Lanjutan Materi Fungsi Logaritma

 $\text{B. Sifat-Sifat Logaritma}$

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

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

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

$\begin{array}{rl}(\text{a})& \log 2=0,3010\\ (\text{b})& \log 3=0,4771\\ (\text{c})& \log 5=0,6990\\ (\text{d})& \log 7=0,8451\end{array}$

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{l}\\ 1.& {}^2\log 3+{}^2\log 8-{}^2\log 24=....\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}{}^2\log 3+{}^2\log 8-{}^2\log 24& ={}^2\log \left({\displaystyle \frac{3\times 8}{24}}\right)\\ & ={}^2\log 1\\ & ={}^2\log 2^0\\ & =0\end{array}\end{array}$

$\begin{array}{l}\\ 2.& {}^2\log 12+{}^2\log 8-{}^2\log 24=....\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}{}^2\log 12+{}^2\log 8-{}^2\log 24& ={}^2\log \left({\displaystyle \frac{12\times 8}{24}}\right)\\ & ={}^2\log 4\\ & ={}^2\log 2^2\\ & =2\end{array}\end{array}$

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

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

$\begin{array}{l}\\ 5.& \text{Jika}{}^{2017}\log {\displaystyle \frac{1}{x}={}^x\log {\displaystyle \frac{1}{y}={}^y\log {\displaystyle \frac{1}{2017}}}}\\ & \text{maka hasil dari}(2x-3y)\\ \\ & \text{Jawab}:\\ \\ & \begin{array}{rl}{}^{2017}\log {\displaystyle \frac{1}{x}}& ={}^x\log {\displaystyle \frac{1}{y}={}^y\log {\displaystyle \frac{1}{2017}}}\\ {}^{2017}\log {\displaystyle \frac{1}{x}}& ={}^y\log {\displaystyle \frac{1}{2017}}\\ {}^{2017}\log {\displaystyle x^{-1}}& ={}^y\log {\displaystyle (2017{)}^{-1}}\\ -{}^{2017}\log x& =-{}^y\log 2017\\ {}^{2017}\log x& ={}^y\log 2017\\ & \text{dipenuhi saat}\\ x& =y=2017\end{array}\\ \\ & (2x-3y)=2x-3x=-x=-2017\end{array}$