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.
Can we prove $a^{\log_bn} = n^{\log_ba}$?
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$\begingroup$
logarithms
exponentiation
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0@fretty Thanks a lot! That works! – 2012-12-20
4 Answers
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Take the log to the base $b$ of both sides.
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1This 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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0You 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: $\log_a(b)=\frac{\ln(b)}{\ln(a)}$