7.\sqrt[\displaystyle 4]{7}}{\sqrt{7}.\sqrt{7}}=\displaystyle \frac{7.\sqrt[\displaystyle 4]{7}}{7}=\sqrt[\displaystyle 4]{7} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 30.&\textrm{Nilai dari}\: \: \displaystyle \left( \frac{1}{2} \right)^{\displaystyle \frac{1}{2}}-\left( \frac{1}{4} \right)^{\displaystyle \frac{1}{4}}=....\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \textrm{b}.&-1\\ \textrm{c}.&0\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle \left( \frac{1}{2} \right)^{4}=\left( \frac{1}{4} \right)^{2}\Leftrightarrow \displaystyle \left( \frac{1}{2} \right)^{\displaystyle \frac{4}{1}}=\left( \frac{1}{4} \right)^{\displaystyle \frac{2}{1}}\\ &\Leftrightarrow \left( \frac{1}{2} \right)^{\displaystyle \frac{1}{2}}=\left( \frac{1}{4} \right)^{\displaystyle \frac{1}{4}}\\ &\Leftrightarrow \left( \frac{1}{2} \right)^{\displaystyle \frac{1}{2}}-\left( \frac{1}{4} \right)^{\displaystyle \frac{1}{4}}=0 \end{aligned} \end{array}$.
EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 5)
$\begin{array}{ll}\\ 21.&(\textbf{UM UGM 05})\textrm{Hasil dari}\\ &\sqrt{0,3+\sqrt{0,08}}=\sqrt{a}+\sqrt{b}\: ,\: \textrm{maka}\: \: \displaystyle \frac{1}{a}+\frac{1}{b}=....\\ &\begin{array}{llll}\\ \textrm{a}.&25\\ \textrm{b}.&20\\ \textrm{c}.&15\\ \textrm{d}.&10\\ \textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\sqrt{0,3+\sqrt{0,08}}&=\sqrt{0,2+0,1+\sqrt{4\times 0,2\times 0,1}}\\ &=\sqrt{0,2+0,1+2\sqrt{\times 0,2\times 0,1}}\\ &=\sqrt{0,2}+\sqrt{0,1}\\ \textrm{maka},\: \: a=0,2&,\: \: b=0,1\\ \textrm{sehingga}\: \displaystyle \frac{1}{a}+\frac{1}{b}&=\displaystyle \frac{1}{0,2}+\frac{1}{0,1}=5+10=15\\ \end{aligned} \end{array}$
$\begin{array}{ll}\\ 22.&(\textbf{SPMB 06})\textrm{Jika bilangan bulat}\: \: a\: \: \: \textrm{dan}\: \: b\: \: \textrm{memenuhi}\\ &\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}=a+b\sqrt{30}\: ,\: \textrm{maka}\: \: ab=....\\ &\begin{array}{llll}\\ \textrm{a}.&-22\\ \textrm{b}.&-11\\ \textrm{c}.&-9\\ \textrm{d}.&2\\ \textrm{e}.&13 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}&=\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}\times \displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}-\sqrt{6}}\\ &=\displaystyle \frac{5-2\sqrt{30}+6}{5-6}\\ &=\displaystyle \frac{11-2\sqrt{30}}{-1}\\ &=-11+2\sqrt{30}\\ \textrm{sehingga}&\: \: \: a=-11,\: \: b=2,\: \: \textrm{maka}\\ ab&=(-11)\times 2\\ &=-22 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 23.&(\textbf{OSK 2013})\textrm{Misal}\: \: a\: \: \textrm{dan}\: \: b\: \: \textrm{bilangan asli}\\ &\textrm{dengan}\: \: a>b.\: \: \textrm{Jika} \: \: \sqrt{94+2\sqrt{2013}}=\sqrt{a}+\sqrt{b}\\ &\textrm{maka nilai} \: \: a-b\: \: \textrm{adalah... .}\\\\ &\textrm{Jawab}:\\ &\begin{aligned} \sqrt{94+2\sqrt{2013}}&=\sqrt{61+33+2\sqrt{61\times 33}}\\ &=\sqrt{61}+\sqrt{33}\\ &=\sqrt{a}+\sqrt{b}\\ \textrm{Sehingga}\: \: a&=61,\: \: b=33,\: \: \textrm{maka}\\ a-b&=61-33\\ &=28 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 24.&\textrm{Daerah hasil dari fungsi eksponen}\: \: y\: =x^{- \frac{2}{3}}\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\\ \textrm{a}.&y< 0\\ \textrm{b}.&y> 0\\ \textrm{c}.&y\geq 0\\ \textrm{d}.&y\leq 0\\ \textrm{e}.&\textrm{Semua bilangan real} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Perhatikanlah gambar berikut} \end{array}$
EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 4)
