Lanjutan Sifat-Sifat Logaritma (Kelas X MIPA)

Sifat-Sifat Logaritma

\begin{array}{|c|}\hline \begin{matrix}\ \bullet \quad a^{^{^{a}\log b}}=b\\ \bullet \quad ^{a}\log b\times d=\: ^{a}\log b+ \: ^{a}\log d\\ \bullet \quad ^{a}\log \displaystyle \frac{b}{d}=\: ^{a}\log b-\: ^{a}\log d\\ \bullet \quad ^{a}\log b=\displaystyle \frac{^{^{x}}\log b}{^{^{x}}\log a}\\ \bullet \quad ^{a}\log b=\displaystyle \frac{1}{^{^{b}}\log a}\\ \bullet \quad ^{a}\log b=m\Rightarrow \: ^{b}\log a=\displaystyle \frac{1}{m}\\ \bullet \quad ^{^{a^{n}}}\log b^{m}=\: \displaystyle \frac{m}{n}\times \: ^{a}\log b\\ \bullet \quad ^{a}\log b\times \: ^{b}\log c\times \: ^{c}\log d=\: ^{a}\log d\\ \bullet \quad ^{a}\log a=1\\ \bullet \quad ^{a}\log 1=0\\ \bullet \quad ^{a}\log a^{n}=n\\ \bullet \quad \log b=\: ^{10}\log b\ \end{matrix}\\\hline \end{array}.

\colorbox{yellow}{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ \fbox{1}.&^{2}\log 12+\: ^{2}\log 8-\: ^{2}\log 24=...\: .\\\\ &\textrm{Solusi}:\\ &\begin{aligned}^{2}\log 12+\: ^{2}\log 8-\: ^{2}\log 24&=\: ^{2}\log \left ( \displaystyle \frac{12\times 8}{24} \right )\\ &=\: ^{2}\log 16\\ &=\: ^{2}\log 2^{4}\\ &=4 \end{aligned} \end{array}.

\color{blue}\begin{array}{ll}\\ \fbox{2}.&\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\\\\ &\color{black}\textrm{Solusi}:\\ &\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}{^{5}\log 2\times \: ^{2}\log 3\times \left ( ^{3}\log 3+\: ^{3}\log 7 \right )}\\ &=\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}.

Untuk cara penyelesaian seperti di atas, tidak harus demikian. Sebagai misal saya memberikan proses yang berbeda misal untuk No. 2 (c), tetapi tetap akan menghasilkan hasil yang sana, yaitu:

\color{magenta}\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}}{^{2}\log 3\times \: ^{3}\log 21}\\ &=\displaystyle \frac{\displaystyle \frac{1}{^{2}\log 5}}{^{2}\log 3\times \: ^{3}\log (3\times 7)}\\ &=\displaystyle \frac{\displaystyle \frac{1}{^{2}\log 5}}{^{2}\log 3\times \: \left ( ^{3}\log 3+\: ^{3}\log 7 \right )}\\ &=\displaystyle \frac{\displaystyle \frac{1}{b}}{c(1+a)}\\ &=\displaystyle \frac{1}{bc(1+a)} \end{aligned}.

\color{blue}\begin{array}{ll}\\ \fbox{3}.&\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}:\\ &\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 \color{black}\blacksquare \end{aligned}\\\hline \end{array} \end{array}.

\color{red}\begin{array}{ll}\\ \fbox{4}.&\textrm{Jika}\: \: ^{2019}\log \displaystyle \frac{1}{x}=\: ^{x}\log \displaystyle \frac{1}{y}=\: ^{y}\log \displaystyle \frac{1}{2019}\: ,\: \textrm{maka hasil dari}\: \: \left ( 2x-3y \right )\\\\ &\textrm{Jawab}:\\\\ &\begin{aligned}^{2019}\log \displaystyle \frac{1}{x}&=\: ^{x}\log \displaystyle \frac{1}{y}=\: ^{y}\log \displaystyle \frac{1}{2019}\\ ^{2019}\log \displaystyle \frac{1}{x}&=\: ^{y}\log \displaystyle \frac{1}{2019}\\ ^{2019}\log \displaystyle x^{-1}&=\: ^{y}\log \displaystyle (2019)^{-1}\\ -\:\: ^{2019}\log x&=-\: \: ^{y}\log 2019\\ ^{2019}\log x&=\: ^{y}\log 2019,&\textnormal{dipenuhi saat}\: \: x=y=2019 \end{aligned}\\\\ &(2x-3y)=2x-3x=-x=-2019 \end{array}.

