Question

Calculate: $2{\log }_4\sqrt{64}-{\log }_3\sqrt{81}$

Answer 1

We know that,

$a\log b=\log b^a$.

Using above property:

$\Rightarrow 2{\log }_4\sqrt{64}-{\log }_3\sqrt{81}$

$={\log }_4(\sqrt{64}{)}^2-{\log }_{3}9$

$={\log }_4{4}^3-{\log }_3{3}^2=3-2=1$