$\begin{array}{ll}\\ 16.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &x^{\displaystyle x^{\displaystyle x^{\displaystyle x^{\displaystyle x^{\cdots }}}}}=2026 \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\sqrt{2026}\\ \textrm{b}.&\sqrt[\displaystyle 2026]{2026}\\ \textrm{c}.&2026^{\sqrt{2026}}\\ \textrm{d}.&\sqrt{2026}^{\sqrt{2026}}\\ \textrm{e}.&\sqrt{2026\sqrt{2026}} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}x^{\displaystyle x^{\displaystyle x^{\displaystyle x^{\displaystyle x^{\cdots }}}}}&=2026\\ x^{2026}&=2026\\ x&=\sqrt[\displaystyle 2026]{2026} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 17.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ & \sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}=3 \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&3\\ \textrm{b}.&6\\ \textrm{c}.&7\\ \textrm{d}.&8\\ \textrm{e}.&9 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Misalkan}\quad A&=\sqrt{+\sqrt{x+\sqrt{+\cdots }}}\\ \sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}&=3\\ \textrm{dikuadratkan}&\\ x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}&=9\\ x+3&=9\\ x&=9-3\\ x&=6 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 18.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &x=\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&7\sqrt[7]{7}\\ \textrm{b}.&7\\ \textrm{c}.&14\\ \textrm{d}.&49\\ \textrm{e}.&\sqrt[3]{81} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}x&=\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\\ x^{3}&=49\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\\ x^{3}&=49x\\ x^{2}&=49\\ x&=\sqrt{49}\\ &=7 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 19.&\textrm{Nilai dari}\\ &\displaystyle \frac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}}\\ &\textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \textrm{b}.&\sqrt[3]{2}+1\\ \textrm{c}.&\sqrt[3]{2}-1\\ \textrm{d}.&\sqrt[3]{4}+1\\ \textrm{e}.&\sqrt[3]{4}-1 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\displaystyle \frac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}}\times \frac{\sqrt[3]{2}-1}{\sqrt[3]{2}-1}\\ &=\displaystyle \frac{\left ( \sqrt[3]{2} \right )^{2}-1}{\sqrt[3]{2}+\sqrt[3]{4}+\sqrt[3]{8}-1-\sqrt[3]{2}-\sqrt[3]{4}}\\ &=\displaystyle \frac{\sqrt[3]{4}-1}{\sqrt[3]{8}-1}\\ &=\displaystyle \frac{\sqrt[3]{4}-1}{2-1}\\ &=\sqrt[3]{4}-1 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 20.&\textrm{Nilai dari}\\ &\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\\ &\textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \textrm{b}.&2\sqrt{2}-1\\ \textrm{c}.&\displaystyle \frac{1}{2}\sqrt{2}\\ \textrm{d}.&\sqrt{\displaystyle \frac{5}{3}}\\ \textrm{e}.&\sqrt{\displaystyle \frac{2}{5}} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\\ &=\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\times \frac{\sqrt{\sqrt{5}+1}}{\sqrt{\sqrt{5}+1}} -\sqrt{3-2\sqrt{2}}\\ &=\displaystyle \frac{\sqrt{7+3\sqrt{5}}+\sqrt{3-\sqrt{5}}}{\sqrt{5}+1}-\left ( \sqrt{2}-1 \right )\\ &=\displaystyle \frac{\left ( \displaystyle \frac{3+\sqrt{5}}{\sqrt{2}} \right )+\left ( \displaystyle \frac{\sqrt{5}-1}{\sqrt{2}} \right )}{\sqrt{5}+1}+1-\sqrt{2}\\ &=\displaystyle \frac{\displaystyle \frac{2+2\sqrt{5}}{\sqrt{2}}}{1+\sqrt{5}}+1-\sqrt{2}\\ &=\displaystyle \frac{2}{\sqrt{2}}+1-\sqrt{2}\\ &=\sqrt{2}+1-\sqrt{2}\\ &=1 \end{aligned} \end{array}$
EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 3)
$\begin{array}{l}\\ 11.&\textrm{Nilai dari}\\ & \displaystyle \frac{2^{2026}+2^{2027}}{3}-2^{2026}\: \: \textrm{adalah}\: ....\\\\ &\begin{array}{lllllllll}\\ \textrm{a}.&-2&&&\\ \textrm{b}.&-1\\ \textrm{c}.&0\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&\displaystyle \frac{2^{2026}+2^{2027}}{3}-2^{2026}\\ &=\displaystyle \frac{2^{2026}+2^{2027}-3.2^{2026}}{3}\\ &=\displaystyle \frac{1.2^{2026}+2^{1}.2^{2026}-3.2^{2026}}{3}\\ &=\displaystyle \frac{(3-3).2^{2026}}{3}\\ &=0 \end{aligned} \end{array}$.
