# Fundamental Laws of Logarithms- Part3

## Fundamental Laws of Logarithms- Part3

7. For a, n > 0 and a ≠ 1; $${{a}^{{{\log }_{a}}n}}=n$$

Proof:

Let $${{\log }_{a}}n=x$$. Then $${{a}^{x}}=n$$.

Therefore, $${{a}^{{{\log }_{a}}n}}=n$$,

Examples:

(1) $${{3}^{{{\log }_{3}}8}}=8$$

(2) $${{2}^{3{{\log }_{2}}6}}={{2}^{{{\log }_{2}}{{6}^{3}}}}={{6}^{3}}$$

(3) $${{5}^{-3{{\log }_{5}}2}}={{5}^{{{\log }_{5}}{{2}^{-3}}}}=\frac{1}{{{2}^{3}}}$$

8. $${{\log }_{{{a}^{q}}}}{{n}^{p}}=\frac{p}{q}{{\log }_{a}}n=y$$, where a, n > 0, a ≠ 1

Proof:

Let $${{\log }_{{{a}^{q}}}}{{n}^{p}}=x\ \ and\ \ {{\log }_{a}}n=y.$$then,

$${{\left( {{a}^{q}} \right)}^{x}}={{n}^{p}}\ and\ {{a}^{y}}=n$$,

$$\left( {{a}^{qx}} \right)={{n}^{p}}\ and\ {{a}^{y}}=n$$,

$$\left( {{a}^{qx}} \right)={{n}^{p}}\ and\ {{\left( {{a}^{y}} \right)}^{p}}={{n}^{p}}$$,

$${{a}^{qx}}={{a}^{yp}}$$,

qx  = yp,

⇒  x = (p/q)y,

⇒ $${{\log }_{{{a}^{q}}}}{{n}^{p}}=\frac{p}{q}{{\log }_{a}}n$$.

Example: What is greater $$x={{\log }_{3}}5$$ (or) $$y={{\log }_{17}}25$$

Solution:

$$y={{\log }_{17}}25$$,

$$\frac{1}{y}={{\log }_{25}}17$$,

$$={{\log }_{{{5}^{2}}}}17$$,

$$=\frac{1}{2}{{\log }_{5}}17$$….(1),

$$x={{\log }_{3}}5$$,

$$\frac{1}{x}={{\log }_{5}}3$$….(2),

From equation (1) and (2),

$$\frac{1}{y}>\frac{1}{x}$$ ,

x > y.

9. $${{a}^{{{\log }_{b}}c}}={{c}^{{{\log }_{b}}a}}$$

Proof: Let $${{a}^{{{\log }_{b}}c}}=p$$$$\Rightarrow {{\log }_{b}}c={{\log }_{a}}p$$,

$$\Rightarrow \frac{\log c}{\log b}=\frac{\log p}{\log a}$$,

$$\Rightarrow \frac{\log a}{\log b}=\frac{\log p}{\log c}$$,

$${{\log }_{b}}a={{\log }_{c}}p$$,

$$p={{c}^{{{\log }_{b}}a}}\Rightarrow {{a}^{{{\log }_{b}}c}}={{c}^{{{\log }_{b}}a}}$$.