Contoh Soal 8 Fungsi Logaritma (Persamaan Logaritma)

$\begin{array}{ll}\\ 36.&\textrm{Persamaan}\: \: ^{x}\log 2+\: ^{x}\log (3x-4)=2\\ &\textrm{mempunyai akar}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2},\: \: \textrm{maka}\\ &\textrm{nilai}\: \: x_{1}+x_{2}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \textrm{b}.&3\\ \textrm{c}.&4\\ \color{red}\textrm{d}.&6\\ \textrm{e}.&8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\color{purple}\textrm{Alternatif 1}\\ &^{x}\log 2+\: ^{x}\log (3x-4)=2\\ &\Leftrightarrow \: ^{x}\log 2(3x-4)=2\\ &\Leftrightarrow \: ^{x}\log 6x-8=2\\ &\Leftrightarrow \: \color{black}6x-8=x^{2}\\ &\Leftrightarrow \: \color{black}x^{2}-6x+8=0,\: \: \color{purple}\textrm{dengan}\: \begin{cases} a &=1 \\ b &=-6 \\ c &=8 \end{cases}\\ &\Leftrightarrow \: x_{1}+x_{2}=-\displaystyle \frac{b}{a}\\ &\Leftrightarrow \: x_{1}+x_{2}=\color{red}-\displaystyle \frac{-6}{1}=6\\ &\color{purple}\textrm{Alternatif 2}\\ &\Leftrightarrow \: \color{black}x^{2}-6x+8=0\\ &\Leftrightarrow \: (x-2)(x-4)\\ &\Leftrightarrow \: x_{1}=2\: \: \textrm{atau}\: \: x_{2}=4\\ &\Leftrightarrow \: x_{1}+x_{2}=\color{red}2+4=6 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 37.&\textrm{Jika}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\: \: \textrm{memenuhi}\\ &(\log x)(2\log x-3)=\log 100\\ &\textrm{maka}\: \: x_{1}\times x_{2}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&100\\ \color{red}\textrm{b}.&10\sqrt{10}\\ \textrm{c}.&\sqrt{10}\\ \textrm{d}.&-\sqrt{10}\\ \textrm{e}.&-10\sqrt{10} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&(\log x)(2\log x-3)=\color{red}\log 100\\ &\Leftrightarrow (\log x)\left ( 2\log x-3 \right )=\color{red}2\\ &\color{black}\Leftrightarrow 2\log ^{2}x-3\log x-2=0\: \color{purple}\begin{cases} a &=2 \\ b &=-3 \\ c &=-2 \end{cases}\\ &\Leftrightarrow \log x_{1}+\log x_{2}=-\displaystyle \frac{-3}{2}=\frac{3}{2}\\ &\Leftrightarrow \log \left ( x_{1}\times x_{2} \right )=1\displaystyle \frac{1}{2}\\ &\Leftrightarrow \color{red}\left ( x_{1}\times x_{2} \right )=10^{1\frac{1}{2}}=10\sqrt{10} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 38.&\textrm{Persamaan}\\ & 10^{\,^{\, ^{2}}\log x^{2} }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ & \textrm{mempunyai dua akar yaitu}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\\ &\textrm{Nilai}\: \: x_{1}\times x_{2}=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \textrm{b}.&-5\\ \color{red}\textrm{c}.&2\\ \textrm{d}.&5\\ \textrm{e}.&10 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&10^{\,^{\, ^{2}}\log x^{2} }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ &10^{\, 2^{\, ^{2}}\log x }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ &\color{black}\textrm{adalah persamaan kuadrat dalam}\: \: \color{red}10^{\,^{\, ^{2}}\log x }\\ &\color{black}\textrm{Misalkan}\: \: \color{red}p=10^{\,^{\, ^{2}}\log x },\: \: \textrm{maka persamaan}\\ &\color{black}\textrm{menjadi}\: \: \color{purple}p^{2}-7p+10=0\: \begin{cases} a & =1 \\ b & =-7 \\ c & =10 \end{cases}\\ & \color{red}\textrm{Karena nilai}\: \: \color{black}p_{1}\times p_{2}=\displaystyle \frac{c}{a}\: \: \textrm{maka}\\ &10^{\,^{\, ^{2}}\log x_{1} }\times 10^{\,^{\, ^{2}}\log x_{2} }=\displaystyle \frac{10}{1}=10\\ &10^{\,^{\, ^{2}}\log x_{1}\: +\: ^{2}\log x_{2} }=10\\ &10^{\,^{\, ^{2}}\log x_{1}\: +\: ^{2}\log x_{2} }=10^{1}\\ &\Leftrightarrow \: ^{2}\log x_{1}\: +\: ^{2}\log x_{2}=1\\ &\Leftrightarrow \: ^{2}\log x_{1}\times x_{2}=1\\ &\Leftrightarrow \: \color{red}x_{1}\times x_{2}=2^{1}=2\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 39.&\textrm{Nilai}\: \: x\: \: \: \: \textrm{yang memenuhi}\\ &^{x}\log (x+12)-3.\: ^{x}\log 4+1=0\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\\ \textrm{b}.&2\\ \color{red}\textrm{c}.&4\\ \textrm{d}.&8\\ \textrm{e}.&16 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&^{x}\log (x+12)-3.\: ^{x}\log 4+1=0\\ &^{x}\log (x+12)-\: ^{x}\log 4^{3}=-1\\ &^{x}\log \displaystyle \frac{x+12}{64}=-1\\ &\displaystyle \frac{x+12}{64}=x^{-1}=\frac{1}{x}\\ &x+12=\displaystyle \frac{64}{x}\\ &\color{purple}x^{2}+12x-64=0\\ &\color{purple}(x+16)(x-4)=0\\ &x=-16\: \: \color{black}\textrm{atau}\: \: \color{red}x=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 40.&\textrm{Nilai}\: \: x\: \: \: \: \textrm{yang memenuhi}\\ &\displaystyle \frac{^{2}\log (2x-3)}{^{2}\log x}-\: ^{x}\log (x+6)+\displaystyle \frac{1}{^{(x+2)}\log x}=1\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-1\: \: \textrm{dan}\: \: 6\\ \textrm{b}.&-2\: \: \textrm{dan}\: \: 6\\ \textrm{c}.&-1\\ \textrm{d}.&-2\\ \color{red}\textrm{e}.&6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&\displaystyle \frac{^{2}\log (2x-3)}{^{2}\log x}-\: ^{x}\log (x+6)+\displaystyle \frac{1}{^{(x+2)}\log x}=1\\ &^{x}\log (2x-3)-\: ^{x}\log (x+6)+\: ^{x}\log (x+2)=1\\ &^{x}\log (2x-3)(x+2)=1+\log (x+6)\\ &^{x}\log \displaystyle \frac{\left ( 2x^{2}+x-6 \right )}{(x+6)}=1\\ &\displaystyle \frac{\left ( 2x^{2}+x-6 \right )}{(x+6)}=x^{1}\\ &\left ( 2x^{2}+x-6 \right )=x^{2}+6x\\ &\color{purple}x^{2}-5x-6=0\\ &\color{purple}(x+1)(x-6)=0\\ &x=-1\: \: \color{black}\textrm{atau}\: \: \color{red}x=6 \end{aligned} \end{array}$

Contoh Soal 6 Fungsi Logaritma (Uraian)

