Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
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
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*}

share|cite|improve this answer

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

share|cite|improve this answer

Using log properties, we have

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

share|cite|improve this answer
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 : $$x^{\log_{b}a} = a^{\log_{b}x}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.