Question

If ${\log }_{a}b=2,{\log }_{b}c=2$ and ${\log }_{3}c=3+{\log }_{3}a,$ then the value of $\frac{c}{ab}$ is

Answer 1

Clearly, $a,b,c>0$ and $a,b\ne 1$

Now, ${\log }_{a}b=2,{\log }_{b}c=2$
$\Rightarrow {\log }_{a}b\cdot {\log }_{b}c=4$
$\Rightarrow \frac{{\log }_{b}c}{{\log }_{b}a}=4$
$\Rightarrow {\log }_{a}c=4$
$\Rightarrow c=a^4...\left(1\right)$

${\log }_{3}c=3+{\log }_{3}a$
$\Rightarrow {\log }_{3}c-{\log }_{3}a=3$
$\Rightarrow {\log }_3\frac{c}{a}=3$
$\Rightarrow c=27a...\left(2\right)$

From $\left(1\right)$ and $\left(2\right),$
$a^4=27a$
$\Rightarrow a(a^3-27)=0$
$\Rightarrow a=3[\because a>0]$
$\Rightarrow c=81$

${\log }_{a}b=2$
$\Rightarrow b=a^2=9$

$\therefore \frac{c}{ab}=3$