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I came across a neat logarithm fact today:

$\large n^{\log_bx} = x^{\log_bn}$

One simple proof is:

$\large \log_bx\cdot \log_bn=\log_bx\cdot \log_bn$

$\large \Rightarrow \log_bx^{log_bn}=log_bn^{log_bx}$

$\large \Rightarrow x^{\log_bn}=n^{\log_bx}$

So you can swap the base with part of the exponent. Does this property have a name? What is the intuition? The other log properties are very intuitive to me, but not this one. Is it more easily derived from the usual four logarithm properties?

Community
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MathStudent
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    The point is that $a^{\log b} = e^{\log a \log b}$. I don't think this really has a name; it doesn't come up very often in my experience. – Qiaochu Yuan Jul 02 '14 at 18:03
  • Oh, that's neat. Sure I've never noticed it before...but keeping in mind that when you have a logarithm in the exponent, you may be able to swap log parameter with base of exponent is neat. – MathStudent Jul 02 '14 at 18:05
  • This is pretty neat. I thought it might be in my short note Harmonious logarithm identities in Mathematical Gazette [volume 93 number 526 (March 2009), pp. 95-97], of which an overview is given here, but it's not. – Dave L. Renfro Jul 02 '14 at 18:21
  • This also follows immediately from the facts that (1) $n=x^{\log_x n}$ and (2) $\log_x n = \log_b n \log_b x$. – MPW Jul 02 '14 at 19:09
  • EDIT: Of course, that's supposed to be $\log_b n / \log_b x$. Can't edit the comment now, unfortunately :( – MPW Jul 02 '14 at 19:26

2 Answers2

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I come across the same property and struggled to grasp the intuition until I try different numbers for n and x for in your equation.

$$n^{\log_bx} = x^{\log_bn}$$

Let's think about n=2, x=8 and assume b=10 and then try to generalize the concept.

$$2^{\log_{10}8} = 8^{\log_{10}2}$$ $$2^{\log_{10}2^3} = 2^{3\log_{10}2}$$ $$2^{3\log_{10}2} = 2^{3\log_{10}2}$$

So we can start to think with $n=x^k$ and it always results as following while log base doesn't matter except being the same:

$$n^{\log_bx} = x^{\log_bn} $$ $$x^{k\log_bx} = x^{\log_bx^k} $$ $$ x^{k\log_bx} = x^{k\log_bx} $$

Alexandra
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$$ n^{\log_b x}=b^{\log_bn\cdot\log_bx}=\dots $$

Vladimir
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