$\begin{array}{ll}\\ 71.&\textbf{(SPMB '05)}\\ &\textrm{Jika}\: \: ^{3}\log 2=p\: \: \textrm{dan}\: \: ^{2}\log 7=q,\\ &\textrm{maka}\: \: ^{14}\log 54=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{p+3}{p+q}\\ \textrm{b}.&\displaystyle \frac{2p}{p+q}\\ \color{red}\textrm{c}.&\displaystyle \frac{p+3}{p(q+1)}\\ \textrm{d}.&\displaystyle \frac{p+q}{p(q+1)}\\ \textrm{e}.&\displaystyle \frac{p(q+1)}{p+q} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&^{14}\log 54\\ &=\displaystyle \frac{^{...}\log 54}{^{...}\log 14}=\frac{^{...}\log (2.27)}{^{...}\log (2.7)},\: \: \color{red}\textrm{pilih basis 2}\\ &\color{blue}\textrm{mengapa tidak pilih basis selain 2}\\ &\color{blue}\textrm{lihat penyebut, di sana terdapat numerus 7}\\ &\color{blue}\textrm{pada soal, pasangan numerus 7 adalah 2},\\ &\textrm{makanya basis 2 dipilih, bukan yang lain}\\ &=\frac{^{2}\log (2.27)}{^{2}\log (2.7)}=\displaystyle \frac{^{2}\log 2+\: ^{2}\log 27}{^{2}\log 2+\: ^{2}\log 7}\\ &=\displaystyle \frac{^{2}\log 2+\: ^{2}\log 3^{3}}{^{2}\log 2+\: ^{2}\log 7}=\displaystyle \frac{^{2}\log 2+\left (3\times \: \displaystyle \frac{1}{^{3}\log 2} \right )}{^{2}\log 2+\: ^{2}\log 7}\\ &=\displaystyle \frac{1+\displaystyle \frac{3}{p}}{1+q}\\ &=\color{red}\displaystyle \frac{p+3}{p(q+1)} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 72.&\textbf{(SPMB '04)}\\ &\textrm{Jika}\: \: a>1\: ,\: \textrm{maka penyelesaian untuk}\\ &\left (^{a}\log (2x+1) \right )\left ( ^{3}\log \sqrt{a} \right )=1\: \: \textrm{adalah}\: ....\\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 1\\ \textrm{b}.&\displaystyle 2\\ \textrm{c}.&\displaystyle 3\\ \color{red}\textrm{d}.&\displaystyle 4\\ \textrm{e}.&\displaystyle 5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\left (^{a}\log (2x+1) \right )\left ( ^{3}\log \sqrt{a} \right )&=1\\ \left ( ^{3}\log \sqrt{a} \right )\left (^{a}\log (2x+1) \right )&=1\\ \left ( ^{3}\log a^{.^{^{\frac{1}{2}}}} \right )\left (^{a}\log (2x+1) \right )&=1\\ \displaystyle \frac{1}{2}\left ( ^{3}\log a \right )\left (^{a}\log (2x+1) \right )&=1\\ ^{3}\log (2x+1)&=2\\ 2x+1&=3^{2}\\ 2x&=9-1\\ 2x&=8\\ x&=\color{red}4 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 73.&\textbf{(SPMB '04)}\\ &\textrm{Nilai}\: \: \displaystyle \frac{\left ( ^{5}\log 10 \right )^{2}-\left ( ^{5}\log 2 \right )^{2}}{^{5}\log \sqrt{20}}\: \: \textrm{adalah}\: ....\\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\\ \textrm{b}.&\displaystyle 1\\ \color{red}\textrm{c}.&\displaystyle 2\\ \textrm{d}.&\displaystyle 4\\ \textrm{e}.&\displaystyle 5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\displaystyle \frac{\left ( ^{5}\log 10 \right )^{2}-\left ( ^{5}\log 2 \right )^{2}}{^{5}\log \sqrt{20}}\\ &=\displaystyle \frac{\left ( ^{5}\log 10+\: ^{5}\log 2 \right )\left ( ^{5}\log 10-\: ^{5}\log 2 \right )}{^{5}\log (20)^{.^{\frac{1}{2}}}}\\ &=\displaystyle \frac{^{5}\log (10.2)\times ^{5}\log \left (\frac{10}{2} \right )}{\displaystyle \frac{1}{2}\times \: ^{5}\log 20}\\ &=2\times \left ( \displaystyle \frac{^{5}\log 20}{^{5}\log 20} \right )\times \: ^{5}\log 5\\ &=2.1.1\\ &=\color{red}2 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 74.