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I am not too sure how to explain this in words. So the question is proofing that

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

So far what I have done was: proof

I cannot think of anything else, I mean if I do the same for $c ^{\log_b a}$, I would be getting $\log_b a^c$ which I believe is not enough to proof. What steps am I missing here?

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Recall how logarithms and exponents are inverse operations:

$$ b^{\log_b x} = x \tag{1}$$

Furthermore, recall how powers interact with logarithms:

$$ \log_bx^k = k\log_b x \tag{2}$$

With that in mind, observe that: $$ \begin{align*} a^{\log_b c} &= b^{\log_b (a^{\log_b c})} & \text{by (1), where } x=a^{\log_b c}\\ &= b^{(\log_b c)(\log_b a)} & \text{by (2), where }x=a \text{ and }k=\log_b c\\ &= b^{(\log_b a)(\log_b c)} \\ &= b^{\log_b (c^{\log_b a})} & \text{by (2), where }x=c \text{ and }k=\log_b a\\ &= c^{\log_b a} & \text{by (1), where }x=c^{\log_b a}\\ \end{align*} $$ as desired.

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  • $\begingroup$ Thanks, I understood that my problem was using x instead of b and of course a mistake in my second step. $\endgroup$ – Sarp Kaya Jul 28 '13 at 8:16
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You must rely on the logarithm properties:

$$\forall\,a,b,x,y>0\;,\;\;a,b\ne 1:\;\;\color{red}{\log_ax=\frac{\log_bx}{\log_ba}\;,\;\;x^y=a^{y\log_ax}}$$

So

$$\begin{align*}a^{\log_bc}&=\underbrace{b^{\log_bc\log_ba}}\\ c^{\log_ba}&=\overbrace{b^{\log_ba\log_bc}}^{||}\end{align*}$$

and we're done...

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I would take logarithms to base $b$ straight away

$$log_b{(a ^{\log_bc})}=\log_b{c}\log_b{a}$$ $$\log_b{(c^{\log_b a})}=\log_b a\log_b c$$

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