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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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    @fretty Thanks a lot! That works!2012-12-20

4 Answers 4

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

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    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.2014-05-27
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$a^{\log_b{n}}=n^{\log_b{a}}$/$\cdot$ $\log_a$

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

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

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

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

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

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    You wrote "$\log_n a = \frac{log_b a}{log_b n}$" twice. Maybe delete "and ..." from your answer. Most helpful, btw.2016-10-18
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use the following formula: $\log_a(b)=\frac{\ln(b)}{\ln(a)}$