&\textbf{(SPMB '03)}\\ &\textrm{Jika diketahui bahwa}\\ &^{4}\log ^{4}\log x-\: ^{4}\log ^{4}\log ^{4}\log 16=2\\ & \textrm{maka}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle ^{2}\log x=8\\ \textrm{b}.&\displaystyle ^{2}\log x=4\\ \color{red}\textrm{c}.&\displaystyle ^{4}\log x=8\\ \textrm{d}.&\displaystyle ^{4}\log x=16\\ \textrm{e}.&\displaystyle ^{16}\log x=8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{4}\log \: ^{4}\log 16=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{4}\log \: ^{4}\log 4^{2}=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{4}\log 2=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{2^{2}}\log 2^{1}=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{4}\log \left ( \displaystyle \frac{1}{2} \right )=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{2^{2}}\log 2^{-1}=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\left ( -\displaystyle \frac{1}{2} \right )=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x+\frac{1}{2}=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x=2-\displaystyle \frac{1}{2}=\frac{3}{2}\\ &\Leftrightarrow \: ^{4}\log x=4^{.\frac{3}{2}}\\ &\Leftrightarrow \: ^{4}\log x=\left (2^{2} \right )^{.^{\frac{3}{2}}}\\ &\Leftrightarrow \: ^{4}\log x=2^{3}\\ &\Leftrightarrow \: ^{4}\log x=\color{red}8\\ \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 75.&\textbf{(UMPTN '92)}\\ &\textrm{Jika}\: \: x\: \: \textrm{memenuhi persamaan}\\ &^{4}\log ^{4}\log x-\: ^{4}\log ^{4}\log ^{4}\log 16=2\\ & \textrm{maka nilai}\: \: ^{16}\log x=\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle 4\\ \textrm{b}.&\displaystyle 2\\ \textrm{c}.&\displaystyle 1\\ \textrm{d}.&\displaystyle -2\\ \textrm{e}.&\displaystyle -4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{4}\log \: ^{4}\log 16=2\\ &\textrm{menyebabkan}\\ & ^{4}\log x=8\Rightarrow x=4^{8}\\ &(\color{purple}\textrm{lihat pembahasan no.23})\\ &\textrm{maka},\\ &\: ^{16}\log x=\: ^{16}\log 4^{8}=\: ^{4^{2}}\log 4^{8}\\ &=\displaystyle \frac{8}{2}\\ &=\color{red}4 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 78.&\textbf{(UMPTN '94)}\\ &\textrm{Hasil kali akar-akar persamaan}\\ &^{3}\log x^{.\left (^{2+\: ^{3}\log x} \right )}=15\\ & \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{1}{9}\\ \textrm{b}.&\displaystyle \frac{1}{3}\\ \textrm{c}.&\displaystyle 1\\ \textrm{d}.&\displaystyle 3\\ \textrm{e}.&\displaystyle 9 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&^{3}\log x^{.\left (^{2+\: ^{3}\log x} \right )}=15\\ &\Leftrightarrow \: \left ({2+\: ^{3}\log x} \right )^{3}\log x-15=0\\ &\Leftrightarrow 2\: ^{3}\log x+\: \left ( ^{3}\log x \right )^{2}-15=0\\ &\Leftrightarrow \: \left ( ^{3}\log x \right )^{2}+2\: ^{3}\log x-15=0\\ &\Leftrightarrow \: \left (^{3}\log x_{1}+5 \right )\left ( ^{3}\log x_{2}-3 \right )=0\\ &\Leftrightarrow \: ^{3}\log x_{1}+5=0\: \: \textrm{atau}\: \: ^{3}\log x_{1}-3=0\\ &\Leftrightarrow \: ^{3}\log x_{1}=-5\: \: \textrm{atau}\: \: ^{3}\log x_{2}=3\\ &\Leftrightarrow \: x_{1}=3^{-5}\: \: \textrm{atau}\: \: x_{2}=3^{3}\\ &\qquad \textrm{maka}\\ &\Leftrightarrow \: x_{1}\times x_{2}=3^{-5}\times 3^{3}=3^{-5+3}=3^{-2}\\ &\Leftrightarrow \qquad =\displaystyle \frac{1}{3^{2}}\\ &\Leftrightarrow \qquad =\color{red}\frac{1}{9} \end{aligned} \end{array}$.
