Math Doubts

Proof of Product law of Logarithms

Formula

$\log_{b}{(m \times n)}$ $\,=\,$ $\log_{b}{m}+\log_{b}{n}$

The product rule is a most commonly used logarithmic identity in logarithms. It states that logarithm of product of quantities is equal to sum of their logs. It can be proved mathematically in algebraic form by the relation between logarithms and exponents, and product rule of exponents.

Write quantities in Exponential form

$m$ and $n$ are two quantities, both quantities are expressed in product form on the basis of another quantity $b$.

The total number of multiplying factors of $b$ is $x$ and the product of them is equal to $m$.

$m$ $\,=\,$ $\underbrace{b \times b \times b \times \ldots \times b}_{\displaystyle x \, factors}$

$\implies m \,=\, b^{\displaystyle x}$

Similarly, the total number of multiplying factors of $b$ is $y$ and the product of them is equal to $n$.

$n$ $\,=\,$ $\underbrace{b \times b \times b \times \ldots \times b}_{\displaystyle y \, factors}$

$\implies n \,=\, b^{\displaystyle y}$

Finally, the quantities are expressed in exponential form as follows.

$(1) \,\,\,\,\,\,$ $m \,=\, b^{\displaystyle x}$

$(2) \,\,\,\,\,\,$ $n \,=\, b^{\displaystyle y}$

Express quantities in logarithmic form

The quantities in exponential form can be written in logarithmic form on the basis of mathematical relation between exponential and logarithmic operations.

$(1) \,\,\,\,\,\,$ $b^{\displaystyle x} \,=\, m$ $\,\, \Leftrightarrow \,\,$ $\log_{b}{m} = x$

$(2) \,\,\,\,\,\,$ $b^{\displaystyle y} \,=\, n$ $\,\,\,\, \Leftrightarrow \,\,$ $\log_{b}{n} = y$

Product of the quantities by multiplication

Multiply the quantities $m$ and $n$ for getting product of them mathematically.

$m \times n$

In fact, the values of the quantities $m$ and $n$ in exponential form are $b^{\displaystyle x}$ and $b^{\displaystyle y}$ respectively.

$\implies m \times n \,=\, b^{\displaystyle x} \times b^{\displaystyle y}$

As per product rule of exponents, the product of exponential terms having same base is equal to base is raised to the power of sum of the exponents.

$\implies m \times n \,=\, b^{\,({\displaystyle x}\,+\,{\displaystyle y})}$

The product of quantities $m$ and $n$ can be written in dot product form or simply $mn$ in mathematics.

$\implies m.n \,=\, b^{\,({\displaystyle x}\,+\,{\displaystyle y})}$

$\implies mn \,=\, b^{\,({\displaystyle x}\,+\,{\displaystyle y})}$

Transform quantity from exponential to logarithm form

Take $s = x+y$ and $p = mn$. Now, write the equation in terms of $s$ and $p$.

$\implies p = b^{\displaystyle s}$

Now express the equation in logarithmic form.

$\implies \log_{b}{p} = s$

Replace the actual values of quantities $s$ and $p$.

$\implies \log_{b}{(m.n)} = x+y$

Get the logarithmic property

The logarithm of product of two quantities $m$ and $n$ to the base $b$ is equal to sum of the quantities $x$ and $y$.

Actually, $x \,=\, \log_{b}{m}$ and $y \,=\, \log_{b}{n}$. Replace them to get the property for the product rule of logarithms.

$\,\,\, \therefore \,\,\,\,\,\, \log_{b}{(m.n)}$ $\,=\,$ $\log_{b}{m}+\log_{b}{n}$

It is proved that logarithm of product of two quantities to a base is equal to sum their logs to the same base. This fundamental is not limited to two quantities and it can be applied to more than two quantities. Therefore, this logarithmic identity is used as a formula in mathematics.

$\log_{\displaystyle b}{(m.n.o \cdots)}$ $=$ $\log_{b} m$ $+$ $\log_{b} n$ $+$ $\log_{b} o$ $\large + \cdots$



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