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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?

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  • $\begingroup$ Simple :take logarithm on both sides of the simplification equation and you get basically the same thing ;; $\endgroup$ – satyatech Dec 17 '16 at 10:38
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Since $a^b=c^{b(\log a)}$, where $c$ is the base of $\log$, we have that $$9^{\log n} = c^{\log n\cdot( \log 9)}=c^{\log 9\cdot( \log n)}=n^{\log 9}.$$

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  • $\begingroup$ Doesn't $log$ have a base of 2? $\endgroup$ – Samu Dec 17 '16 at 10:35
  • $\begingroup$ @Samu If your $\log$ has base $2$ then replace $c$ with $2$. See my edited answer. $\endgroup$ – Robert Z Dec 17 '16 at 10:41
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Hint:

Take logarithms. For LHR we have

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

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