$\begin{array}{l}\\ 79.&\textrm{Diketahui bahwa}\\ & ^{^{2}}\log 3=p \: \: \textrm{dan}\: \: ^{^{3}}\log 11=q,\\ &\textrm{maka nilai}\: \: ^{^{44}}\log 66=....\\\\ &\textrm{Jawab}:\\ &\begin{aligned}^{^{44}}\log 66&=\displaystyle \frac{^{^{^{...}}}\log 66}{^{^{^{...}}}\log 44}\\ &=\displaystyle \frac{^{^{^{...}}}\log \left (2\times 3\times 11 \right )}{^{^{^{...}}}\log \left (2^{2}\times 11 \right )}\\ &=\displaystyle \frac{^{^{^{3}}}\log 2+\: ^{^{^{3}}}\log 3+\:^{^{^{3}}}\log 11}{^{^{^{3}}}\log 2^{2}+\: ^{^{^{3}}}\log 11}\\ &=\displaystyle \frac{\frac{1}{^{^{^{2}}}\log 3}+\: ^{^{^{3}}}\log 3+\:^{^{^{3}}}\log 11}{\frac{2}{^{^{^{3}}}\log 2^{2}}+\: ^{^{^{3}}}\log 11}\\ &=\displaystyle \frac{\frac{1}{p}+1+q}{\frac{2}{p}+q}\\ &=\displaystyle \frac{\frac{1}{p}+1+q}{\frac{2}{p}+q}\times \displaystyle \frac{p}{p}\\ &=\color{red}\frac{1+p+pq}{2+pq} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 80.&\textbf{(AIME 1984)}\\ &\textrm{Diketahui bahwa}\: \: x\: \: \textrm{dan}\: \: y\\ & \textrm{adalah bilangan real yang memenuhi}\\\\ &\left\{\begin{matrix} ^{^{8}}\log x+\: ^{^{4}}\log y^{2}=5\\ \\ ^{^{8}}\log y+\: ^{^{4}}\log x^{2}=7 \end{matrix}\right.\\\\ &\textrm{Tentukanlah nilai dari}\: \: xy\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&^{^{^{8}}}\log x+\: ^{^{4}}\log y^{2}=5\\ &\Leftrightarrow \: ^{^{^{2^{3}}}}\log x+\: ^{^{^{2^{2}}}}\log y^{2}=5....(1)\\ &^{^{^{8}}}\log y+\: ^{^{4}}\log x^{2}=7\\ &\Leftrightarrow \: ^{^{^{2^{3}}}}\log y+\: ^{^{^{2^{2}}}}\log x^{2}=7....(2)\\ &\textrm{selanjutnya},&\\ &\frac{1}{3}\:. ^{^{^{2}}}\log x+\: ^{^{^{2}}}\log y=5\: \: |\times \frac{1}{3}|\\ &\Rightarrow \frac{1}{9}\: .^{^{^{2}}}\log x+\: \frac{1}{3}.\: ^{^{^{2}}}\log y=\frac{5}{3}....(3)\\ &^{^{^{2}}}\log x+\:\frac{1}{3}\: . ^{^{^{2}}}\log y=7\: \: |\times 1|\\ &\Rightarrow ^{^{^{2}}}\log x+\:\frac{1}{3}\: . ^{^{^{2}}}\log y=7....(4)\\ &\textrm{saat persamaan}\: \: (3)-(4)\\ &=-\frac{8}{9}.\: ^{^{^{2}}}\log x=\frac{5}{3}-7=-\frac{16}{3}\\ &\color{purple}\textrm{maka}\\ &^{^{^{2}}}\log x=\left ( -\frac{16}{3} \right )\left ( -\frac{9}{8} \right )\\ &^{^{^{2}}}\log x=6\Leftrightarrow x=2^{6}\Leftrightarrow x=64\\ &\color{purple}\textrm{Pada persamaan 1 selanjutnya}\\ &\frac{1}{3}.\: ^{^{^{2}}}\log x+\: ^{^{^{2}}}\log y=5\\ &\Leftrightarrow \: \: \frac{1}{3}.