What is the math behind this transformation on exponents that are logarithms?

I understand that $$a^{\log_b(n)} = n^{\log_b(a)}.$$

What is the math behind this transformation that allows you to swap the $a$ and $n$?

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Take the base-$b$ logarithms of both sides for starters. –  Ｊ. Ｍ. Apr 28 '12 at 19:22

By definition, $a=b^{\log_b(a)}$ and $n=b^{\log_b(n)}$. Therefore $$a^{\log_b(n)}=(b^{\log_b(a)})^{\log_b(n)}=b^{\log_b(a)\cdot\log_b(n)}=(b^{\log_b(n)})^{\log_b(a)}=n^{\log_b(a)}.$$

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Use change of base on $\log_b(n)$ and write it as $\frac{\log_a(n)}{\log_a(b)}$. Thus, $a^{\frac{\log_a(n)}{\log_a(b)}}=n^{\frac{1}{\log_a(b)}}=n^{\log_b(a)}$.

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I do not follow the step $a^{\frac{\log_a(n)}{\log_a(b)}}=n^{\frac{1}{\log_a(b)}}$. Also why is the reciprocal of a log the same as swapping the base and argument? –  WilliamKF Apr 28 '12 at 19:33
I hope this helps: $a^{\frac{\log_a(n)}{\log_a(b)}}=\left(a^{\log_a(n)}\right)^{\frac{1}{\log_a(b)}‌​}=n^{\frac{1}{\log_a(b)}}=n^{\log_b(a)}$. In general, $\log_a(b) = \frac{\log_b(b)}{\log_b(a)} = \frac{1}{\log_b(a)}$. –  Eric Apr 29 '12 at 21:01