# Fundamental Laws of Logarithms – Part 1

## Fundamental Laws of Logarithms – Part 1

1.For m, n > 0, a ≠ 1; $${{\log }_{a}}\left( mn \right)={{\log }_{a}}m+{{\log }_{a}}n$$

Proof: Let $${{\log }_{a}}m=x$$ and $${{\log }_{a}}n=y$$,

$${{\log }_{a}}m=x\Rightarrow m={{a}^{x}}$$…(1),

$${{\log }_{a}}n=y\Rightarrow n={{a}^{y}}$$….(2),

From equation (1) and (2),

$$mn={{a}^{x}}{{a}^{y}}$$,

$$mn={{a}^{x+y}}$$,

$${{\log }_{a}}\left( mn \right)=x+y$$,

$${{\log }_{a}}\left( mn \right)={{\log }_{a}}m+{{\log }_{a}}n$$,

Hence proved.

In general, if $${{x}_{1}},{{x}_{2}},…,{{x}_{n}}$$ are positive real numbers, then,

$${{\log }_{a}}\left( {{x}_{1}}{{x}_{2}}…{{x}_{n}} \right)={{\log }_{a}}{{x}_{1}}+{{\log }_{a}}{{x}_{2}}+….+{{\log }_{a}}{{x}_{n}}$$,

2. For m, n, a>0, a ≠ 1; $${{\log }_{a}}\left( \frac{m}{n} \right)={{\log }_{a}}m-{{\log }_{a}}n$$.

Proof: Let $${{\log }_{a}}m=x$$ and $${{\log }_{a}}n=y$$

$${{\log }_{a}}m=x\Rightarrow m={{a}^{x}}$$…(1),

$${{\log }_{a}}n=y\Rightarrow n={{a}^{y}}$$….(2),

From equation (1) and (2),

$$\frac{m}{n}=\frac{{{a}^{x}}}{{{a}^{y}}}$$,

$$\frac{m}{n}={{a}^{x-y}}$$,

$${{\log }_{a}}\left( \frac{m}{n} \right)=x-y={{\log }_{a}}m-{{\log }_{a}}n$$,

$${{\log }_{a}}\left( \frac{m}{n} \right)={{\log }_{a}}m-{{\log }_{a}}n$$,

Hence proved.

Example: Solve $${{\log }_{e}}\left( \frac{a+b}{2} \right)=\left( {{\log }_{e}}a+{{\log }_{e}}b \right)$$

Solution:

$${{\log }_{e}}\left( \frac{a+b}{2} \right)=\left( {{\log }_{e}}a+{{\log }_{e}}b \right)$$,

$${{\log }_{e}}\left( \frac{a+b}{2} \right)={{\log }_{e}}\left( ab \right)$$,

$$\left( \frac{a+b}{2} \right)=\left( ab \right)$$,

$$a+b=2ab$$.