$\begin{array}{l}\\ 26.&\textrm{Diketahui bahwa}\\ & ^{^{2}}\log 3=p \: \: \textrm{dan}\: \: ^{^{3}}\log 11=q,\\ &\textrm{maka nilai}\: \: ^{^{44}}\log 66=....\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}^{^{44}}\log 66&=\displaystyle \frac{^{^{^{...}}}\log 66}{^{^{^{...}}}\log 44}\\ &=\displaystyle \frac{^{^{^{...}}}\log \left (2\times 3\times 11 \right )}{^{^{^{...}}}\log \left (2^{2}\times 11 \right )}\\ &=\displaystyle \frac{^{^{^{3}}}\log 2+\: ^{^{^{3}}}\log 3+\:^{^{^{3}}}\log 11}{^{^{^{3}}}\log 2^{2}+\: ^{^{^{3}}}\log 11}\\ &=\displaystyle \frac{\frac{1}{^{^{^{2}}}\log 3}+\: ^{^{^{3}}}\log 3+\:^{^{^{3}}}\log 11}{\frac{2}{^{^{^{3}}}\log 2^{2}}+\: ^{^{^{3}}}\log 11}\\ &=\displaystyle \frac{\frac{1}{p}+1+q}{\frac{2}{p}+q}\\ &=\displaystyle \frac{\frac{1}{p}+1+q}{\frac{2}{p}+q}\times \displaystyle \frac{p}{p}\\ &=\frac{1+p+pq}{2+pq} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 27.&\textbf{(AIME 1984)}\\ &\textrm{Diketahui bahwa}\: \: x\: \: \textrm{dan}\: \: y\\ & \textrm{adalah bilangan real yang memenuhi}\\\\ &\left\{\begin{matrix} ^{^{8}}\log x+\: ^{^{4}}\log y^{2}=5\\ \\ ^{^{8}}\log y+\: ^{^{4}}\log x^{2}=7 \end{matrix}\right.\\\\ &\textrm{Tentukanlah nilai dari}\: \: xy\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}&^{^{^{8}}}\log x+\: ^{^{4}}\log y^{2}=5\\ &\Leftrightarrow \: ^{^{^{2^{3}}}}\log x+\: ^{^{^{2^{2}}}}\log y^{2}=5....(1)\\ &^{^{^{8}}}\log y+\: ^{^{4}}\log x^{2}=7\\ &\Leftrightarrow \: ^{^{^{2^{3}}}}\log y+\: ^{^{^{2^{2}}}}\log x^{2}=7....(2)\\ &\textrm{selanjutnya},&\\ &\frac{1}{3}\:. ^{^{^{2}}}\log x+\: ^{^{^{2}}}\log y=5\: \: |\times \frac{1}{3}|\\ &\Rightarrow \frac{1}{9}\: .^{^{^{2}}}\log x+\: \frac{1}{3}.\: ^{^{^{2}}}\log y=\frac{5}{3}....(3)\\ &^{^{^{2}}}\log x+\:\frac{1}{3}\: . ^{^{^{2}}}\log y=7\: \: |\times 1|\\ &\Rightarrow ^{^{^{2}}}\log x+\:\frac{1}{3}\: . ^{^{^{2}}}\log y=7....(4)\\ &\textrm{saat persamaan}\: \: (3)-(4)\\ &=-\frac{8}{9}.\: ^{^{^{2}}}\log x=\frac{5}{3}-7=-\frac{16}{3}\\ &\color{purple}\textrm{maka}\\ &^{^{^{2}}}\log x=\left ( -\frac{16}{3} \right )\left ( -\frac{9}{8} \right )\\ &^{^{^{2}}}\log x=6\Leftrightarrow x=2^{6}\Leftrightarrow x=64\\ &\color{purple}\textrm{Pada persamaan 1 selanjutnya}\\ &\frac{1}{3}.\: ^{^{^{2}}}\log x+\: ^{^{^{2}}}\log y=5\\ &\Leftrightarrow \: \: \frac{1}{3}.\: ^{^{^{2}}}\log 2^{6}+\: ^{^{^{2}}}\log y=5\\ &\Leftrightarrow \: \: \frac{1}{3}.6+\: ^{^{^{2}}}\log y=5\\ &\Leftrightarrow \: \: 2+\: ^{^{^{2}}}\log y=5\\ &\Leftrightarrow \: \: ^{^{^{2}}}\log y=5-2=3\Leftrightarrow y=2^{3}=8\\ &\textrm{Jadi},\: \: x.y=64.8=512 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 28.&\textrm{Tentukanlah nilai dari}\\ &\textrm{a}.\quad \left ( 2^{\: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: ^{^{^{9}}}\log 5} \right )\left ( 5^{\: ^{^{^{\frac{1}{5}}}}\log 2} \right )\\ &\textrm{b}.\quad \left ( ^{27}\log 125 \right )\left ( ^{25}\log \displaystyle \frac{1}{64} \right )\left ( ^{64}\log \frac{1}{9} \right )\\ &\textrm{c}.\quad \left ( ^{625}\log 19 \right )\left ( ^{7}\log \displaystyle \frac{1}{25} \right )\left ( ^{19}\log 7 \right )\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{array}{|l|}\hline \begin{aligned}\textrm{a}.\quad &\left ( 2^{\: \: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: ^{^{^{9}}}\log 5} \right )\left ( 5^{\: ^{^{^{\frac{1}{5}}}}\log 2} \right )\\ &=\left ( 2^{\: \: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: \: ^{^{^{3^{2}}}}\log 5} \right )\left ( 5^{\: \: ^{^{^{5^{-1}}}}\log 2} \right )\\ &=\left ( 2^{\: \: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: \: ^{^{^{3}}}\log 5^{^{\frac{1}{2}}}} \right )\left ( 5^{\: \: ^{^{^{5}}}\log 2^{-1}} \right )\\ &=\left ( 2^{\: ^{^{2}}\log 6} \right )\left ( 3^{\: ^{^{3}}\log \sqrt{5}} \right )\left ( 5^{\: ^{^{5}}\log \frac{1}{2}} \right )\\ &=6\times \sqrt{5}\times \frac{1}{2}\\ &=3\sqrt{5} \end{aligned}\\\hline \end{array}\\ &\color{blue}\begin{array}{|l|}\hline \begin{aligned}\textrm{b}.\quad &\left ( ^{27}\log 125 \right )\left ( ^{25}\log \displaystyle \frac{1}{64} \right )\left ( ^{64}\log \frac{1}{9} \right )\\ &=\left ( ^{^{3^{3}}}\log 5^{3} \right )\left ( ^{^{5^{2}}}\log 4^{-3} \right )\left ( ^{^{4^{3}}}\log 3^{-2} \right )\\ &=\displaystyle \frac{3}{3}.\left ( -\frac{3}{2} \right ).\left ( -\frac{2}{3} \right ).\: ^{^{3}}\log 5.\: ^{^{5}}\log 4.\: ^{^{4}}\log 3\\ &=1\end{aligned}\\\hline \end{array}\\ &\color{purple}\textrm{Pembahasan diserahkan kepada}\\ &\color{purple}\textrm{Pembaca yang budiman untuk poin c} \end{array}$

