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This actually triggered me in my mind from here. After some playing around I notice that the relation $a^{\log_b x} = x^{\log_b a}$ is true for any valid value of $a,b$ and $x$. I am very inquisitive to see how this holds ?

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Did you read Bill's answer in the question you linked? The first sentence there holds the answer to this question too. –  Aryabhata Nov 18 '10 at 18:38
    
Sory,dunno why I am not getting that first line itself :( Am I such a stupid ? :( –  Quixotic Nov 18 '10 at 18:44
    
What is the definition of logarithm? log_b(x) is the number which gives x when b is raised to that right? So x = b^(log_b(x)). Now use (b^m)^n = b^(mn). –  Aryabhata Nov 18 '10 at 18:52
    
No Moron it's just I am being moron today, somehow it got into my mind that $r^s = b^{\log_b(r^s)} = b^{s\log_b(r)}$ holds good only when $b = e$, silly me. –  Quixotic Nov 18 '10 at 18:57
    
Bill also pointed out (in parentheses) that by taking log to base b on both sides also gives you a proof and might be clearer. –  Aryabhata Nov 18 '10 at 18:58

4 Answers 4

up vote 8 down vote accepted

For any positive $r$ and any $s$, you have $$r^s = b^{\log_b(r^s)} = b^{s\log_b(r)}.$$ So, taking $r=a$ and $s=\log_b(x)$, we have: \begin{align*} a^{\log_b(x)} &= b^{\log_b(x)\log_b(a)}\\ &= b^{\log_b(a)\log_b(x)}\\ & = b^{\log_b(x^{\log_b(a)})}\\ &= x^{\log_b(a)}. \end{align*}

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HINT $\rm\ \ \ log(A^{\log X})\ =\ log\ X\ \ log\ A\ =\ log(X^{\log A})\:,\ $ where $\rm\ log := log_b$

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Using log properties, we have

$a^{\log_b(x)} = b^{\log_b\left(a^{\log_b(x)}\right)} = b^{\log_b(x)\log_b(a)} = \left(b^{\log_b(x)}\right)^{\log_b(a)} = x^{\log_b(a)}$

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Did not see Arturo's answer until this was posted. –  Hans Parshall Nov 18 '10 at 18:47

Otherwise :

Changing the base of logarithm, we have :

$$\log_{b}a= \displaystyle\frac{\log_{x}a}{\log_{x}b}$$ and $$\log_{x}b= \displaystyle\frac{\log_{b}b}{\log_{b}x} = \frac{1}{\log_{b}x} $$

By combine these two ecuations,

$\log_{b}a= (\log_{b}x)(\log_{x}a) \Leftrightarrow (\log_{b}a)(\log_{x}x)= (\log_{b}x)(\log_{x}a) \Leftrightarrow \log_{x}x^{\log_{b}a} = \log_{x}a^{\log_{b}x}$

By last, canceling $\log_{x}$ on both sides, we have : $$x^{\log_{b}a} = a^{\log_{b}x}$$

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