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:

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

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

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

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

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

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

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