$\begin{array}{l}\\ 12.&\textrm{Nilai dari}\\ &\displaystyle \frac{2^{2026}+2^{2027}+2^{2028}}{2^{2029}+2^{2030}+2^{2031}}\: \: \textrm{adalah}\: ....\\\\ &\begin{array}{lllllllll}\\ \textrm{a}.&0,125&&&\\ \textrm{b}.&0,\overline{333}\\ \textrm{c}.&0,45\\ \textrm{d}.&0,5\\ \textrm{e}.&0,\overline{666} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\displaystyle \frac{2^{2026}+2^{2027}+2^{2028}}{2^{2029}+2^{2030}+2^{2031}}\\ &=\displaystyle \frac{1.2^{2026}+2^{1}.2^{2026}+2^{2}.2^{2026}}{2^{3}.2^{2026}+2^{4}.2^{2026}+2^{5}.2^{2026}}\\ &=\displaystyle \frac{(1+2+4).2^{2026}}{(8+16+32).2^{2026}}\\ &=\displaystyle \frac{7}{56}=\frac{1}{8}=0,125 \end{aligned} \end{array}$.
$\begin{array}{l}\\ 13.&\textrm{Jika nilai dari}\\ &a^{\displaystyle a}=3\: ,\: \textrm{maka nilai}\quad a^{\displaystyle a^{\displaystyle a+1}}\quad \textrm{adalah}\: ....\\\\ &\begin{array}{lllllllll}\\ \textrm{a}.&9&&&\\ \textrm{b}.&18\\ \textrm{c}.&27\\ \textrm{d}.&81\\ \textrm{e}.&243 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&\displaystyle a^{\displaystyle a^{\displaystyle a+1}}=a^{\displaystyle a^{\displaystyle a}.a}=a^{\displaystyle 3.a}=\left( a^{\displaystyle a} \right)^{3}=3^{3}=27 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 14.&\textrm{Hitunglah}\:\\\\ &\quad\quad\qquad \displaystyle \sqrt[8]{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{...}}}}}\\\\ &\textrm{nyatakan jawabannya dalam bentuk }\: \displaystyle \frac{a\pm b\sqrt{c}}{d}\\ &\textrm{dengan a, b, c, dan d bilangan-bilangan bulat}\\ \end{array}$
Pembahasan:
$\begin{aligned}x^{8}&=2207-\displaystyle \underset{x^{8}}{\underbrace{\displaystyle \frac{1}{2207-\frac{1}{2207-\frac{1}{2207-...}}}}}\\ x^{8}&=2207-\displaystyle \frac{1}{x^{8}}\\ x^{8}+\displaystyle \frac{1}{x^{8}}&=2207\\ \left ( x^{4}+\displaystyle \frac{1}{x^{4}} \right )^{2}&=2207+2\\ \left ( x^{4}+\displaystyle \frac{1}{x^{4}} \right )&=\sqrt{2209}=47 \end{aligned}$
$\begin{aligned}x^{4}+\displaystyle \frac{1}{x^{4}}&=47\\ \left ( x^{2}+\displaystyle \frac{1}{x^{2}} \right )^{2}&=47+2\\ x^{2}+\displaystyle \frac{1}{x^{2}}&=\sqrt{49}=7\\ \left ( x+\displaystyle \frac{1}{x} \right )^{2}&=7+2\\ x+\displaystyle \frac{1}{x}&=\sqrt{9}=3\\ x^{2}-3x+1&=0,\\ &\textrm{persamaan kuadrat dalam x,}\\ & \textbf{gunakan rumus abc}\\ x_{1,2}=&\displaystyle \frac{3\pm \sqrt{5}}{2}=\displaystyle \frac{3\pm 1\sqrt{5}}{2}=\displaystyle \frac{a\pm b\sqrt{c}}{d}\\ &\textbf{Sehingga},\quad \begin{cases} & a=3 \\ & b=1 \\ & c=5 \\ & d=2 \end{cases} \end{aligned}$.
