Correct. Use the power property: $\displaystyle{\log}_{{b}}{{M}}^{{n}}={n}\cdot{\log}_{{b}}{M}\usepackage{cmbright}\Huge\$

Then use the definition of logarithm:

$\displaystyle{\log}_{{{3}}}{{81}}^{{2}}={2}\cdot{\log}_{{{3}}}{81}={2}\cdot{4}={8}\usepackage{cmbright}\Huge\$