Answer
$2\log_b{x} + \log_b{y}$
Work Step by Step
RECALL:
(1) $\log{(b^c)}=c \cdot \log{b}$
(2) $\log_b{(xy)} = \log_b{x} + \log_b{y}$
Use rule (2) above to obtain:
$=log_b{(x^2)}+\log_b{y}$
Use rule (1) above to obtain:
$=2\log_b{x} + \log_b{y}$