$\begin{array}{ll}\\ 15.&\textrm{Diketahui}\\ &x=\displaystyle \frac{1+p+p^{2}+p^{3}+\cdots +p^{n-1}}{1+p+p^{2}+p^{3}+\cdots +p^{n-2}+p^{n-1}+p^{n}} \\ &y=\displaystyle \frac{1+q+q^{2}+q^{3}+\cdots +q^{n-1}}{1+q+q^{2}+q^{3}+\cdots +q^{n-2}+q^{n-1}+q^{n}}\\\\ &\textrm{dan}\: \: p>q>0\\\\ &\textrm{Tunjukkan bahwa}\: \: x<y \\\\\\ &\textbf{Bukti}:\\ &\begin{aligned}&\textrm{Perhatikan bahwa}:\: \: p>q>0\\ &\textrm{sehingga}\\ &\displaystyle \frac{1}{p}< \frac{1}{q},\: \: \displaystyle \frac{1}{p^{2}}< \frac{1}{q^{2}},\cdots , \displaystyle \frac{1}{p^{n}}< \frac{1}{q^{n}}\\ &\textrm{Jika bentuk di atas dijumlahkan, maka}\\ &\displaystyle \frac{1}{p}+\frac{1}{p^{2}}+\cdots +\frac{1}{p^{n}}< \frac{1}{q}+\frac{1}{q^{2}}+\cdots +\frac{1}{q^{n}}\\ &\Leftrightarrow \displaystyle \frac{p^{n-1}+\cdots +p^{2}+p+1}{p^{n}}< \displaystyle \frac{q^{n-1}+\cdots +q^{2}+q+1}{q^{n}}\\ &\Leftrightarrow \displaystyle \frac{p^{n}}{1+p+p^{2}+\cdots +p^{n-1}}>\displaystyle \frac{q^{n}}{1+q+q^{2}+\cdots +q^{n-1}}\\ &\Leftrightarrow \displaystyle \frac{p^{n}}{1+p+p^{2}+\cdots +p^{n-1}}+1>\displaystyle \frac{q^{n}}{1+q+q^{2}+\cdots +q^{n-1}}+1\\ &\Leftrightarrow \displaystyle \frac{1+p+p^{2}+\cdots +p^{n-1}+p^{n}}{1+p+p^{2}+\cdots +p^{n-1}}>\displaystyle \frac{1+q+q^{2}+\cdots +q^{n-1}+q^{n}}{1+q+q^{2}+\cdots +q^{n-1}}\\ &\Leftrightarrow \displaystyle \frac{1+p+p^{2}+\cdots +p^{n-1}}{1+p+p^{2}+\cdots +p^{n-1}+p^{n}}<\displaystyle \frac{1+q+q^{2}+\cdots +q^{n-1}}{1+q+q^{2}+\cdots +q^{n-1}+q^{n}}\\ &\Leftrightarrow x<y\qquad \blacksquare \end{aligned} \end{array}$.
EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 2)
$\begin{array}{l}\\ 06.&\textrm{Jumlah akar-akar persamaan}\\ & 5^{x+1}+5^{2-x}-30=0\: \: \textrm{adalah}\: ....\\\\ &\begin{array}{lllllllll}\\ \textrm{a}.&-2&&&\\ \textrm{b}.&-1\\ \textrm{c}.&0\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}5^{x+1}+5^{2-x}-30&=0\\ \left (5^{x} \right ).5^{1}+\displaystyle \frac{5^{2}}{5^{x}}-30&=0\\ 5\left ( 5^{x} \right )^{2}+25-30\left ( 5^{x} \right )&=0\\ \textrm{Persamaan kuadrat}&\: \textrm{dalam}\: \: 5^{x},\: \textrm{maka}\\ 5(5^{x})^{2}-30(5^{x})+25&=0\begin{cases} a & =5 \\ b & =-30 \\ c & =25 \end{cases}\\ (5^{x_{1}}).\left ( 5^{x_{2}} \right )&=\displaystyle \frac{c}{a}\\ 5^{x_{1}+x_{2}}&=\displaystyle \frac{25}{5}=5\\ 5^{x_{1}+x_{2}}&=5^{1}\\ x_{1}+x_{2}&=1 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 07.&\textrm{Jumlah akar-akar persamaan}\\ &2020^{x^{2}-7x+7}=2021^{x^{2}-7x+7}\: \: \textrm{adalah}\: ....\\\\ &\begin{array}{lllllllll}\\ \textrm{a}.&-7\\ \textrm{b}.&-5\\ \textrm{c}.&-3\\ \textrm{d}.&5\\ \textrm{e}.&7 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}2020^{x^{2}-7x+7}&=2021^{x^{2}-7x+7}\\ \textrm{Karena basis}&\: \textrm{tidak sama},\\ \textrm{maka harusl}&\textrm{ah pangkatnya}=0,\\ x^{2}-7x+7&=0\\ \textrm{dan jumlah}\: &\textrm{akar-akarnya adalah}:\\ x_{1}+x_{2}&=-\displaystyle \frac{b}{a}, \: \: \textrm{dari persamaan}\\ x^{2}-7x+7&=0\begin{cases} a &=1 \\ b &=-7 \\ c &=7 \end{cases}\\ \textrm{maka}\: \: x_{1}+x_{2}&=-\displaystyle \frac{b}{a}=-\frac{-7}{1}=7 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 08.&\textrm{Nilai dari}\: \: \displaystyle \frac{2^{2020}+2^{2018}}{2^{2018}+2^{2016}} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \textrm{b}.&5\\ \textrm{c}.&10\\ \textrm{d}.&20\\ \textrm{e}.&40 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\displaystyle \frac{2^{2020}+2^{2018}}{2^{2018}+2^{2016}}&=\displaystyle \frac{2^{4}.2^{2016}+2^{2}.2^{2016}}{2^{2}.2^{2018}+2^{2016}}\\ &=\displaystyle \frac{2^{2016}\left ( 2^{4}+2^{2} \right )}{2^{2016}\left ( 2^{2}+1 \right )}\\ &=\displaystyle \frac{16+4}{4+1}\\ &=\displaystyle \frac{20}{5}\\ &=4 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 09.&(\textbf{UM IPB})\textrm{Jika}\: \: ab=a^{b} \: \: \textrm{dan}\: \: \displaystyle \frac{a}{b}=a^{3b}\\ &\textrm{maka nilai}\: \: a\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&0,5\\ \textrm{c}.&1\\ \textrm{d}.&0,25\\ \textrm{e}.&0,75 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Diketahui}&\\ ab&=a^{b}\\ b&=\displaystyle \frac{a^{b}}{a}=a^{b-1}.....\textbf{1}\\ \textrm{maka}&\\ \displaystyle \frac{a}{b}&=a^{3b}...............\textbf{2}\\ \textbf{1}&\: \: ke\: \: \textbf{2}\\ \displaystyle \frac{a}{a^{b-1}}&=a^{3b}\\ a^{2-b}&=a^{3b}\\ 2-b&=3b\\ -4b&=-2\\ b&=\displaystyle \frac{1}{2}................\textbf{3}\\ \textbf{3}&\: \: ke\: \: \textbf{1}\\ a\left ( \displaystyle \frac{1}{2} \right )&=a^{\frac{1}{2}}\\ \displaystyle \frac{1}{4}a^{2}&=a\\ a^{2}-4a&=0\\ a(a-4)&=0\\ a=0\: \: &\textrm{atau}\: \: a=4 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 10.