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:

  1. \displaystyle {{\log }_{a}}m+{{\log }_{a}}n={{\log }_{a}}\left( mn \right)
  2. \displaystyle {{\log }_{a}}m-{{\log }_{a}}n={{\log }_{a}}\left( \frac{m}{n} \right)
  3. \displaystyle {{\log }_{a}}{{m}^{n}}=n{{\log }_{a}}m
  4. \displaystyle {{\log }_{{{a}^{b}}}}m=\frac{1}{b}{{\log }_{a}}m
  5. \displaystyle {{\log }_{n}}m=\frac{{{\log }_{a}}m}{{{\log }_{a}}n}
  6. \displaystyle {{\log }_{n}}m\times {{\log }_{m}}n=1
  7. \displaystyle {{\log }_{p}}m\times {{\log }_{q}}p\times {{\log }_{n}}q={{\log }_{n}}m
  8. \displaystyle \log 1=0
  9. \displaystyle {{\log }_{m}}m=1
  10. If \displaystyle {{\log }_{a}}x=ythen \displaystyle x={{a}^{y}}


\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

Post Author: E-Maths