Discussion Forum

The value of $log_{2\sqrt{3}}(1728)$ is :

Answer:

Explanation:

Let $log_{2\sqrt{3}}(1728)$ $=x$.

Then, $(2\sqrt{3})^{x}$ $=1728$

$=(12)^{3}$ $=\left[\left(2\sqrt{3}\right)^{2}\right]^{3}$

$=(2\sqrt{3})^{6}$.

$\therefore x=6$

i.e., $log_{2\sqrt{3}}(1728)$ = 6.