&\textrm{Jika}\: \: \displaystyle 3.x^{^{^{^{ \frac{3}{2}}}}}=4\: ,\: \textrm{maka}\: \: x=\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle 1,1\\ \textrm{b}.&\displaystyle 1,2\\ \textrm{c}.&1,3\\ \textrm{d}.&\displaystyle 1,4\\ \textrm{e}.&1,5\\\\ &&&(\textbf{SAT Test Math Level 2})\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\displaystyle 3.x^{^{^{^{ \frac{3}{2}}}}}&=4\\ \left (3.x^{^{^{^{ \frac{3}{2}}}}} \right )^{2}&=4^{2}\\ 3^{2}.x^{3}&=4^{2}\\ x^{3}&=\displaystyle \frac{4^{2}}{3^{2}}\\ x^{3}&=\displaystyle \frac{4^{2}}{3^{2}}\times \frac{3}{3}\\ x^{3}&\leq \displaystyle \frac{4^{2}}{3^{2}}\times \frac{4}{3}\\ x^{3}&\leq \left ( \displaystyle \frac{4^{3}}{3^{3}} \right )\\ x^{3}&\leq \left ( \displaystyle \frac{4}{3} \right )^{3}\\ x&\leq \displaystyle \frac{4}{3}\\ x&\leq 1,\overline{333}\\ x&\approx 1,3 \end{aligned} \end{array}$
EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 1)
$\begin{array}{ll}\\ 01.&\textrm{Jika bentuk}\: \: \displaystyle \frac{ab^{-1}}{a^{-1}-b^{-1}}\\ & \textrm{dinyatakan dalam pangkat positif}=\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle \frac{a^{2}}{a-b}&&&\\\\ \textrm{b}.&\displaystyle \frac{a^{2}}{a-1}\\\\ \textrm{c}.&\displaystyle \frac{b-a}{ab}\\\\ \textrm{d}.&\displaystyle \frac{a^{2}}{b-a}\\\\ \textrm{e}.&\displaystyle \frac{1}{a-b}\\\\ &&&&(\textbf{SAT Test Math Level 2})\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\displaystyle \frac{ab^{-1}}{a^{-1}-b^{-1}}&=\displaystyle \frac{ab^{-1}}{a^{-1}-b^{-1}}\times \frac{b}{b}\\ &=\displaystyle \frac{a}{a^{-1}b-1}\\ &=\displaystyle \frac{a}{a^{-1}b-1}\times \frac{a}{a}\\ &=\displaystyle \frac{a^{2}}{b-a} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 02.&\textrm{Jika terdapat hubungan berikut}\\ &\textrm{a}.\quad 2^{p}=3^{q}=6^{r},\: \: \textrm{tunjukkan bahwa}\: \: pr+qr-pq=0\\ &\textrm{b}.\quad 2^{x}=3^{2y}=6^{z},\: \: \textrm{tunjukkan bahwa }\: \: 2xy-2yz-xz=0\\ &\textrm{c}.\quad 3^{15a}=5^{5b}=15^{3c},\: \: \textrm{tunjukkan bahwa }\: \: 5ab-bc-3ac=0\\\\ &\textrm{Bukti}\\ &\textrm{Yang akan ditunjukkan adalah no. 02 yang poin c, yaitu:}\\ &\begin{aligned}3^{15a}=5^{5b}=15^{3c}&\begin{cases} 3=5^{\frac{5b}{15a}} & \\ 3^{\frac{15a}{5b}}=b &\left ( a^{b}=c^{d}\rightarrow a=c^{\frac{d}{b}}\: \: \textrm{atau}\: \: a^{\frac{b}{d}}=c \right ) \end{cases}\\ 3^{15a}&=15^{3c}\\ 3^{15a}&=(3\times 5)^{3c}\\ 3^{15a}&=(3\times 3^{\frac{15a}{5b}})^{3c}\\ 3^{15a}&=3^{3c+\frac{9c}{b}}\\ a^{f(x)}&=a^{g(x)}\\ f(x)&=g(x)\\ 15a&=3c+\frac{9ac}{b}\\ 15ab&=3bc+9ac\\ 5ab&=bc+3ac\\ 5ab-bc-3ac&=0\quad \color{black}\blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 03.&\textrm{Diketahui bahwa}\quad 5^{p}=9^{q}=2025, \: \textrm{nilai}\: \: \displaystyle \frac{pq}{p+q}=....\\ &\textrm{Jawab}:\\ &\begin{aligned}&5^{p}=9^{q}=2025=45^{2} \quad \textrm{dengan}\quad\begin{cases} 5=9^{\frac{q}{p}} & \\ 5^{\frac{p}{q}}=9 \end{cases}\\ &\Leftrightarrow 5^{p}=(5\times 9)=5^{2}\times 9^{2}\\ &\Leftrightarrow 5^{p}=5^{2}\times \left(5^{\frac{p}{q}} \right)^{2}\\ &\Leftrightarrow 5^{p}=5^{2+\displaystyle \frac{2p}{q}},\quad \textrm{ingat}\quad a^{f(x)}=a^{g(x)}\Rightarrow f(x)=g(x)\\ &\Leftrightarrow p=2+\displaystyle \frac{2p}{q}\\ &\Leftrightarrow pq=2q+2p=2(p+q)\\ &\Leftrightarrow \displaystyle \frac{pq}{p+q}=2 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 04.