Simplify $\displaystyle 2loga+3logb$:

I thought it you could do add the 2 and 3, but multiply a + b to give $\displaystyle 5logab$.

But the answer says: $\displaystyle 2loga+3logb = log{a^2}+log{b^2} = log{a^2}{b^3}$.

I perfectly understand how the mark scheme did it, moving the powers of 2 and 3 form the log laws, but i was just wondering though:

Is it wrong adding the first bit I did of the 2 and the 3 before the logs, and multiplying the a and b?

Thanks.