$\begin{array}{ll}\\ 29.&\textrm{Tentukanlah nilai}\: \: a+b\: \: \textrm{dimana}\: \: a\: \: \textrm{dan}\: \: b\\ &\textrm{adalah bilangan riil positif}.\\ &^{7}\log \left ( 1+a^{2} \right )-\: ^{7}\log 25=\: ^{7}\log \left ( 2ab-15 \right )-\: ^{7}\log \left ( 25+b^{2} \right )\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}&^{7}\log \displaystyle \frac{\left ( 1+a^{2} \right )}{25}=\: ^{7}\log \displaystyle \frac{\left ( 2ab-15 \right )}{\left ( 25+b^{2} \right )}\\ &\textrm{diambil}\: \textrm{persamaannya, maka}\\ &\displaystyle \frac{\left ( 1+a^{2} \right )}{25}=\displaystyle \frac{\left ( 2ab-15 \right )}{\left ( 25+b^{2} \right )}\\ &\displaystyle \left ( 1+a^{2} \right )\left ( 25+b^{2} \right )=25\left ( 2ab-15 \right )\\ &\begin{cases} \left ( 1+a^{2} \right ) & \text{ faktor dari} \: \: 25,\: \: a> 0,\: a\in \mathbb{R} \\ &\textrm{atau}\\ \left ( 25+b^{2} \right ) & \text{ faktor dari }\: \: 25,\: \: b> 0,\: b\in \mathbb{R}\: \: \: \textrm{juga} \end{cases}\\ \end{aligned}\\\\ &\color{purple}\begin{array}{|c|l|c|l|c|c|c|}\hline \textrm{No}&a&\left ( 1+a^{2} \right )&b&\left ( 25+b^{2} \right )&\textrm{Keterangan}&a+b\\\hline 1&2&5&10&125&\textrm{Memenuhi}&12\\\hline 2&7&50&\cdots &\cdots &\textrm{tidak}&\cdots \\\hline 3&\cdots &\cdots &5&50&\textrm{tidak}&\cdots \\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 30.&\textrm{Jika}\: \: 60^{a}=3\: \: \textrm{dan}\: \: 60^{b}=5\\ &\textrm{maka hasil dari}\: \: 12^{^{\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )}}\\\\ &\textrm{Jawab}:\\ &\color{blue}\textrm{Perhatikanlah bahwa}\\ &\color{blue}\begin{aligned}&\begin{cases} 60^{a}=3 & \Rightarrow \: ^{60}\log 3=a \\ 60^{b}=5 & \Rightarrow \: ^{60}\log 5=b \end{cases} \end{aligned} \\ &\color{blue}\textrm{Selanjutnya}\\ &\color{blue}\textrm{Untuk}\: \: (1-a-b),\: \textrm{maka}\\ &\color{blue}\begin{aligned}1-a-b&=1-\: ^{60}\log 3-\: ^{60}\log 5\\ &=\: ^{60}\log 60-\: ^{60}\log 3-\: ^{60}\log 5\\ &=\: ^{60}\log \displaystyle \frac{60}{3\times 5}\\ &=\: ^{60}\log 4\\ &=\: ^{60}\log 2^{2} \end{aligned} \\ &\color{blue}\textrm{Untuk}\: \: 2(1-b),\: \textrm{maka}\\ &\color{blue}\begin{aligned}2(1-b)&=2\left ( 1-\: ^{60}\log 5 \right )\\ &=2\left ( ^{60}\log 60-\: ^{60}\log 5 \right )\\ &=2\left ( ^{60}\log \displaystyle \frac{60}{5} \right )\\ &=2\left ( ^{60}\log 12 \right )\\ &=\: ^{60}\log 12^{2} \end{aligned} \\ &\color{blue}\textrm{Untuk}\: \: \left ( \displaystyle \frac{1-x-y}{2(1-b)} \right ),\: \textrm{maka}\\ &\color{blue}\begin{aligned}\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )&=\displaystyle \frac{\: ^{60}\log 2^{2}}{\: ^{60}\log 12^{2}}\\ &=\displaystyle \frac{2\: ^{60}\log 2}{2\: ^{60}\log 12}\\ &=\displaystyle \frac{\: ^{60}\log 2}{\: ^{60}\log 12}\\ &=\: ^{12}\log 2 \end{aligned}\\ &\color{blue}\textrm{Jadi},\: \: 12^{^{\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )}}=12^{^{\: ^{12}\log 2}}=2 \end{array}$

$\begin{array}{ll}\\ 31.&\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}:\\\\ &\color{blue}\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}\\\\ &\color{blue}\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)=\displaystyle \frac{\log x\times \log (2z)}{\log 2}\: ......(4) \end{aligned}\\\\ &\color{blue}\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}=2y\: ....(5) \end{aligned} \end{array}$

$.\: \: \qquad\color{purple}\begin{aligned}&\textrm{dari persamaan}\: \: (4)\: \: \textrm{dan}\: \: (5)\\ &\log (2y)=\displaystyle \frac{\log x\times \log (2z)}{\log 2}\\ &\log \left ( \displaystyle \frac{x}{z} \right )=\displaystyle \frac{\log x\times \log (2z)}{\log 2}\\ &\log 2\left ( \log x-\log z \right )=\log x\times \log (2z)\\ &\log 2\times \log x-\log 2\times \log z=\log x\times \left ( \log 2+\log z \right )\\ &\log 2\times \log x-\log 2\times \log z=\log x\times \log 2+\log x\times \log z\\ &-\log 2\times \log z=\log x\times \log z\\ &\log 2^{-1}=\log x\\ &\displaystyle \frac{1}{2}=x\: .....(6) \end{aligned}$

$.\: \: \qquad\color{blue}\begin{array}{|c|c|}\hline \textrm{persamaan}\: \: (2)&\textrm{Menentukan nilai}\: \: z\\\hline \begin{aligned} \log 2y\times \log (4x)&=\log x\times \log (8yz)\\ \log 2y\times \log (4(2yz))&=\log x\times \log (8yz)\\ \log 2y\times \log (8yz)&=\log x\times \log (8yz)\\ \log (2y)&=\log x\\ 2y&=x\\ y&=\displaystyle \frac{1}{2}x\\ &=\displaystyle \frac{1}{2}\times \frac{1}{2}\\ &=\displaystyle \frac{1}{4}\: .....(7)\\ & \end{aligned}&\begin{aligned}&\\ x&=2yz\\ \displaystyle \frac{1}{2}&=2\left ( \displaystyle \frac{1}{4} \right )z\\ 1&=z\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array}$

$.\: \: \qquad\color{purple}\begin{aligned}&\textrm{maka nilai untuk}\: \: xy^{5}z\: \: \textrm{adalah}\\ &xy^{5}z=\left ( \displaystyle \frac{1}{2} \right ).\left ( \displaystyle \frac{1}{4} \right )^{5}.1\\ &=\displaystyle \frac{1}{2\times 4^{5}}\\ &=\displaystyle \frac{1}{2\times \left ( 2^{2} \right )^{5}}=\displaystyle \frac{1}{2^{1+10}}\\ &=\displaystyle \frac{1}{2^{11}}=\displaystyle \frac{1}{2^{^{\frac{11}{1}}}}=\displaystyle \frac{1}{2^{^{\frac{p}{q}}}}\\ &\begin{cases} p & =11 \\ q & =1 \end{cases}\quad \textrm{dan jelas bahwa} \: \: p\: \: \textrm{dan}\: \: q\: \: \textrm{saling prima}\\ &\textrm{Jadi},\\ &p+q=11+1=12 \end{aligned}$

DAFTAR PUSTAKA

1. Idris, M., Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
2. Sembiring, S. 2002. Olimpiade Matematika untuk SMU. Bandung: YRAMA WIDYA.

