Logarithmic simplification

Question

I don't understand the last simplification $9^{\log n} = n^{\log 9}$: $$3^{\log(n^2)} = 3^{2\log n} = 9^{\log n} = n^{\log 9}$$ Can someone please show me how they did it? Is there a rule that I have missed?

Answer 1

Since $a^b=c^{b(\log a)}$, where $c$ is the base of $\log$, we have that $$9^{\log n} = c^{\log n\cdot( \log 9)}=c^{\log 9\cdot( \log n)}=n^{\log 9}.$$

Answer 2

Hint:

Take logarithms. For LHR we have

$$ \log(9^{\log n})=\log n \log 9 $$ do the same for RHR