# Can we prove $a^{\log_bn} = n^{\log_ba}$?

Can we prove $$a^{\log_bn} = n^{\log_ba}?$$ I forget how to prove this theorem. I picked up one numbers for test, and they worked.

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Hint: Start with the fact that $(log_bn)(log_ba) = (log_ba) (log_bn)$. – fretty Dec 20 '12 at 16:27
@fretty Thanks a lot! That works! – xwb1989 Dec 20 '12 at 16:30

Take the log to the base $b$ of both sides.

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Thanks a lot! That works! – xwb1989 Dec 20 '12 at 16:31
This is deceptively simple. It's a good hint, but with it, one must be careful not to assume what one is trying to prove. See fretty's comment on the question for a safer starting point. – Gamma Function May 27 '14 at 9:42

$a^{\log_b{n}}=n^{\log_b{a}}$/$\cdot$ $\log_a$

$\log_a a^{\log_b{n}}=\log_a n^{\log_b{a}}$

$\log_b{n}=\log_b{a} \log_a n$

$\log_b{n}=\frac{\log a}{\log b}\cdot\frac{\log n}{\log a}$

$\log_b{n}=\frac{\log n}{\log b}$

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use $$\log_a(b)=\frac{\ln(a)}{\ln(b)}$$

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$$a^{\log_b n}=n^{\log_n a \log_b n}=n^{\log_b a},\quad \text{using}\quad\log_n a=\frac{\log_b a}{\log_b n}\quad \text{and}\quad\log_n a=\frac{\log_b a}{\log_b n}.$$

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