\: ^{^{^{2}}}\log 2^{6}+\: ^{^{^{2}}}\log y=5\\ &\Leftrightarrow \: \: \frac{1}{3}.6+\: ^{^{^{2}}}\log y=5\\ &\Leftrightarrow \: \: 2+\: ^{^{^{2}}}\log y=5\\ &\Leftrightarrow \: \: ^{^{^{2}}}\log y=5-2=3\Leftrightarrow y=2^{3}=8\\ &\textrm{Jadi},\: \: x.y=64.8=\color{red}512 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 81.&\textrm{Tentukanlah nilai dari}\\ &\textrm{a}.\quad \left ( 2^{\: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: ^{^{^{9}}}\log 5} \right )\left ( 5^{\: ^{^{^{\frac{1}{5}}}}\log 2} \right )\\ &\textrm{b}.\quad \left ( ^{27}\log 125 \right )\left ( ^{25}\log \displaystyle \frac{1}{64} \right )\left ( ^{64}\log \frac{1}{9} \right )\\ &\textrm{c}.\quad \left ( ^{625}\log 19 \right )\left ( ^{7}\log \displaystyle \frac{1}{25} \right )\left ( ^{19}\log 7 \right )\\\\ &\textrm{Jawab}:\\ &\begin{array}{|l|}\hline \begin{aligned}\textrm{a}.\quad &\left ( 2^{\: \: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: ^{^{^{9}}}\log 5} \right )\left ( 5^{\: ^{^{^{\frac{1}{5}}}}\log 2} \right )\\ &=\left ( 2^{\: \: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: \: ^{^{^{3^{2}}}}\log 5} \right )\left ( 5^{\: \: ^{^{^{5^{-1}}}}\log 2} \right )\\ &=\left ( 2^{\: \: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: \: ^{^{^{3}}}\log 5^{^{\frac{1}{2}}}} \right )\left ( 5^{\: \: ^{^{^{5}}}\log 2^{-1}} \right )\\ &=\left ( 2^{\: ^{^{2}}\log 6} \right )\left ( 3^{\: ^{^{3}}\log \sqrt{5}} \right )\left ( 5^{\: ^{^{5}}\log \frac{1}{2}} \right )\\ &=6\times \sqrt{5}\times \frac{1}{2}\\ &=\color{red}3\sqrt{5} \end{aligned}\\\hline \end{array}\\ &\begin{array}{|l|}\hline \begin{aligned}\textrm{b}.\quad &\left ( ^{27}\log 125 \right )\left ( ^{25}\log \displaystyle \frac{1}{64} \right )\left ( ^{64}\log \frac{1}{9} \right )\\ &=\left ( ^{^{3^{3}}}\log 5^{3} \right )\left ( ^{^{5^{2}}}\log 4^{-3} \right )\left ( ^{^{4^{3}}}\log 3^{-2} \right )\\ &=\displaystyle \frac{3}{3}.\left ( -\frac{3}{2} \right ).\left ( -\frac{2}{3} \right ).\: ^{^{3}}\log 5.\: ^{^{5}}\log 4.\: ^{^{4}}\log 3\\ &=\color{red}1\end{aligned}\\\hline \end{array}\\ &\color{blue}\textrm{Pembahasan diserahkan kepada}\\ &\color{blue}\textrm{Pembaca yang budiman untuk poin c} \end{array}$