\color{green}\begin{array}{ll}\\ \fbox{5}.&\textrm{Bila diberikan}\: \: ^{2}\log \left ( ^{8}\log x \right )=\: ^{8}\log \left ( ^{2}\log x \right ),\: \textrm{maka hasil dari}\: \: \left ( ^{2}\log x \right )^{2}\\\\ &\textrm{Jawab}:\\\\ &\begin{aligned}^{2}\log \left ( ^{8}\log x \right )&=\: ^{8}\log \left ( ^{2}\log x \right )\\ ^{2}\log \left ( ^{8}\log x \right )&=\: ^{2^{3}}\log \left ( ^{2}\log x \right )\\ ^{2}\log \left ( ^{8}\log x \right )&=\: ^{2}\log \left ( ^{2}\log x \right )^{\frac{1}{3}}\\ ^{8}\log x&=\left ( ^{2}\log x \right )^{\frac{1}{3}}\\ \left (^{2^{3}}\log x \right )^{3}&=\: ^{2}\log x\\ \left ( \displaystyle \frac{1}{3}\: .^{2}\log x \right )^{3}&=\: ^{2}\log x\\ \displaystyle \frac{1}{27}\: .\left (^{2}\log x \right )^{3}&=\: ^{2}\log x\\ \left ( ^{2}\log x \right )^{2}.\left ( ^{2}\log x \right )&=27.\: ^{2}\log x\\ \left ( ^{2}\log x \right )^{2}&=27 \end{aligned} \end{array}.

\color{blue}\begin{array}{ll}\\ \fbox{6}.&\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}\\\\ &\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 memenuhi}&\cdots \\\hline 3&\cdots &\cdots &5&50&\textrm{tidak memenuhi}&\cdots \\\hline \end{array} \end{array}.

\color{green}\begin{array}{ll}\\ \fbox{7}.&\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}:\\ &\\ &\begin{array}{|c|c|c|}\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ &\begin{cases} 60^{a}=3 & \Rightarrow \: ^{60}\log 3=a \\ 60^{b}=5 & \Rightarrow \: ^{60}\log 5=b \end{cases}\\ & \end{aligned}}\\\hline (1-a-b)&2(1-b)&\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )\\\hline \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}&\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}&\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}\\\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ 12^{^{\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )}}&=12^{^{\: ^{12}\log 2}}=2\\ & \end{aligned}}\\\hline \end{array} \end{array}.

\color{green}\begin{array}{ll}\\ \fbox{8}.&\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)\quad\Rightarrow \quad \log (2y)\times \log (2x)=\log x\times \log (4y)............1\\ ^{x}\log (2y)&=\: ^{4x}\log (8yz)\quad\Rightarrow \quad \log (2y)\times \log (4x)=\log x\times \log (8yz)............2\\ ^{2x}\log (4y)&=\: ^{4x}\log (8yz)\quad\Rightarrow \quad \log (4y)\times \log (4x)=\log (2x)\times \log (8yz)............ 3\\ & \end{aligned} \end{array}.
\color{green}\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)&\textnormal{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{green}\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)&\textnormal{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}.
\color{green}\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}.
\color{blue}\begin{array}{|c|c|c|c|}\hline \textrm{persamaan}\: \: 6&\textrm{persamaan}\: \: 5&\textrm{persamaan}\: \: 2&\textrm{Menentukan nilai}\: \: z\\\hline \begin{aligned}&\\ x=\displaystyle \frac{1}{2}&\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}&\\ x&=2yz\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\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}.
\color{green}\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}}}}\quad \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}.

Sumber Referensi

  1. Idris, M., Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
  2. Kurnianingsih, S., Kuntarti, dan Sulistiyono. 2007. Matematika SMA dan MA untuk Kelas X Semester 1 Standar Isi 2006. Jakarta: ESIS.
  3. Sembiring, S. 2002. Olimpiade Matematika untuk SMU. Bandung: YRAMA WIDYA.
  4. Sembiring, S., Suparmin, S. 2015. Pena Emas OSN Matematika SMA. Bandung: YRAMA WIDYA.

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Lanjutan: Eksponen dan Logaritma

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