Product Law of Logarithms


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


$\large m$ and $\large n$ are two variables. On the basis of another variable $\large b$, the values of both $\large m$ and $\large n$ are written as $\large b^x$ and $\large b^y$ in exponential notation.

$\large b^x = m \,\,\,$ and $\,\,\, \large b^y = n$

Step: 1

The exponential relations can be written in logarithmic form as per the relation between exponential and logarithmic operations.

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

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

Step: 2

Multiply both variables $\large m$ and $\large n$ to obtain product of them.

$\large m \times n = b^x \times b^y$

According to the product of exponents, the product of two exponential terms having same base can be written as follows.

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

The product of $\large m$ and $\large n$ is represented as $m.n$ or simply, $mn$ in mathematics.

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

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

Step: 3

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

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

Now express the expression in logarithmic form.

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

Now replace the actual values of literals $\large s$ and $\large p$.

$\implies \large \log_{b} (mn) = x+y$

Step: 4

According to step one, $\large x = \log_{b} m$ and $\large y = \log_{b} n$. Replace, $\large x$ and $\large y$ in above logarithmic equation.

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

Therefore, it is proved that the logarithm of a product of two numbers is transformed as the sum of logarithms of each number. It can also be used in reverse operation.

The property of the logarithm is in product form. Hence, the logarithmic identity is called as the product rule of logarithm.

The fundamental product law of logarithm is not limited to two numbers. It can be used to more two numbers as well. It can be proved by using above procedure and it is written in general form as follows.

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


The fundamental product rule of logarithm can be verified in mathematics by using numerical method.

Take $m = 2$ and $n = 3$

$\log_{\displaystyle 10} 2 = 0.3010$ and $\log_{\displaystyle 10} 3 = 0.4771$. Now add both of them.

$\log_{\displaystyle 10} 2 + \log_{\displaystyle 10} 3 = 0.3010 + 0.4771$

$\log_{\displaystyle 10} 2 + \log_{\displaystyle 10} 3 = 0.7781$

Now calculate logarithm of product of both numbers

$\log_{\displaystyle 10} (2 \times 3) = \log_{\displaystyle 10} (6)$

$\implies \log_{\displaystyle 10} 6 = 0.7781$

$\therefore \,\,\,\,\, \log_{\displaystyle 10} (2 \times 3)$ $=$ $\log_{\displaystyle 10} 2 + \log_{\displaystyle 10} 3$ $=$ $0.7781$

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