How to write $\log_{15} 81$ in terms of $A$, if $A$ is equal to $\log_{15} 5$?

At first it seemed really simple but I've been stuck with this exercise for many minutes, I've tried a lot of things but I just get nowhere, would someone be kind enough to give me a hint? The answer is $4(1-A)$ but to me that doesn't really make any sense as to how I would get to that because if I try to go through it backwards I get: $$4-4A=4-4(\log_{15} 5)$$ $$4-4(\log_{15} 5)=4-\log_{15} 625$$ And I don't know what to do next. Thanks.

Note that $$\log_{15}81=\log_{15}(3^4)=\log_{15}\left(\frac{15}5\right)^4=\log_{15}(15^4)-\log_{15}(5^4)=4-4A=4(1-A)$$ as required.