$\begin{array}{ll}\\ 82.&\textrm{Tentukanlah nilai}\: \: a+b\: \: \textrm{dimana}\: \: a\: \: \textrm{dan}\: \: b\\ &\textrm{adalah bilangan riil positif}.\\ &^{7}\log \left ( 1+a^{2} \right )-\: ^{7}\log 25=\: ^{7}\log \left ( 2ab-15 \right )-\: ^{7}\log \left ( 25+b^{2} \right )\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&^{7}\log \displaystyle \frac{\left ( 1+a^{2} \right )}{25}=\: ^{7}\log \displaystyle \frac{\left ( 2ab-15 \right )}{\left ( 25+b^{2} \right )}\\ &\textrm{diambil}\: \textrm{persamaannya, maka}\\ &\displaystyle \frac{\left ( 1+a^{2} \right )}{25}=\displaystyle \frac{\left ( 2ab-15 \right )}{\left ( 25+b^{2} \right )}\\ &\displaystyle \left ( 1+a^{2} \right )\left ( 25+b^{2} \right )=25\left ( 2ab-15 \right )\\ &\begin{cases} \left ( 1+a^{2} \right ) & \text{ faktor dari} \: \: 25,\: \: a> 0,\: a\in \mathbb{R} \\ &\textrm{atau}\\ \left ( 25+b^{2} \right ) & \text{ faktor dari }\: \: 25,\: \: b> 0,\: b\in \mathbb{R}\: \: \: \textrm{juga} \end{cases}\\ \end{aligned}\\\\ &\color{red}\begin{array}{|c|l|c|l|c|c|c|}\hline \textrm{No}&a&\left ( 1+a^{2} \right )&b&\left ( 25+b^{2} \right )&\textrm{Keterangan}&a+b\\\hline 1&2&5&10&125&\textrm{Memenuhi}&12\\\hline 2&7&50&\cdots &\cdots &\textrm{tidak}&\cdots \\\hline 3&\cdots &\cdots &5&50&\textrm{tidak}&\cdots \\\hline \end{array} \end{array}$
$\begin{array}{ll}\\ 83.&\textrm{Jika}\: \: 60^{a}=3\: \: \textrm{dan}\: \: 60^{b}=5\\ &\textrm{maka hasil dari}\: \: 12^{^{\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )}}\\\\ &\textrm{Jawab}:\\ &\textrm{Perhatikanlah bahwa}\\ &\color{blue}\begin{aligned}&\begin{cases} 60^{a}=3 & \Rightarrow \: ^{60}\log 3=a \\ 60^{b}=5 & \Rightarrow \: ^{60}\log 5=b \end{cases} \end{aligned} \\ &\color{blue}\textrm{Selanjutnya}\\ &\color{blue}\textrm{Untuk}\: \: (1-a-b),\: \textrm{maka}\\ &\begin{aligned}1-a-b&=1-\: ^{60}\log 3-\: ^{60}\log 5\\ &=\: ^{60}\log 60-\: ^{60}\log 3-\: ^{60}\log 5\\ &=\: ^{60}\log \displaystyle \frac{60}{3\times 5}\\ &=\: ^{60}\log 4\\ &=\: ^{60}\log 2^{2} \end{aligned} \\ &\textrm{Untuk}\: \: 2(1-b),\: \textrm{maka}\\ &\begin{aligned}2(1-b)&=2\left ( 1-\: ^{60}\log 5 \right )\\ &=2\left ( ^{60}\log 60-\: ^{60}\log 5 \right )\\ &=2\left ( ^{60}\log \displaystyle \frac{60}{5} \right )\\ &=2\left ( ^{60}\log 12 \right )\\ &=\: ^{60}\log 12^{2} \end{aligned} \\ &\textrm{Untuk}\: \: \left ( \displaystyle \frac{1-x-y}{2(1-b)} \right ),\: \textrm{maka}\\ &\begin{aligned}\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )&=\displaystyle \frac{\: ^{60}\log 2^{2}}{\: ^{60}\log 12^{2}}\\ &=\displaystyle \frac{2\: ^{60}\log 2}{2\: ^{60}\log 12}\\ &=\displaystyle \frac{\: ^{60}\log 2}{\: ^{60}\log 12}\\ &=\: ^{12}\log 2 \end{aligned}\\ &\color{blue}\textrm{Jadi},\: \: 12^{^{\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )}}=12^{^{\: ^{12}\log 2}}=\color{red}2 \end{array}$
