# Fundamental Laws of Logarithms – Part 2

## Fundamental Laws of Logarithms – Part 2

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

Proof: let $${{\log }_{a}}m=x$$,

$${{a}^{x}}=m\ (or)\ {{\left( {{a}^{x}} \right)}^{n}}={{m}^{n}}$$,

$${{a}^{xn}}={{m}^{n}}$$,

$${{\log }_{a}}{{m}^{n}}=nx$$,

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

Hence Proved.

4. For a > 1, if $${{\log }_{a}}{{x}_{2}}>{{x}_{1}}\ \ then\ \ 0<{{x}_{2}}<{{a}^{{{x}_{1}}}}$$

Proof: $${{\log }_{a}}{{x}_{2}}>{{x}_{1}}$$,

$${{\log }_{a}}{{x}_{2}}>{{x}_{1}}{{\log }_{a}}a$$,

$${{\log }_{a}}{{x}_{2}}>{{\log }_{a}}{{a}^{{{x}_{1}}}}$$,

$${{x}_{2}}>{{a}^{{{x}_{1}}}}$$  (since a > 1),

Hence proved.

5. For 0 < a < 1, if $${{\log }_{a}}{{x}_{2}}>{{x}_{1}}\ \ then\ \ 0<{{x}_{2}}<{{a}^{{{x}_{1}}}}$$

Proof: $${{\log }_{a}}{{x}_{2}}>{{x}_{1}}$$,

$${{\log }_{a}}{{x}_{2}}>{{x}_{1}}{{\log }_{a}}a$$,

$${{\log }_{a}}{{x}_{2}}>{{\log }_{a}}{{a}^{{{x}_{1}}}}$$,

$$0<{{x}_{2}}<{{a}^{{{x}_{1}}}}$$,

(since 0 < a < 1),

Hence proved.

6. If m, a, b > 0 and a ≠ 1, b ≠ 1, then $${{\log }_{a}}m=\frac{{{\log }_{b}}m}{{{\log }_{b}}a}$$

Proof: Let $${{\log }_{a}}m=x$$. Then, $${{a}^{x}}=m$$,

Taking log on both sides,

$${{\log }_{b}}{{a}^{x}}={{\log }_{b}}m$$,

$$x{{\log }_{b}}a={{\log }_{b}}m$$,

(since $$x={{\log }_{a}}m\,$$),

$${{\log }_{a}}m{{\log }_{b}}a={{\log }_{b}}m$$,

$${{\log }_{a}}m=\frac{{{\log }_{b}}m}{{{\log }_{b}}a}$$,

Put b = m,

$${{\log }_{a}}m=\frac{{{\log }_{m}}m}{{{\log }_{m}}a}$$,

$${{\log }_{a}}m=\frac{1}{{{\log }_{m}}a}$$.

Example: Find the value of $${{\log }_{2}}\left( 2\sqrt[3]{9}-2 \right)+{{\log }_{2}}\left( 12\sqrt[3]{3}+4+4\sqrt[3]{9} \right)$$

solution:

$${{\log }_{2}}\left( 2\sqrt[3]{9}-2 \right)+{{\log }_{2}}\left( 12\sqrt[3]{3}+4+4\sqrt[3]{9} \right)$$,

$${{\log }_{2}}\left( {{\left( 72 \right)}^{1/3}}-{{\left( 8 \right)}^{1/3}} \right)+{{\log }_{2}}\left( {{\left( 72 \right)}^{2/3}}+{{\left( 8 \right)}^{2/3}}+{{\left( 72 \right)}^{1/3}}{{\left( 8 \right)}^{1/3}} \right)$$,

$${{\log }_{2}}\left( \left( {{\left( 72 \right)}^{1/3}}-{{\left( 8 \right)}^{1/3}} \right)\left( {{\left( 72 \right)}^{2/3}}+{{\left( 8 \right)}^{2/3}}+{{\left( 72 \right)}^{1/3}}{{\left( 8 \right)}^{1/3}} \right) \right)$$,

$${{\log }_{2}}\left( 72-8 \right)={{\log }_{2}}64$$,

$${{\log }_{2}}{{2}^{6}}=6$$.