Contoh Soal 5 Fungsi Logaritma

$\begin{array}{ll}\\ 21.&\textbf{(SPMB '04)}\\ &\textrm{Jika}\: \: a>1\: ,\: \textrm{maka penyelesaian untuk}\\ &\left (^{a}\log (2x+1) \right )\left ( ^{3}\log \sqrt{a} \right )=1\: \: \textrm{adalah}\: ....\\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 1\\ \textrm{b}.&\displaystyle 2\\ \textrm{c}.&\displaystyle 3\\ \color{red}\textrm{d}.&\displaystyle 4\\ \textrm{e}.&\displaystyle 5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\left (^{a}\log (2x+1) \right )\left ( ^{3}\log \sqrt{a} \right )&=1\\ \left ( ^{3}\log \sqrt{a} \right )\left (^{a}\log (2x+1) \right )&=1\\ \left ( ^{3}\log a^{.^{^{\frac{1}{2}}}} \right )\left (^{a}\log (2x+1) \right )&=1\\ \displaystyle \frac{1}{2}\left ( ^{3}\log a \right )\left (^{a}\log (2x+1) \right )&=1\\ ^{3}\log (2x+1)&=2\\ 2x+1&=3^{2}\\ 2x&=9-1\\ 2x&=8\\ x&=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 22.&\textbf{(SPMB '04)}\\ &\textrm{Nilai}\: \: \displaystyle \frac{\left ( ^{5}\log 10 \right )^{2}-\left ( ^{5}\log 2 \right )^{2}}{^{5}\log \sqrt{20}}\: \: \textrm{adalah}\: ....\\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\\ \textrm{b}.&\displaystyle 1\\ \color{red}\textrm{c}.&\displaystyle 2\\ \textrm{d}.&\displaystyle 4\\ \textrm{e}.&\displaystyle 5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&\displaystyle \frac{\left ( ^{5}\log 10 \right )^{2}-\left ( ^{5}\log 2 \right )^{2}}{^{5}\log \sqrt{20}}\\ &=\displaystyle \frac{\left ( ^{5}\log 10+\: ^{5}\log 2 \right )\left ( ^{5}\log 10-\: ^{5}\log 2 \right )}{^{5}\log (20)^{.^{\frac{1}{2}}}}\\ &=\displaystyle \frac{^{5}\log (10.2)\times ^{5}\log \left (\frac{10}{2} \right )}{\displaystyle \frac{1}{2}\times \: ^{5}\log 20}\\ &=2\times \left ( \displaystyle \frac{^{5}\log 20}{^{5}\log 20} \right )\times \: ^{5}\log 5\\ &=2.1.1\\ &=2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 23.&\textbf{(SPMB '03)}\\ &\textrm{Jika diketahui bahwa}\\ &^{4}\log ^{4}\log x-\: ^{4}\log ^{4}\log ^{4}\log 16=2\\ & \textrm{maka}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle ^{2}\log x=8\\ \textrm{b}.&\displaystyle ^{2}\log x=4\\ \color{red}\textrm{c}.&\displaystyle ^{4}\log x=8\\ \textrm{d}.&\displaystyle ^{4}\log x=16\\ \textrm{e}.&\displaystyle ^{16}\log x=8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{4}\log \: ^{4}\log 16=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{4}\log \: ^{4}\log 4^{2}=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{4}\log 2=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{2^{2}}\log 2^{1}=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{4}\log \left ( \displaystyle \frac{1}{2} \right )=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{2^{2}}\log 2^{-1}=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\left ( -\displaystyle \frac{1}{2} \right )=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x+\frac{1}{2}=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x=2-\displaystyle \frac{1}{2}=\frac{3}{2}\\ &\Leftrightarrow \: ^{4}\log x=4^{.\frac{3}{2}}\\ &\Leftrightarrow \: ^{4}\log x=\left (2^{2} \right )^{.^{\frac{3}{2}}}\\ &\Leftrightarrow \: ^{4}\log x=2^{3}\\ &\Leftrightarrow \: ^{4}\log x=8\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 24.&\textbf{(UMPTN '92)}\\ &\textrm{Jika}\: \: x\: \: \textrm{memenuhi persamaan}\\ &^{4}\log ^{4}\log x-\: ^{4}\log ^{4}\log ^{4}\log 16=2\\ & \textrm{maka nilai}\: \: ^{16}\log x=\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle 4\\ \textrm{b}.&\displaystyle 2\\ \textrm{c}.&\displaystyle 1\\ \textrm{d}.&\displaystyle -2\\ \textrm{e}.&\displaystyle -4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{4}\log \: ^{4}\log 16=2\\ &\textrm{menyebabkan}\\ & ^{4}\log x=8\Rightarrow x=4^{8}\\ &(\color{purple}\textrm{lihat pembahasan no.23})\\ &\textrm{maka},\\ &\: ^{16}\log x=\: ^{16}\log 4^{8}=\: ^{4^{2}}\log 4^{8}\\ &=\displaystyle \frac{8}{2}\\ &=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 25.&\textbf{(UMPTN '94)}\\ &\textrm{Hasil kali akar-akar persamaan}\\ &^{3}\log x^{.\left (^{2+\: ^{3}\log x} \right )}=15\\ & \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{1}{9}\\ \textrm{b}.&\displaystyle \frac{1}{3}\\ \textrm{c}.&\displaystyle 1\\ \textrm{d}.&\displaystyle 3\\ \textrm{e}.&\displaystyle 9 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&^{3}\log x^{.\left (^{2+\: ^{3}\log x} \right )}=15\\ &\Leftrightarrow \: \left ({2+\: ^{3}\log x} \right )^{3}\log x-15=0\\ &\Leftrightarrow 2\: ^{3}\log x+\: \left ( ^{3}\log x \right )^{2}-15=0\\ &\Leftrightarrow \: \left ( ^{3}\log x \right )^{2}+2\: ^{3}\log x-15=0\\ &\Leftrightarrow \: \left (^{3}\log x_{1}+5 \right )\left ( ^{3}\log x_{2}-3 \right )=0\\ &\Leftrightarrow \: ^{3}\log x_{1}+5=0\: \: \textrm{atau}\: \: ^{3}\log x_{1}-3=0\\ &\Leftrightarrow \: ^{3}\log x_{1}=-5\: \: \textrm{atau}\: \: ^{3}\log x_{2}=3\\ &\Leftrightarrow \: x_{1}=3^{-5}\: \: \textrm{atau}\: \: x_{2}=3^{3}\\ &\qquad \textrm{maka}\\ &\Leftrightarrow \: x_{1}\times x_{2}=3^{-5}\times 3^{3}=3^{-5+3}=3^{-2}\\ &\Leftrightarrow \qquad =\displaystyle \frac{1}{3^{2}}\\ &\Leftrightarrow \qquad =\frac{1}{9} \end{aligned} \end{array}$

