Question

If a > 0 , c > 0 , b = $\sqrt{a}c$ , $a\ne 1$ $c\ne 1$ $ac\ne 1$ $n>1$ prove that $\frac{lo{g}_{a}n}{lo{g}_{c}n}=\frac{lo{g}_{a}n-lo{g}_{b}n}{lo{g}_{a}n-lo{g}_{b}n}$

Answer 1

R . H . S
$=\frac{\left(1/lo{g}_{n}a\right)-\left(1/lo{g}_{n}b\right)}{\left(1/lo{g}_{n}b\right)-\left(1/lo{g}_{n}c\right)}$
$=\frac{lo{g}_{n}b-lo{g}_{n}a}{lo{g}_{n}c-lo{g}_{n}b)}$
$=\frac{lo{g}_n\left(b/a\right)}{lo{g}_n\left(c/d\right))}\cdot \frac{lo{g}_{n}c}{lo{g}_{n}a}$
$=1\cdot \frac{lo{g}_{n}c}{(lo{g}_{n}a}=\frac{lo{g}_{a}n}{lo{g}_{c}n}$
$[\because b=\sqrt{a}c\Rightarrow b^2=ac\Rightarrow b/a=c/b]$