Logarithms are very useful for lengthy calculation.
There are two types of logs: (1) log (2) ln (Natural log)
(1) The base is 10 in log
(2) The base is e in natural log
Here, we will discuss the properties of log.
Some important properties with example are given below:
- \(\displaystyle {{\log }_{a}}m+{{\log }_{a}}n={{\log }_{a}}\left( mn \right)\)
- \(\displaystyle {{\log }_{a}}m-{{\log }_{a}}n={{\log }_{a}}\left( \frac{m}{n} \right)\)
- \(\displaystyle {{\log }_{a}}{{m}^{n}}=n{{\log }_{a}}m\)
- \(\displaystyle {{\log }_{{{a}^{b}}}}m=\frac{1}{b}{{\log }_{a}}m\)
- \(\displaystyle {{\log }_{n}}m=\frac{{{\log }_{a}}m}{{{\log }_{a}}n}\)
- \(\displaystyle {{\log }_{n}}m\times {{\log }_{m}}n=1\)
- \(\displaystyle {{\log }_{p}}m\times {{\log }_{q}}p\times {{\log }_{n}}q={{\log }_{n}}m\)
- \(\displaystyle \log 1=0\)
- \(\displaystyle {{\log }_{m}}m=1\)
- If \(\displaystyle {{\log }_{a}}x=y\)then \(\displaystyle x={{a}^{y}}\)
NOTE:
\(\displaystyle {{\log }_{a}}m+{{\log }_{a}}n\ne {{\log }_{a}}\left( m+n \right)\)
\(\displaystyle {{\log }_{a}}m-{{\log }_{a}}n\ne {{\log }_{a}}\left( m-n \right)\)
Example 1:
Simplify \(\displaystyle \log \left( \sqrt[3]{4}\,\times \sqrt[4]{3} \right)\)
Solution 1:
\(\displaystyle \log \left( \sqrt[3]{4}\times \sqrt[4]{3} \right)\)
\(\displaystyle \begin{array}{l}=\log \left( \sqrt[3]{4} \right)+\log \left( \sqrt[4]{3} \right)\\=\log {{\left( 4 \right)}^{{}^{1}\!\!\diagup\!\!{}_{3}\;}}+\log {{\left( 3 \right)}^{{}^{1}\!\!\diagup\!\!{}_{4}\;}}\\=\frac{1}{3}\log 4+\frac{1}{4}\log 3\\=\frac{1}{3}\log {{2}^{2}}+\frac{1}{4}\log 3\\=\frac{2}{3}\log 2+\frac{1}{4}\log 3\end{array}\)
Example 2:
Simplify \(\displaystyle 2{{\log }_{10}}5+{{\log }_{10}}8-{{\log }_{10}}2\)
Solution 2:
\(\displaystyle 2{{\log }_{10}}5+{{\log }_{10}}8-{{\log }_{10}}2\)
\(\displaystyle \begin{array}{l}={{\log }_{10}}{{5}^{2}}+{{\log }_{10}}8-{{\log }_{10}}2\\={{\log }_{10}}\left( \frac{25\times 8}{2} \right)\\={{\log }_{10}}\left( 100 \right)\\={{\log }_{10}}{{10}^{2}}\\=2{{\log }_{10}}10\\=2\times 1\\=2\end{array}\)
Now, solve the problems bases on log. If any doubt, visit our discussion forum http://onlinepadho.com/forum/
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