Contoh Soal 4 Fungsi Logaritma

$\begin{array}{ll}\\ 16.&\textbf{(UMPTN '01)}\\ &\textrm{Jika}\: \: ^{10}\log x=b\: ,\: \textrm{maka}\: \: ^{10x}\log 100=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{b+1}\\ \color{red}\textrm{b}.&\displaystyle \frac{2}{b+1}\\ \textrm{c}.&\displaystyle \frac{1}{b}\\ \textrm{d}.&\displaystyle \frac{2}{b}\\ \textrm{e}.&\displaystyle \frac{2}{10b} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&^{10x}\log 100\\ &=\displaystyle \frac{\log 100}{\log 10x}\\ &=\displaystyle \frac{^{10}\log 100}{^{10}\log 10x},\quad \color{magenta}\textrm{pilih basis 10}\\ &\color{purple}\textrm{alasannya: supaya sama dengan soal}\\ &=\displaystyle \frac{^{10}\log 10^{2}}{^{10}\log 10+\: ^{10}\log x}\\ &=\displaystyle \frac{2}{1+b}\: \: \textrm{atau}\\ &=\frac{2}{b+1} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 17.&\textbf{(UM UGM '03)}\\ &\textrm{Jika}\: \: ^{4}\log 6=m+1\: ,\: \textrm{maka}\: \: ^{9}\log 8=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{3}{4m-2}\\ \color{red}\textrm{b}.&\displaystyle \frac{3}{4m+2}\\ \textrm{c}.&\displaystyle \frac{3}{2m+4}\\ \textrm{d}.&\displaystyle \frac{3}{2m-4}\\ \textrm{e}.&\displaystyle \frac{3}{2m+2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\color{purple}\textrm{Sebelumnya perhatikanlah}\\ &^{4}\log 6=m+1\\ &\Leftrightarrow \: ^{2^{2}}\log (2.3)^{1}=m+1\\ &\Leftrightarrow \: \displaystyle \frac{1}{2}\times \: ^{2}\log (2.3)=m+1\\ &\Leftrightarrow \: \displaystyle \frac{1}{2}\times \: \left (^{2}\log 2 +\: ^{2}\log 3 \right )=m+1\\ &\Leftrightarrow \: \displaystyle \frac{1}{2}\times \: \left (1 +\: ^{2}\log 3 \right )=m+1\\ &\Leftrightarrow \: 1 +\: ^{2}\log 3=2m+2\\ &\Leftrightarrow \: ^{2}\log 3=2m+1\\ &\color{purple}\textrm{Selanjutnya adalah}:\\ &^{9}\log 8=\: \displaystyle \frac{1}{^{8}\log 9}\\ &=\: \displaystyle \frac{1}{^{2^{3}}\log 3^{2}}\\ &=\: \displaystyle \frac{1}{\displaystyle \frac{2}{3}\: ^{2}\log 3}\\ &=\: \displaystyle \frac{3}{2\: ^{2}\log 3}\\ &=\: \displaystyle \frac{3}{2(2m+1)}\\ &=\: \displaystyle \frac{3}{4m+2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&\textbf{(UMPTN '00)}\\ &\textrm{Jika}\: \: ^{3}\log 5=p\: \: \textrm{dan}\: \: ^{3}\log 4=q,\\ &\textrm{maka}\: \: ^{4}\log 15=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{pq}{1+p}\\ \textrm{b}.&\displaystyle \frac{p+q}{pq}\\ \color{red}\textrm{c}.&\displaystyle \frac{p+1}{pq}\\ \textrm{d}.&\displaystyle \frac{p+1}{q+1}\\ \textrm{e}.&\displaystyle \frac{pq}{1-p} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&^{4}\log 15\\ &=\displaystyle \frac{^{...}\log 15}{^{...}\log 4},\: \: \color{purple}\textrm{pilih basis 5}\\ &\color{magenta}\textrm{mengapa tidak pilih basis selain 5}\\ &\color{magenta}\textrm{lihat penyebut, di sana terdapat numerus 4}\\ &\color{magenta}\textrm{pada soal, pasangan numerus 4 adalah 5},\\ &\color{black}\textrm{makanya basis 5 dipilih, bukan yang lain}\\ &=\displaystyle \frac{^{5}\log 15}{^{5}\log 4}=\displaystyle \frac{^{5}\log (3.5)}{^{5}\log 4}\\ &=\displaystyle \frac{^{5}\log 3+\: ^{5}\log 5}{^{5}\log 4}=\displaystyle \frac{\displaystyle \frac{1}{^{3}\log 5}+\: ^{5}\log 5}{^{5}\log 4}\\ &=\displaystyle \frac{\displaystyle \frac{1}{p}+1}{q}=\displaystyle \frac{1+p}{pq},\: \: \textrm{atau}\\ &=\displaystyle \frac{p+1}{pq} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 19.&\textbf{(UMPTN '94)}\\ &\textrm{Jika}\: \: ^{6}\log 5=a\: \: \textrm{dan}\: \: ^{5}\log 4=b,\\ &\textrm{maka}\: \: ^{4}\log 0,24=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{a+2}{ab}\\ \textrm{b}.&\displaystyle \frac{2a+1}{ab}\\ \textrm{c}.&\displaystyle \frac{a-2}{ab}\\ \textrm{d}.&\displaystyle \frac{2a+1}{2ab}\\ \color{red}\textrm{e}.&\displaystyle \frac{1-2a}{ab} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&^{4}\log 0,24\\ &=\displaystyle \frac{^{...}\log 0,24}{^{...}\log 4}=\frac{^{...}\log \displaystyle \frac{6}{25}}{^{...}\log 4},\: \: \color{purple}\textrm{pilih basis 5}\\ &\color{magenta}\textrm{mengapa tidak pilih basis selain 5}\\ &\color{magenta}\textrm{lihat penyebut, di sana terdapat numerus 4}\\ &\color{magenta}\textrm{pada soal, pasangan numerus 4 adalah 5},\\ &\color{black}\textrm{makanya basis 5 dipilih, bukan yang lain}\\ &=\frac{^{5}\log \displaystyle \frac{6}{25}}{^{5}\log 4}=\displaystyle \frac{^{5}\log 6-\: ^{5}\log 25}{^{5}\log 4}\\ &=\displaystyle \frac{\displaystyle \frac{1}{^{6}\log 5}-\: ^{5}\log 5^{2}}{^{5}\log 4}=\displaystyle \frac{\displaystyle \frac{1}{a}-2}{b}=\frac{1-2a}{ab} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 20.&\textbf{(SPMB '05)}\\ &\textrm{Jika}\: \: ^{3}\log 2=p\: \: \textrm{dan}\: \: ^{2}\log 7=q,\\ &\textrm{maka}\: \: ^{14}\log 54=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{p+3}{p+q}\\ \textrm{b}.&\displaystyle \frac{2p}{p+q}\\ \color{red}\textrm{c}.&\displaystyle \frac{p+3}{p(q+1)}\\ \textrm{d}.&\displaystyle \frac{p+q}{p(q+1)}\\ \textrm{e}.&\displaystyle \frac{p(q+1)}{p+q} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&^{14}\log 54\\ &=\displaystyle \frac{^{...}\log 54}{^{...}\log 14}=\frac{^{...}\log (2.27)}{^{...}\log (2.7)},\: \: \color{purple}\textrm{pilih basis 2}\\ &\color{magenta}\textrm{mengapa tidak pilih basis selain 2}\\ &\color{magenta}\textrm{lihat penyebut, di sana terdapat numerus 7}\\ &\color{magenta}\textrm{pada soal, pasangan numerus 7 adalah 2},\\ &\color{black}\textrm{makanya basis 2 dipilih, bukan yang lain}\\ &=\frac{^{2}\log (2.27)}{^{2}\log (2.7)}=\displaystyle \frac{^{2}\log 2+\: ^{2}\log 27}{^{2}\log 2+\: ^{2}\log 7}\\ &=\displaystyle \frac{^{2}\log 2+\: ^{2}\log 3^{3}}{^{2}\log 2+\: ^{2}\log 7}=\displaystyle \frac{^{2}\log 2+\left (3\times \: \displaystyle \frac{1}{^{3}\log 2} \right )}{^{2}\log 2+\: ^{2}\log 7}\\ &=\displaystyle \frac{1+\displaystyle \frac{3}{p}}{1+q}\\ &=\displaystyle \frac{p+3}{p(q+1)} \end{aligned} \end{array}$