&\textrm{Bentuk sederhana dari}\\ &\sqrt{33+\sqrt{800}}-\sqrt{27-2\sqrt{162}}=....\\\\ &\qquad\qquad\qquad(\textbf{SIMAK UI 2012 Mat IPA})\\ &\begin{array}{llllllll}\\ \textrm{a}.&2-\sqrt{2}\\ \textrm{b}.&8-\sqrt{2}\\ \textrm{c}.&-2+\sqrt{2}\\ \textrm{d}.&2+5\sqrt{2}\\ \textrm{e}.&8+5\sqrt{2} \end{array}\\\\ &\textrm{Jawab}:\qquad \textbf{b}\\ &\textrm{misalkan},\\ &\begin{aligned}x&=\sqrt{33+\sqrt{800}}-\sqrt{27-2\sqrt{162}}\\ &=\left ( \sqrt{33+20\sqrt{2}}-\sqrt{27-2.9\sqrt{2}}\: \right )\\ &=\sqrt{33+20\sqrt{2}}-\sqrt{27-18\sqrt{2}}\\ x^{2}&=33+20\sqrt{2}+27-18\sqrt{2}-2\sqrt{\left ( 33+20\sqrt{2} \right )\left ( 27-18\sqrt{2} \right )}\\ &=60+2\sqrt{2}-2\sqrt{33.27-33.18\sqrt{2}+27.20\sqrt{2}-20.18.2}\\ &=60+2\sqrt{2}-2\sqrt{891-720+540\sqrt{2}-594\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-54\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-2.27\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-2\sqrt{27.27}\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-2\sqrt{162.9}}\\ &=60+2\sqrt{2}-2\sqrt{162+9-2\sqrt{162.9}}\\ &=60+2\sqrt{2}-2\left ( \sqrt{162}-\sqrt{9} \right )\\ &=60+2\sqrt{2}-2\left ( 9\sqrt{2}-3 \right )\\ x^{2}&=66-16\sqrt{2}\\ x&=\sqrt{66-2.8\sqrt{2}}\\ &=\sqrt{64+2-2\sqrt{64.2}}\\ &=\sqrt{64}-\sqrt{2}\\ &=8-\sqrt{2} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 05.&\textrm{Jika}\: \: \displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\displaystyle \frac{a\sqrt{2}+b\sqrt{3}+c\sqrt{5}}{12}\: ,\\\\ &\textrm{maka}\: \: a+b+c=....\\ &\qquad\qquad\qquad\qquad (\textbf{UM UGM 2016 Mat Das})\\ &\begin{array}{llllllll}\\ \textrm{a}.&0\\ \textrm{b}.&1\\ \textrm{c}.&2\\ \textrm{d}.&3\\ \textrm{e}.&4 \end{array}\\\\ &\textrm{Jawab}:\qquad \textbf{e}\\ &\begin{aligned}\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}&=\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\times \displaystyle \frac{\left ( \sqrt{2}+\sqrt{3}-\sqrt{5} \right )}{\left ( \sqrt{2}+\sqrt{3}-\sqrt{5} \right )}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\left ( \sqrt{2}+\sqrt{3} \right )^{2}-\left ( \sqrt{5} \right )^{2}}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2+3+2\sqrt{2.3}-5}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2\sqrt{6}}\times \displaystyle \frac{\sqrt{6}}{\sqrt{6}}\\ &=\displaystyle \frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{12}\\ &=\displaystyle \frac{2\sqrt{3}+3\sqrt{2}-\sqrt{30}}{12}\\ &=\displaystyle \frac{3\sqrt{2}+2\sqrt{3}-\sqrt{30}}{12}\\ &\quad \begin{cases} a &=3 \\ b &=2 \\ c &=-1 \end{cases}\\ a+b+c&=3+2+(-1)\\ &=4 \end{aligned} \end{array}$.
DAFTAR PUSTAKA
- Baskoro, B.D. 2012. Aljabar dan Trigonometri Cespleng Olimpiade Matematika. Yogyakarta: BERLIAN
- Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bnadung: SEWU.
- Kanginan, M., Terzalgi, Y. 2013. Matematika untuk SMA-MA/SMK Kelas X Wajib. Bandung: SEWU.
- Tung, Khoe Yao. 2012. Pintar Matematika SMA Kelas XII IPA untuk Olimpiade dan Pengayaan Pelajaran. Yogyakarta: ANDI.