Math Doubts

Fundamental Power Logarithmic identity

Formula

$\log_{b}{(m^{\displaystyle n})}$ $\,=\,$ $n\log_{b}{m}$

Introduction

The quantities in exponential notation are appeared in some cases in logarithms. It is not possible to find the log of the exponential form quantities directly unless the exponential quantity is evaluated. So, it requires a special mathematical law to evaluate the logarithm of the quantities in exponential notation.

However, there is a mathematical property to find the log of exponential form quantities fundamentally. According to this property, the logarithm of a quantity in exponential form is equal to the product of the exponent and the logarithm of the base of exponential quantity. It is called the fundamental power rule of the logarithms.

Math form

Let $b$, $m$ and $n$ represent three quantities in algebraic form, and the $m$ raised to the power of $n$ forms a quantity and It is written in exponential notation as $m^{\displaystyle n}$ in mathematics.

The log of $m$ raised to the power of $n$ to the base $b$ is written in logarithmic system as follows.

$\log_{b}{(m^{\displaystyle n})}$

The logarithm of $m$ raised to the $n$-th power is equal to the product of the exponent $n$ and the log of the base quantity $m$ to the base $b$.

$\therefore\,\,\,$ $\log_{b}{(m^{\displaystyle n})}$ $\,=\,$ $n \times \log_{b}{(m)}$

It is an algebraic form of the fundamental logarithmic power law and it is used as a formula to find the logarithm of any quantity in exponential notation.

Verification

Evaluate $\log_{2}{\big(2^5\big)}$

Let us verify the power logarithmic identity arithmetically by an example.

$\implies$ $\log_{2}{\big(2^5\big)}$ $\,=\,$ $\log_{2}{(2 \times 2 \times 2 \times 2 \times 2)}$

$\implies$ $\log_{2}{\big(2^5\big)}$ $\,=\,$ $\log_{2}{\Big(\underbrace{2 \times 2 \times 2 \times 2 \times 2}_{\displaystyle 5 \ factors}\Big)}$

$\,\,\,\therefore\,\,\,\,\,\,$ $\log_{2}{\big(2^5\big)}$ $\,=\,$ $5$

It is calculated that the binary logarithm of $2$ raised to the power of $5$ is equal to $5$. Now, calculate the product of the exponent $5$ and the binary logarithm of $2$.

$\implies$ $5 \times \log_{2}{(2)}$ $\,=\,$ $5 \times 1$

$\,\,\,\therefore\,\,\,\,\,\,$ $5 \times \log_{2}{(2)}$ $\,=\,$ $5$

It is verified that the logarithm of an exponential quantity is equal to the product of exponent and logarithm of the base of exponential quantity.

$\,\,\,\therefore\,\,\,\,\,\,$ $\log_{2}{\big(2^5\big)}$ $\,=\,$ $5 \times \log_{2}{(2)}$

Proof

Learn how to derive the fundamental power logarithmic identity in algebraic form.

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