$\log_{b} m = \log_{p} m \log_{b} p$

Express the logarithm of a number on the basis of two numbers to derive this logarithmic property in mathematics.

For example, $m$ is a literal number and its logarithm to base $b$ is $x$ and also logarithm of $m$ to base $p$ is $y$. They are expressed in logarithmic form in mathematics as follows.

$\log_{b} m = x$ and $\log_{p} m = y$

They are written in exponential notation respectively as follows.

$m = b^{\displaystyle x}$ and $m = p^{\displaystyle y}$

Now, write the logarithm of one base to another base. In this case, $b$ and $p$ are taken as bases. So, the logarithm of $p$ to base $b$ and assume it is equal to $z$.

$\log_{b} p = z$ and its exponential notation is $p = b^{\displaystyle z}$.

Consider the equation $m = p^{\displaystyle y}$

Now change the base $p$ in above equation in terms of $b$ and it can be done by the equation $p = b^{\displaystyle z}$.

$m = p^{\displaystyle y} = {\Big(b^{\displaystyle z} \Big)}^y$

According to the power rule of an exponential term, it can be simplified.

$\implies m = p^{\displaystyle y} = b^{\displaystyle yz}$

$\therefore \,\,\,\,\,\, m = b^{\displaystyle yz}$

Take $yz = t$.

$\implies m = b^{\displaystyle t}$

Express this equation in logarithm form.

$\implies t = \log_{b} m$

But $t = yz$. So, replace it in the above equation.

$\implies yz = \log_{b} m$

$\implies \log_{b} m = yz$

Now, replace the values of $y$ and $z$ by their respective logarithmic functions.

$\therefore \,\,\,\,\,\, \log_{b} m = \log_{p} m \log_{b} p$.

The logarithm of a number to a base is equal to the product of logarithm of the number to any other number and logarithm of the same other number to base.

During the derivation of this theorem, an assumed base is changed by the actual base to express logarithmic property in change base form. Hence, the logarithmic identity is called as the change of base logarithmic identity but it is in product form. Therefore, this fundamental law can be called as the change of base logarithmic identity in product form.

$\log_{2} 64 = 6$

You can take any number to change the base $2$. In this example, $8$ is considered to change the base.

Find the value of log of $64$ to base $8$ and also find log of $8$ to base $2$.

$\log_{8} 64 = 2$ and $\log_{2} 8 = 3$

Now, multiply them to obtain their product.

$\log_{8} 64 \times \log{2} 8 = 2 \times 3 = 6$

$\therefore \,\,\,\,\,\, \log_{2} 64 = \log_{8} 64 \times \log_{2} 8 = 6$

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