Contoh Soal 3 Fungsi Logaritma

$\begin{array}{ll}\\ 11.&\textrm{Nilai dari}\\ & \displaystyle \frac{1}{6}.\, ^{2}\log 25-\: \frac{1}{3}.\, ^{2}\log 10\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -3\\ \textrm{b}.&\displaystyle -2\\ \textrm{c}.&-2 \displaystyle \frac{1}{2}\\ \textrm{d}.& -\displaystyle \frac{1}{2}\\ \color{red}\textrm{e}.&- \displaystyle \frac{1}{3} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&=\: \displaystyle \frac{1}{6}.\, ^{2}\log 25-\: \frac{1}{3}.\, ^{2}\log 10\\ &=\displaystyle \frac{1}{3}.\frac{1}{2}.\, ^{2}\log 25-\: \frac{1}{3}.\, ^{2}\log 10\\ &=\frac{1}{3}\left ( ^{2}\log 25^{\frac{1}{2}} -\: ^{2}\log 10\right )\\ &=\frac{1}{3}\left ( ^{2}\log 5-\: ^{2}\log 10 \right )\\ &=\frac{1}{3}\left ( ^{2}\log \frac{5}{10} \right )\\ &=\frac{1}{3}\left ( ^{2}\log \frac{1}{2} \right )\\ &=\frac{1}{3}\left ( ^{2}\log 2^{-1} \right )\\ &=-\frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 12.&\textrm{Nilai dari}\\ & ^{2}\log \left ( ^{2}\log \sqrt{\sqrt{2}} \right )\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -4\\ \color{red}\textrm{b}.&\displaystyle -2\\ \textrm{c}.&-1 \displaystyle \frac{1}{2}\\ \textrm{d}.& -\displaystyle \frac{1}{2}\\ \textrm{e}.&- \displaystyle \frac{1}{4} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&=\: ^{2}\log \left ( ^{2}\log \sqrt{\sqrt{2}} \right )\\ &=\: ^{2}\log \left ( ^{2}\log 2^{\frac{1}{4}} \right )\\ &=\: ^{2}\log \frac{1}{4}\\ &=\: ^{2}\log \left (2 \right )^{-2}\\ &=-2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 13.&\textbf{(UMPTN '99)}\\ &\textrm{Diketahui}\: \: \log 2=0,3010\: \: \textrm{dan}\: \: \log 3=0,4771\\ &\textrm{maka}\: \: \log \left ( \sqrt[3]{2}\times \sqrt{3} \right )\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 0,1505\\ \textrm{b}.&\displaystyle 0,1590\\ \textrm{c}.&\displaystyle 0,2007\\ \color{red}\textrm{d}.&\displaystyle 0,3389\\ \textrm{e}.&\displaystyle 0,3891 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\log \left ( \sqrt[3]{2}\times \sqrt{3} \right )\\ &=\log \sqrt[3]{2}+\log \sqrt{3}\\ &=\log 2^{\frac{1}{3}}+\log 3^{\frac{1}{2}}\\ &=\displaystyle \frac{1}{3}\log 2+\displaystyle \frac{1}{2}\log 3\\ &=\displaystyle \frac{1}{3}(0,3010)+\displaystyle \frac{1}{2}(0,4771)\\ &=0,1003+0,2386\\ &=0,3389 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 14.&\textbf{(UMPTN '98)}\\ &\textrm{Nilai}\: \: ^{a}\log \displaystyle \frac{1}{b}\times \: ^{b}\log \displaystyle \frac{1}{c^{2}}\times \: ^{c}\log \displaystyle \frac{1}{a^{3}}\: =\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle -6\\ \textrm{b}.&\displaystyle 6\\ \textrm{c}.&\displaystyle \frac{b}{a^{2}c}\\ \textrm{d}.&\displaystyle \frac{a^{2}c}{b}\\ \textrm{e}.&\displaystyle -\frac{1}{6} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&^{a}\log \displaystyle \frac{1}{b}\times \: ^{b}\log \displaystyle \frac{1}{c^{2}}\times \: ^{c}\log \displaystyle \frac{1}{a^{3}}\\ &=\: ^{a}\log \displaystyle b^{-1}\times \: ^{b}\log \displaystyle c^{-2}\times \: ^{c}\log \displaystyle a^{-3}\\ &=(-1).(-2).(-3)\times \: ^{a}\log \displaystyle a\times \: ^{b}\log \displaystyle c\times \: ^{c}\log \displaystyle a\\ &=-6\times \: ^{a}\log a\\ &=-6\times 1\\ &=-6 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 15.&\textbf{(UMPTN '01)}\\ &\textrm{Jika}\: \: \displaystyle \frac{^{2}\log a}{^{3}\log b}=m\: \: \textrm{dan}\: \: \displaystyle \frac{^{3}\log a}{^{2}\log b}=n\\ &\textrm{dengan}\: \: a> 1,\: b> 1,\: \textrm{maka}\: \: \displaystyle \frac{m}{n}=....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle ^{2}\log 3\\ \textrm{b}.&\displaystyle ^{3}\log 2\\ \textrm{c}.&\displaystyle ^{4}\log 9\\ \textrm{d}.&\displaystyle \left ( ^{3}\log 2 \right )^{2}\\ \color{red}\textrm{e}.&\displaystyle \left ( ^{2}\log 3 \right )^{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\displaystyle \frac{m}{n}&=\displaystyle \frac{\displaystyle \frac{^{2}\log a}{^{3}\log b}}{\displaystyle \frac{^{3}\log a}{^{2}\log b}}\\ &=\displaystyle \frac{^{2}\log a\times \: ^{2}\log b}{^{3}\log b\times \: ^{3}\log a}\\ &=\displaystyle \frac{^{2}\log a\times \: \displaystyle \frac{1}{^{b}\log 2}}{^{3}\log b\times \: \displaystyle \frac{1}{^{a}\log 3}}\\ &=\displaystyle \frac{^{2}\log a\times \: ^{a}\log 3}{^{3}\log b\times \: ^{b}\log 2}\\ &=\displaystyle \frac{^{2}\log 3}{^{3}\log 2}=\displaystyle \frac{^{2}\log 3}{\displaystyle \frac{1}{^{2}\log 3}}\\ &=\left ( ^{2}\log 3 \right )^{2} \end{aligned} \end{array}$

Contoh Soal 2 Fungsi Logaritma

$\begin{array}{ll}\\ 6.&\textrm{Nilai dari}\: \: ^{\sqrt{2}}\log 16\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 10\\ \textrm{b}.&\displaystyle 9\\ \color{red}\textrm{c}.& \displaystyle 8\\ \textrm{d}.& \displaystyle 6\\ \textrm{e}.& \displaystyle 4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&=\: ^{\sqrt{2}}\log 16\\ &=\: ^{\displaystyle 2^{\frac{1}{2}}}\log 2^{4}\\ &=\displaystyle \frac{4}{\frac{1}{2}}\times \: ^{2}\log 2\\ &=8\end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Nilai dari}\: \: ^{\sqrt{5}}\log \sqrt{125}\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -3\\ \textrm{b}.&\displaystyle -2\\ \textrm{c}.& \displaystyle 2\\ \color{red}\textrm{d}.& \displaystyle 3\\ \textrm{e}.& \displaystyle 5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&=\: ^{\sqrt{5}}\log \sqrt{125}\\ &=\: ^{\sqrt{5}^{1}}\log \left ( \sqrt{5} \right )^{3}\\ &=\displaystyle \frac{3}{1}\times \: ^{\sqrt{5}}\log \sqrt{5}\\ &=3\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Nilai dari}\: \: ^{\sqrt{\sqrt{2}}}\log \sqrt{8\sqrt{8}}\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 4\\ \textrm{b}.&\displaystyle 6\\ \textrm{c}.& \displaystyle 8\\ \color{red}\textrm{d}.& \displaystyle 9\\ \textrm{e}.& \displaystyle 12 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&=\: ^{\sqrt{\sqrt{2}}}\log \sqrt{8\sqrt{8}}\\ &=\: ^{\sqrt[4]{2}}\log \left ( 8\left ( 8 \right )^{\frac{1}{2}} \right )^{\frac{1}{2}}\\ &=\: ^{2^{\frac{1}{4}}}\log 8^{\left (\frac{1}{2}+\frac{1}{4} \right )}\\ &=\: ^{2^{\frac{1}{4}}}\log 2^{3\left ( \frac{3}{4} \right )}\\ &=\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}\times \: ^{2}\log 2\\ &=9 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 9.&\textrm{Nilai dari}\\ & ^{6}\log 8 +\: ^{6}\log \displaystyle \frac{9}{2}\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 4\\ \textrm{b}.&\displaystyle 3\\ \textrm{c}.&3 \displaystyle \frac{1}{2}\\ \textrm{d}.& 2\displaystyle \frac{1}{2}\\ \color{red}\textrm{e}.& \displaystyle 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&=\: ^{6}\log 8 +\: ^{6}\log \displaystyle \frac{9}{2}\\ &=\: ^{6}\log 8\times \frac{9}{2}\\ &=\: ^{6}\log 36\\ &=\: ^{6}\log 6^{2}\\ &=2\\ &\\ &\\ & \end{aligned} \end{array}$

$\begin{array}{ll}\\ 10.&\textrm{Nilai dari}\\ & ^{2}\log \displaystyle \frac{4}{3}+\: 2.\, ^{2}\log \sqrt{12}\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 6\\ \color{red}\textrm{b}.&\displaystyle 4\\ \textrm{c}.&3 \displaystyle \frac{1}{2}\\ \textrm{d}.& 2\displaystyle \frac{1}{2}\\ \textrm{e}.& \displaystyle 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&=\: ^{2}\log \displaystyle \frac{4}{3}+\: 2.\, ^{2}\log \sqrt{12}\\ &=\: ^{2}\log \frac{4}{3}+\: ^{2}\log \left ( \sqrt{12} \right )^{2}\\ &=\: ^{2}\log \frac{4}{3}+\: ^{2}\log 12\\ &=\: ^{2}\log \frac{4}{3}\times 12\\ &=\: ^{2}\log 16\\ &=\: ^{2}\log 2^{4}\\ &=4 \end{aligned} \end{array}$

