According to the basic rules of $\log$, I'm solving both $\log$ terms as for first: base is $3$, $N$ is $9$ so exponent is calculated as $2$, and same for other term. But I'm confused with this '$x$'. Adding both terms $\log$ according to my logic will result in $4$. But I know I'm doing something wrong here, how to treat this $x$? Is B the correct answer?
HINT
Recall that
$$\log_3 (9\cdot x)=\log_3 9+\log_3 x$$
then what about $\log_2 (4\cdot x)$?