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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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  • $\begingroup$ Hint: You can use the change of base formula: $\log_b(x) = \frac{\ln x}{\ln b}$. $\endgroup$ – JavaMan Mar 4 '13 at 3:41
  • $\begingroup$ Should this question be a duplicate of this question simply because it was posted after, or should it be the other way around due to the seemingly better content? Perhaps, I would think, they should be duplicates of this question, which likely has the best quality of content. $\endgroup$ – Simply Beautiful Art Feb 7 '17 at 2:54
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$\large{a^{\log_b c}=e^{\ln a \cdot\log_b c}= e^{\ln a\cdot\ln c/\ln b}}=c^{\ln a/\ln b}=c^{\log_b a}$.

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

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

$$\log_b c = \log_b a \log_a c$$

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

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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    $\begingroup$ 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 >_< $\endgroup$ – salxander Mar 4 '13 at 3:50
  • $\begingroup$ The way to think of it is: you are trying to show that $a^{\log_b c} = c^{\log_b a}$, start at the change of base formula that relates $a$ and $c$ and work backwards to find your identity. This was super helpful, thank you @newToProgramming $\endgroup$ – albertjorlando Feb 13 at 17:52
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$a^{\log_b c} =c^{\log_b a}$

let $k= a^{log_b c}$

$\log k=\log( a^{\log_b c})$

$\log k=\log_b c \log a$

$\log k=\frac{\log c}{\log b} \log a$

$\log k=\frac{\log a}{\log b} \log c$

$\log k=\log_b a \log c$

$\log k=\log c^{\log_b a}$

substituting value of k

$\log a^{\log_b c}=\log c^{\log_b a} $

therefore

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

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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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