Question

Prove that
$\frac{ log_{a}(x) }{ log_{ab}(x) } = 1 + log_{a}(b)$
for permissible values of letters involved in the result.

Answer 1

using below two formulas
$log_{a}(x ) \div log_{ab}(x) = log_{x}(ab) \div log_{x}(a)$
we know that
$log_{a}(b) = 1 \div log_{b}(a)$
and
$log_{a}(b) = log_{x}(b) \div log_{x}(a)$

now (using above formula)
$\frac{ log_{x}(ab) }{ log_{x}(a) } = log_{a}(ab)$

we know
log ab= log a + log b

similarly
$log_{a}(ab ) = log_{a}(a) + log_{a}(b)$

we know that
$log_{a}(a) = 1$
we get

$log_{a}(a) + log_{a}(b) = 1 + log_{a}(b)$

here we got the final answer
hence proved