Question

If ${\log }_{2\sqrt{3}}1728=x$, then the value of $x$ is

Answer 1

${\log }_{2\sqrt{3}}1728={\log }_{2\sqrt{3}}(64\times 27)={\log }_{2\sqrt{3}}\left(2^6{\left(\sqrt{3}\right)}^6\right)=6{\log }_{2\sqrt{3}}2\sqrt{3}=6,[\because {\log }_{a}a=1]$