Question

Prove the following Identities :
i. $\frac{{\log }_{a}n}{{\log }_{ab}n}=1+{\log }_{a}b$
ii. ${\log }_{ab}x=\frac{{\log }_{a}x{\log }_{b}x}{{\log }_{a}x+{\log }_{b}x}$

Answer 1

i) Consider, $\frac{{\log }_{a}n}{{\log }_{ab}n}$

$=\frac{\frac{1}{{\log }_{n}a}}{\frac{1}{{\log }_{n}ab}}$ ($\because {\log }_{a}b=\frac{1}{{\log }_{b}a}$)

$=\frac{{\log }_{n}ab}{{\log }_{n}a}$
$=\frac{{\log }_{n}a+{\log }_{n}b}{{\log }_{n}a}$

$=1+\frac{{\log }_{n}b}{{\log }_{n}a}$

$=1+{\log }_{a}b$
ii) Consider, ${\log }_{ab}x$

$=\frac{1}{{\log }_{x}ab}$

$=\frac{1}{{\log }_{x}a+{\log }_{x}b}$

$=\frac{1}{\frac{1}{{\log }_{a}x}+\frac{1}{{\log }_{b}x}}$ ($\because {\log }_{a}b=\frac{1}{{\log }_{b}a}$)

$=\frac{{\log }_{b}x{\log }_{b}x}{{\log }_{b}x+{\log }_{a}x}$