$\begin{array}{ll}\\ 84.&\textrm{Diberikan bilangan riil positif}\: \: x,\: y,\: \textrm{dan}\: z\\ & \textrm{yang memenuhi persamaan}\\ &2\: ^{x}\log (2y)=2\: ^{2x}\log (4z)=2\: ^{4x}\log (8yz)\neq 0.\\ &\textrm{Jika nilai}\: \: xy^{5}z\: \: \textrm{dapat dinyatakan dengan}\: \: \displaystyle \frac{1}{2^{\displaystyle \frac{p}{q}}}\\ & \textrm{dengan}\: \: p\: \: \textrm{dan}\: \: q\: \: \textrm{bilangan asli yang saling prima},\\ &\textrm{maka nilai dari}\: \: p+q=....\\\\ &\textrm{Jawab}:\\\\ &\begin{aligned}&2\: ^{x}\log (2y)=2\: ^{2x}\log (4z)=2\: ^{4x}\log (8yz)\neq 0\\ &\textrm{maka}\\ &^{x}\log (2y)=\: ^{2x}\log (4y)\\ &\Rightarrow \quad \log (2y)\times \log (2x)=\log x\times \log (4y)...(1)\\ &^{x}\log (2y)=\: ^{4x}\log (8yz)\\ &\Rightarrow \quad \log (2y)\times \log (4x)=\log x\times \log (8yz)....(2)\\ &^{2x}\log (4y)=\: ^{4x}\log (8yz)\\ &\Rightarrow \quad \log (4y)\times \log (4x)=\log (2x)\times \log (8yz)....(3) \end{aligned}\\\\ &\begin{aligned}&\textrm{Perhatikan persamaan}\: \: (2),\: \textrm{yaitu}:\\ &\log (2y)\times \log (4x)=\log x\times \log (8yz)\\ &\log (2y)\times \left (\log (2x)+\log 2 \right )=\log x\times \log (8yz)\\ &\log (2y)\times \log (2x)+\log (2y)\times \log 2=\log x\times \log (8yz)\\ &\log x\times \log (4y)+\log (2y)\times \log 2=\log x\times \log (8yz)\\ &\quad\textrm{persamaan di atas, persamaan}\: \: (1)\: \: \textrm{disubstitusikan}\\ &\log (2y)=\displaystyle \frac{\log x\times \log (8yz)-\log x\times \log (4y)}{\log 2}\\ &\log (2y)=\displaystyle \frac{\log x\times \left ( \log \displaystyle \frac{8yz}{4y} \right )}{\log 2}\\ &\log (2y)=\color{red}\displaystyle \frac{\log x\times \log (2z)}{\log 2}\: \color{black}......(4) \end{aligned}\\\\ &\begin{aligned}&\textrm{Perhatikan juga persamaan}\: \: (3),\: \textrm{yaitu}:\\ &\log (4y)\times \log (4x)=\log (2x)\times \log (8yz)\\ &\left (\log (2y)+\log 2 \right )\times \log (4x)=\log (2x)\times \log (8yz)\\ &\log (2y)\times \log (4x)+\log 2\times \log (4x)=\log (2x)\times \log (8yz)\\ &\log x\times \log (8yz)+\log 2\times \log (4x)=\log (2x)\times \log (8yz)\\ &\quad\textrm{di atas, persamaan}\: \: (2)\: \: \textrm{disubstitusikan}\\ &\log 2\times \log (4x)=\log (2x)\times \log (8yz)-\log x\times \log (8yz)\\ &\log 2\times \log (4x)=\log (8yz)\times \left ( \log (2x)-\log x \right )\\ &\log 2\times \log (4x)=\log (8yz)\times \left ( \log \displaystyle \frac{2x}{x} \right )\\ &\log 2\times \log (4x)=\log (8yz)\times \log 2\\ &\log 4x=\log (8yz)\\ &4x=8yz\\ &\displaystyle \frac{x}{z}=\color{red}2y\: \color{black}....(5) \end{aligned} \end{array}$
