Answer
$\log_2 A$ + 2$\log_2 B$
Work Step by Step
$Expand$ $the$ $expression$:
$\log_2 (AB^2)$
Apply the First Law of Logarithms
$\log_2 (A\times B^2)$ = $\log_2 A$ + $\log_2 B^2$
Apply the Third Law of Logarithms to $\log_2 B^2$
$\log_2 B^2$ = 2$\log_2 B$
$\log_2 A$ + $\log_2 B^2$ = $\log_2 A$ + 2$\log_2 B$