Answer
$\log_2{A} + 2\log_2{B}$
Work Step by Step
RECALL:
$\log_b{(PQ)} = \log_b{P} + \log_b{Q}$
Use the rule above to obtain:
$=\log_2{A} + \log_a{(B^2)}$
Write $B^2$ as $B\cdot B$ to obtain:
$=\log_2{A} + \log_2{(B\cdot B)}$
Use the rule above to obtain:
$=\log_2{A} + \log_2{B} + \log_2{B}$
