$\log_{b} m = \dfrac{\log_{a} m}{\log_{a} b}$

$m$ and $a$ are two literals and the logarithm of $m$ to base $a$ is expressed in logarithmic form as $\log_{a} m$.

As per the base change rule of the logarithm in product form, the logarithm of $m$ to base $a$ can be expressed in product form by considering another literal number $b$.

$\log_{a} m = \log_{b} m \times \log_{a} b$

$\implies \log_{b} m \times \log_{a} b = \log_{a} m$

$\therefore \,\,\,\,\,\, \log_{b} m = \dfrac{\log_{a} m}{\log_{a} b}$

The logarithm of $m$ to base $b$ can be expressed in division form by considering $a$ as a common base to both logarithmic functions. It is called as the change of base law of logarithm in division form.

The literal $a$ can be any number. So, it can be a number $10$ to express the logarithmic functions as the common logarithms in the ratio.

$\implies \log_{b} m = \dfrac{\log_{10} m}{\log_{10} b}$

$\therefore \,\,\,\,\,\, \log_{b} m = \dfrac{\log m}{\log b}$

For example, the logarithm of $27$ to base $3$ is $3$ in logarithmic system.

$\log_{3} 27 = 3$

Now, calculate logarithm of $27$ and logarithm of $3$ by taking any number as a common base. In this example, common logarithmic system is used by taking number $10$ as base to both logarithmic terms.

$\log_{10} 27 = 1.43236$

$\log_{10} 3 = 0.47712$

Now, calculate the quotient of log of $27$ by log of $3$.

$\dfrac{\log_{10} 27}{\log_{10} 3} = \dfrac{1.43236}{0.47712}$

$\implies \dfrac{\log_{10} 27}{\log_{10} 3} = 3$

The quotient of them is $3$.

$\therefore \,\,\,\,\,\, \log_{3} 27 = \dfrac{\log_{10} 27}{\log_{10} 3} = 3$

Therefore, the example has successfully verified the change base logarithmic law in division form.

Copyright © 2012 - 2017 Math Doubts, All Rights Reserved