# Logarithm proof problem: $a^{\log_b c} = c^{\log_b a}$

I have been hit with a homework problem that I just have no idea how to approach. Any help from you all is very much appreciated. Here is the problem

Prove the equation: $a^{\log_b c} = c^{\log_b a}$

Any ideas?

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Hint: You can use the change of base formula: $\log_b(x) = \frac{\ln x}{\ln b}$. – JavaMan Mar 4 '13 at 3:41

If you apply the logarithm with base a to both sides you obtain,

$log_a\space a^{log_b c} = \log_a\space c^{log_b a}$

$log_b\space c = log_b\space a *log_a\space c$

$\frac{log_b\space c}{log_b\space a} = log_a\space c$

however this last equality is the change of base formula and hence is true. Reversing the steps leads to the desired equality.

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This is getting me closer to understanding it. This is actually for an algorithms class and the professor kinda sprung this one on us without teaching us a whole lot about logarithm properties. Haven't touched this stuff since clac >_< – salxander Mar 4 '13 at 3:50

$\large{a^{\log_b c}=e^{\ln a \cdot\log_b c}= e^{\ln a\cdot\ln c/\ln b}}=c^{\ln a/\ln b}=c^{\log_b a}$.

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Hint:$a^{\log_b c} = c^{\log_b a}$ take log_a then $$log_aa^{\log_b c}=log_ac^{\log_b a}$$the we have${\log_b c}=(log_{a}{c})log _ba$

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In log there is a property that :$x^{\large {\log _x y}}=y$,$\log_x y=\log_w y\times\log_x w$ where $w$ can be any hold any possible value that is valid for a log base. so $$a^{\large{\log_b c}}\implies a^{\large{\log_a c\times \log_b a}}$$ since $a^{\large{\log_a c}}=c$ so $$a^{\large{\log_a c\times \log_b a}}\implies c^{\large{\log_b a}}$$

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