# Logarithm Product Law

The logarithm of the product of two or more numbers is equal to summation of their logarithms.

For example, $m$ and $n$ are two numbers.

## Formula

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

### Proof

Assume $m$ is a number. It is expressed as multiplying factors of another number $b$ and the total number of multiplying factors are $x$.

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

The number $m$ is converted as multiplying factors on the basis of $b$. So, the relation between them is logarithm of $m$ to the base $b$ and it is equal to $x$.

$\log_{\displaystyle b} m = x$

Assume, $n$ is a number. It is expressed as multiplying factors of another number $b$ but the total number of multiplying factors are $y$.

$n = b^{\displaystyle y}$

The number $n$ is transformed as multiplying factors on the basis of $b$. So, the relation between them is logarithm of $n$ to the base $b$ and it is equal to $y$.

$\log_{\displaystyle b} n = y$

Multiply both $m$ and $n$ numbers.

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

The product of them is denoted by $m.n$. The bases of two multiplying exponential form expressions are equal. So, they can be transformed as a single exponential form term.

$m.n = b^{\displaystyle x+y}$

Assume, the product of them is denoted by $t$ and summation of $x$ and $y$ is denoted by $s$. It means, $t = m.n$ and $s = x + y$.

$t = b^{\displaystyle s}$

It means, the number $t$ is converted as $s$ multiplicative factors on the basis of $b$. So, express it in logarithm form.

$\log_{\displaystyle b} t = s$

Now replace the original values of letters $s$ and $t$.

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

Actually, $x$ is logarithm of $m$ to base $b$ and $y$ is logarithm of $n$ to base $b$. Substitute them to obtain the required logarithmic formula.

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

The logarithm of the product of two numbers $m$ and $n$ is equal to the summation of logarithm of $m$ and logarithm of $n$.

#### Verification

Let us verify this logarithmic identity by substituting two numbers before using it as a logarithmic formula in mathematics.

Assume $m = 2$ 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

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

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

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