$.\: \: \qquad\begin{aligned}&\textrm{dari persamaan}\: \: (4)\: \: \textrm{dan}\: \: (5)\\ &\log (2y)=\displaystyle \frac{\log x\times \log (2z)}{\log 2}\\ &\log \left ( \displaystyle \frac{x}{z} \right )=\displaystyle \frac{\log x\times \log (2z)}{\log 2}\\ &\log 2\left ( \log x-\log z \right )=\log x\times \log (2z)\\ &\log 2\times \log x-\log 2\times \log z=\log x\times \left ( \log 2+\log z \right )\\ &\log 2\times \log x-\log 2\times \log z=\log x\times \log 2+\log x\times \log z\\ &-\log 2\times \log z=\log x\times \log z\\ &\log 2^{-1}=\log x\\ &\displaystyle \frac{1}{2}=\color{red}x\: \color{black}.....(6) \end{aligned}$
$.\: \: \qquad\begin{array}{|c|c|}\hline \textrm{persamaan}\: \: (2)&\textrm{Menentukan nilai}\: \: z\\\hline \begin{aligned} \log 2y\times \log (4x)&=\log x\times \log (8yz)\\ \log 2y\times \log (4(2yz))&=\log x\times \log (8yz)\\ \log 2y\times \log (8yz)&=\log x\times \log (8yz)\\ \log (2y)&=\log x\\ 2y&=x\\ y&=\displaystyle \frac{1}{2}x\\ &=\displaystyle \frac{1}{2}\times \frac{1}{2}\\ &=\color{red}\displaystyle \frac{1}{4}\: \color{black}.....(7)\\ & \end{aligned}&\begin{aligned}&\\ x&=2yz\\ \displaystyle \frac{1}{2}&=2\left ( \displaystyle \frac{1}{4} \right )z\\ 1&=z\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array}$
$.\: \: \qquad\begin{aligned}&\textrm{maka nilai untuk}\: \: xy^{5}z\: \: \textrm{adalah}\\ &xy^{5}z=\left ( \displaystyle \frac{1}{2} \right ).\left ( \displaystyle \frac{1}{4} \right )^{5}.1\\ &=\displaystyle \frac{1}{2\times 4^{5}}\\ &=\displaystyle \frac{1}{2\times \left ( 2^{2} \right )^{5}}=\displaystyle \frac{1}{2^{1+10}}\\ &=\displaystyle \frac{1}{2^{11}}=\displaystyle \frac{1}{2^{^{\frac{11}{1}}}}=\displaystyle \frac{1}{2^{^{\frac{p}{q}}}}\\ &\begin{cases} p & =11 \\ q & =1 \end{cases}\quad \textrm{dan jelas bahwa} \: \: p\: \: \textrm{dan}\: \: q\: \: \textrm{saling prima}\\ &\textrm{Jadi},\\ &p+q=11+1=\color{red}12 \end{aligned}$
DAFTAR PUSTAKA
- Idris, M., Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
- Sembiring, S. 2002. Olimpiade Matematika untuk SMU. Bandung: YRAMA WIDYA.
