Question

Solve the given exponential equations.
(i) $(\sqrt{6}{)}^{x-2}=1$ (ii) $3^{4x}=\frac{1}{81}$

Answer 1

$\left(i\right)(\sqrt{6}{)}^{x-2}=16^{\left(\frac{x-2}{2}\right)}=6^0\therefore (\frac{x-2}{2})=0$

$\therefore x=2$

$\left(ii\right)3^{4x}=\left(\frac{1}{81}\right)81x={81}^{-1}\therefore x=-1$