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What is the fastest way to check if $x^y > y^x$ if I were writing a computer program to do that?

The issue is that $x$ and $y$ can be very large.

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You can test if $\ln(y) + \ln \big( \ln(x) \big) > \ln(x) + \ln \big( \ln(y) \big)$. – jibounet Oct 7 '13 at 10:39
You want fast or robustly correct? – lhf Oct 7 '13 at 10:40
@lhf, Fast and in checking among n such pairs of numbers, n/10 could go wrong. Could you please help me deduce what accuracy I'm looking for? – learner Oct 7 '13 at 10:44
Are you doing – Mårten W Oct 7 '13 at 10:50
@MårtenW: His problem is more specific. The link you posted is about x^a < y^b. He is considers the special case a=y, b=x – Christian Fries Oct 7 '13 at 12:14
up vote 69 down vote accepted

If both $x$ and $y$ are positive then you can just check: $$ \frac{\log(x)}{x} \gt \frac{\log(y)}{y}$$ so if both $x$ and $y$ are greater than $e \approx 2.7183$ then you can just check: $$x \lt y$$

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But how x < y? How can we omit the logs? Intuitively it seems right, but is it really so? – learner Oct 7 '13 at 11:18
@learner, the point is that the function $log(x)/x$ has a single maximum at $x=e$ and is decreasing after that. – lhf Oct 7 '13 at 11:19
I don't know why this answer is the accepted one, but multiplication is usually faster than division, so I assume $y log(x) > x log(y)$ is faster than this one. – Christian Fries Oct 7 '13 at 12:15
@ChristianFries, this answer is the accepted one because of the massive optimization given in the second half of the answer. Even if you don't use the monotonicity of $\frac{\log{x}}{x}$, the fact that it's a single function can make repeated queries much faster. – jwg Oct 7 '13 at 12:29
If $x$ and $y$ are both between $0$ and $e$ then the test becomes $x \gt y$. The hard part is when one is below and one above: for example $2^4 = 4^2$ and $\frac{\log 2}{2} = \frac{\log 4}{4}$ – Henry Oct 7 '13 at 17:16

You might get by testing whether $y \log x > x \log y$, especially if the numbers are only moderately large.

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