Contoh Soal 1 Fungsi Logaritma

$\begin{array}{ll}\\ 1.&\textrm{Nilai dari}\: \: ^{2}\log \displaystyle \frac{1}{32}\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -7\\ \color{red}\textrm{b}.&\displaystyle -5\\ \textrm{c}.& \displaystyle -3\\ \textrm{d}.& \displaystyle -2\\ \textrm{e}.& 5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&=\: ^{2}\log \displaystyle \frac{1}{32}\\ &=\: ^{2^{1}}\log 2^{-5}\\ &=\displaystyle \frac{-5}{1}\times \: ^{2}\log 2\\ &=-5\\ &\\ & \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Nilai dari}\: \: ^{0,333...}\log 0,111....\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{3}\\ \textrm{b}.&\displaystyle \frac{1}{2}\\ \color{red}\textrm{c}.& 2\\ \textrm{d}.& 3\\ \textrm{e}.& 6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&=\: ^{0,333...}\log 0,111...\\ &=\: ^{\frac{1}{3}}\log \frac{1}{9}\\ &=\: ^{\frac{1}{3}}\log \left (\frac{1}{3} \right )^{2}\\ &=2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Nilai dari}\: \: ^{5}\log 25\sqrt{5}\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{5}{2}\\ \textrm{b}.&\displaystyle \frac{3}{2}\\ \textrm{c}.& \displaystyle \frac{1}{2}\\ \textrm{d}.& \displaystyle 2\\ \textrm{e}.& 3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&=\: ^{5}\log 25\sqrt{5}\\ &=\: ^{5^{1}}\log 5^{2}.5^{\frac{1}{2}}\\ &=\: ^{5^{1}}\log 5^{\frac{5}{2}}\\ &=\displaystyle \frac{\frac{5}{2}}{1}\times \: ^{5}\log 5\\ &=\displaystyle \frac{5}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Nilai dari}\: \: ^{\sqrt{3}}\log 81\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 12\\ \textrm{b}.&\displaystyle 10\\ \textrm{c}.& \displaystyle 9\\ \color{red}\textrm{d}.& \displaystyle 8\\ \textrm{e}.& 6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&=\: ^{\sqrt{3}}\log 81\\ &=\: ^{\displaystyle 3^{\frac{1}{2}}}\log 3^{4}\\ &=\displaystyle \frac{4}{\frac{1}{2}}\times \: ^{3}\log 3\\ &=8 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Nilai dari}\: \: ^{\frac{1}{3}}\log \displaystyle \frac{1}{243}\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 6\\ \color{red}\textrm{b}.&\displaystyle 5\\ \textrm{c}.& \displaystyle 4\\ \textrm{d}.& \displaystyle 3\\ \textrm{e}.& \displaystyle 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&=\: ^{\frac{1}{3}}\log \displaystyle \frac{1}{243}\\ &=\: ^{\left (\frac{1}{3} \right )^{1}}\log \displaystyle \left (\frac{1}{3} \right )^{5}\\ &=\displaystyle \frac{5}{1}\times \: ^{\frac{1}{3}}\log \frac{1}{3}\\ &=5 \end{aligned} \end{array}$

Lanjutan Materi Fungsi Logaritma

 $\color{blue}\textrm{B. Sifat-Sifat Logaritma}$

Jika syarat logaritma memenuhi untuk bilangan yang diposisikan sebagai basis dan numerus, maka akan berlaku sifat-sifat loaritma berikut:

$\color{purple}\begin{aligned}(1)\quad&a^{{^{a}}\textrm{log b}}=b\\ (2)\quad&^{a}\log (b.c)=\: ^{a}\log b+\: ^{a}\log c\\ (3)\quad&^{a}\log \left ( \displaystyle \frac{b}{c} \right )=\: ^{a}\log b-\: ^{a}\log c\\ (4)\quad&^{a}\log b=\: \displaystyle \frac{^{x}\log b}{^{x}\log c}\\ (5)\quad&^{a}\log b=\: \displaystyle \frac{1}{^{b}\log a}\\ (6)\quad&^{a}\log b=n\Rightarrow \: ^{b}\log a=\displaystyle \frac{1}{n}\\ (7)\quad&^{a^{m}}\log b^{n}=\displaystyle \frac{n}{m}\times \: ^{a}\log b\\ (8)\quad&^{a}\log b\times \: ^{b}\log c\times \: ^{c}\log p=\: ^{a}\log p\\ (9)\quad&^{a}\log a=1\\ (10)\quad&^{a}\log a^{n}=\: n\\ (11)\quad&^{a}\log 1=\: 0\\ (12)\quad&^{.}\log b=\: ^{10}\log b\\ \end{aligned}$

ada yang tak kalah penting untuk diketahui walaupun kadang sebagian orang menganggap tidak perlu dituliskan, di sini saya tuliskan, yaitu:

$\color{blue}\begin{aligned}(\textrm{a})\quad&\log 2=0,3010\\ (\textrm{b})\quad&\log 3=0,4771\\ (\textrm{c})\quad&\log 5=0,6990\\ (\textrm{d})\quad&\log 7=0,8451\\ \end{aligned}$

