Question

${\log }_{2\sqrt{3}}1728$

Answer 1

Here we can see that $2\sqrt{3}$ can be written as $\sqrt{12}$

Hence,

${\log }_{2\sqrt{3}}1728={\log }_{\sqrt{12}}1728$

$\Rightarrow {\log }_{2\sqrt{3}}1728={\log }_{\sqrt{12}}(12{)}^3$

This can be further simplified as

$\Rightarrow {\log }_{2\sqrt{3}}1728={\log }_{\sqrt{12}}(\sqrt{12}{)}^6$

$\Rightarrow {\log }_{2\sqrt{3}}1728=6\times {\log }_{\sqrt{12}}\sqrt{12}$

We know that ${\log }_{x}x=1$

Therefore,

$\Rightarrow {\log }_{2\sqrt{3}}1728=6\times 1$

$\Rightarrow {\log }_{2\sqrt{3}}1728=6$