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?

$$\log_3 (9\cdot x)=\log_3 9+\log_3 x$$
then what about $$\log_2 (4\cdot x)$$?