$\LARGE\color{blue}\fbox{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&^{2}\log 3+\: ^{2}\log 8-\: ^{2}\log 24=...\: .\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}^{2}\log 3+\: ^{2}\log 8-\: ^{2}\log 24&=\: ^{2}\log \left ( \displaystyle \frac{3\times 8}{24} \right )\\ &=\: ^{2}\log 1\\ &=\: ^{2}\log 2^{0}\\ &=0 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&^{2}\log 12+\: ^{2}\log 8-\: ^{2}\log 24=...\: .\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}^{2}\log 12+\: ^{2}\log 8-\: ^{2}\log 24&=\: ^{2}\log \left ( \displaystyle \frac{12\times 8}{24} \right )\\ &=\: ^{2}\log 4\\ &=\: ^{2}\log 2^{2}\\ &=2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Diketahui}\: \: ^{3}\log 7=a,\: \: ^{5}\log 2=b,\: \: \textrm{dan}\: \: ^{2}\log 3=c\\ &\textrm{Nyatakanlah logaritma berikut dalam bentuk}\: \: a,\: b,\: \textrm{dan}\: \: c,\: \: \textrm{yaitu}:\\ &\textrm{a}.\quad ^{7}\log 3\\ &\textrm{b}.\quad ^{4}\log 5\\ &\textrm{c}.\quad ^{21}\log 5\\ &\textrm{d}.\quad ^{6}\log 7\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{array}{|l|l|}\hline \begin{aligned}&\\ ^{7}\log 3&=\displaystyle \frac{1}{^{3}\log 7}\\ &=\displaystyle \frac{1}{a}\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}^{4}\log 5&=\displaystyle \frac{1}{^{5}\log 4}\\ &=\displaystyle \frac{1}{^{5}\log 2^{2}}\\ &=\displaystyle \frac{1}{2\: ^{5}\log 2}\\ &=\displaystyle \frac{1}{2b} \end{aligned}\\\hline \begin{aligned}&\\ ^{21}\log 5&=\displaystyle \frac{^{...}\log 5}{^{...}\log 21}\\ &=\displaystyle \frac{^{2}\log 5}{^{2}\log 21}\\ &=\displaystyle \frac{\displaystyle \frac{1}{^{5}\log 2}}{\displaystyle \frac{^{3}\log 21}{^{3}\log 2}}\\ &=\displaystyle \frac{\displaystyle \frac{1}{^{5}\log 2}}{\displaystyle \frac{^{3}\log 3\times 7}{^{3}\log 2}}\\ &=\displaystyle \frac{1}{bc(1+a)} \end{aligned}&\begin{aligned}&\\ ^{6}\log 7&=\displaystyle \frac{^{3}\log 7}{^{3}\log 6}\\ &=\displaystyle \frac{^{3}\log 7}{^{3}\log 2\times 3}\\ &=\displaystyle \frac{^{3}\log 7}{^{3}\log 2+\: ^{3}\log 3}\\ &=\displaystyle \frac{^{3}\log 7}{\displaystyle \frac{1}{^{2}\log 3}+\: ^{3}\log 3}\\ &=\displaystyle \frac{a}{\displaystyle \frac{1}{c}+1}\\ &=\displaystyle \frac{ac}{1+c}\\ &\\ & \end{aligned}\\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Diketahui bahwa}\: \: \: ^{4}\log 5=a\\ &\textrm{a}.\quad \textrm{Carilah nilai}\: \: \: ^{4}\log 10\\ &\textrm{b}.\quad \textrm{Tunjukkan bahwa}\: \: \: ^{0,1}\log 1,25=\displaystyle \frac{2-2a}{2a+1}\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{array}{|l|l|}\hline \begin{aligned}&\\ ^{4}\log 10&=\: ^{4}\log (2\times 5)\\ &=\: ^{4}\log 2+\: ^{4}\log 5\\ &=\: ^{2^{2}}\log 2+\: a\\ &=\: \displaystyle \frac{1}{2}.\: ^{2}\log 2+\: a\\ &=\: \displaystyle \frac{1}{2}+a\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}&\\ &^{0,1}\log 1,25\\ &=\displaystyle \frac{^{4}\log 1,25}{^{4}\log 0,1}\\ &=\displaystyle \frac{^{4}\log \displaystyle \frac{125}{100}}{^{4}\log \displaystyle \frac{1}{10}}\\ &=\displaystyle \frac{^{4}\log 125-\: ^{4}\log 100}{^{4}\log 10^{-1}}\\ &=\displaystyle \frac{^{4}\log 5^{3}-\: ^{4}\log 10^{2}}{-\: ^{4}\log 10}\\ &=\displaystyle \frac{3.\: ^{4}\log 5-\: 2.\: ^{4}\log 10}{-\: ^{4}\log 10}\\ &=\displaystyle \frac{3a-2\left ( \displaystyle \frac{1}{2}+a \right )}{-\left ( \displaystyle \frac{1}{2}+a \right )}\\ &=\displaystyle \frac{a-1}{-a-\displaystyle \frac{1}{2}}\times \displaystyle \frac{-2}{-2}\\ &=\displaystyle \frac{2-2a}{2a+1}\qquad \blacksquare \end{aligned}\\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Jika}\: \: ^{2017}\log \displaystyle \frac{1}{x}=\: ^{x}\log \displaystyle \frac{1}{y}=\: ^{y}\log \displaystyle \frac{1}{2017}\\ &\textrm{maka hasil dari}\: \: \left ( 2x-3y \right )\\\\ &\textrm{Jawab}:\\\\ &\color{blue}\begin{aligned}^{2017}\log \displaystyle \frac{1}{x}&=\: ^{x}\log \displaystyle \frac{1}{y}=\: ^{y}\log \displaystyle \frac{1}{2017}\\ ^{2017}\log \displaystyle \frac{1}{x}&=\: ^{y}\log \displaystyle \frac{1}{2017}\\ ^{2017}\log \displaystyle x^{-1}&=\: ^{y}\log \displaystyle (2017)^{-1}\\ -\:\: ^{2017}\log x&=-\: \: ^{y}\log 2017\\ ^{2017}\log x&=\: ^{y}\log 2017\\ &\textrm{dipenuhi saat}\\ x&=y=2017 \end{aligned}\\\\ &(2x-3y)=2x-3x=-x=-2017 \end{array}$

Fungsi Logaritma

$\color{blue}\textrm{A. Pendahuluan}$

Logaritma merupakan invers(balikan) dari perpangkatan

Secara definisi:

$\LARGE\color{purple}\boxed{a^{c}=b\Rightarrow ^{a}\log b=c}$, tetapi di sini diberikan syarat bahwa bilangan basis/dasar perpangkatannya harus berupa bilangan real positif dan tidak sama dengan satu serta bilangan pangkatnya(ekponen) harus berupa bilangan real positif juga.

Perhatikanlah ringkasannya

$^{a}\log b=c\: \: \: \begin{cases} a & \textrm{syaratnya}:a> 0,\: a\neq 1 \\ &\color{magenta}\textrm{selanjutnya disebut basis}\\ b & \textrm{syaratnya}:b>0 \\ & \color{magenta}\textrm{selanjutnya disebut}\: \: \color{blue}\textbf{numerus}\\ c&\textrm{tidak ada syarat apapun}\\ &\color{magenta}\textrm{selanjutnya disebut hasil logaritma} \end{cases}$

Contoh berikut adalah mengubah bentuk perpangkatan ke dalam logaritma yang memenuhi persyaratan

$\color{purple}\begin{aligned} (1)\quad&2^{4}=16\Rightarrow \: ^{2}\log 16=4\\ (2)\quad&2^{3}=8\Rightarrow \: ^{2}\log 8=3\\ (3)\quad&2^{2}=4\Rightarrow \: ^{2}\log 4=2\\ (4)\quad&2^{1}=2\Rightarrow \: ^{2}\log 2=1\\ (5)\quad&2^{0}=1\Rightarrow \: ^{2}\log 1=0\\ (6)\quad&2^{-1}=\displaystyle \frac{1}{2}=0,5\Rightarrow \: ^{2}\log \displaystyle \frac{1}{2}=-1\\ (7)\quad&2^{-2}=\displaystyle \frac{1}{4}=0,25\Rightarrow \: ^{2}\log \displaystyle \frac{1}{4}=-2\\ (8)\quad&2^{-3}=\displaystyle \frac{1}{8}=0,125\Rightarrow \: ^{2}\log \displaystyle \frac{1}{8}=-3\\ (9)\quad&2^{-4}=\displaystyle \frac{1}{16}=0,0625\Rightarrow \: ^{2}\log \displaystyle \frac{1}{16}=-4\\ \end{aligned}$

Berikut contoh kebalikan di atas yang tidak memenuhi definisi logaritma yang ada, yaitu:

$\color{blue}\begin{aligned} (1)\quad&(-2)^{4}=16\Rightarrow \: ^{(-2)}\log 16=\cdots \\ (2)\quad&(-2)^{3}=-8\Rightarrow \: ^{(-2)}\log (-8)=\cdots \\ (3)\quad&(-2)^{2}=4\Rightarrow \: ^{(-2)}\log 4=\cdots \\ (4)\quad&(-2)^{1}=-2\Rightarrow \: ^{(-2)}\log (-2)=\cdots \\ (5)\quad&(-2)^{0}=1\Rightarrow \: ^{(-2)}\log 1=\cdots \\ (6)\quad&(-2)^{-1}=-\displaystyle \frac{1}{2}\Rightarrow \: ^{(-2)}\log \left (-\displaystyle \frac{1}{2} \right )=\cdots \\ (7)\quad&(-2)^{-2}=\displaystyle \frac{1}{4}\Rightarrow \: ^{(-2)}\log \displaystyle \frac{1}{4}=\cdots \\ (8)\quad&(-2)^{-3}=-\displaystyle \frac{1}{8}\Rightarrow \: ^{(-2)}\log \left (-\displaystyle \frac{1}{8} \right )=\cdots \\ (9)\quad&(-2)^{-4}=\displaystyle \frac{1}{16}\Rightarrow \: ^{(-2)}\log \displaystyle \frac{1}{16}=\cdots \\ \end{aligned}$