# Fastest way to check if $x^y > y^x$?

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.

• You can test if $\ln(y) + \ln \big( \ln(x) \big) > \ln(x) + \ln \big( \ln(y) \big)$. Commented Oct 7, 2013 at 10:39
• You want fast or robustly correct?
– lhf
Commented Oct 7, 2013 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? Commented Oct 7, 2013 at 10:44
• Are you doing http://projecteuler.net/problem=99? Commented Oct 7, 2013 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 Commented Oct 7, 2013 at 12:14

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: $$x \lt y$$

• @learner, the point is that the function $log(x)/x$ has a single maximum at $x=e$ and is decreasing after that.
– lhf
Commented Oct 7, 2013 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. Commented Oct 7, 2013 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
Commented Oct 7, 2013 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 $\frac{\log 2}{2} = \frac{\log 4}{4}$ Commented Oct 7, 2013 at 17:16
• @Henry: Sorry for hijacking this comment thread (I will delete this comment in a moment), but I have a question to Henry: I noticed that you just removed your answer to my question on integer division. Was the answer wrong?
– Sid
Commented Dec 20, 2016 at 12:31

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

We reduce the question to: is the quotient $$\frac{x^y} {y^x}$$ less or greater than $$1$$.

Noticing that $$x > 0, y > 0,$$ we take the xy-th root of the quotient:

Is the ratio $$(x^y)^{\frac1{xy}} / (y^x)^{\frac1{xy}}$$ less or greater than $$1^{\frac1{xy}}$$.

We reduce the complicated powers, the power of base 1 is 1:

is the ratio $$x^{\frac1x} / y^{\frac1y}$$ less or greater than 1.

is the function $$f(x) = x^{\frac1x}$$ increasing or decreasing at the interval $$[x,y]$$?

· If already known, via the derivate, that the function $$f(x) = x^{\frac1x}$$ has only one maximum, so an absolute maximum for $$x = e$$ , the point $$(e, e^{\frac1e})$$,

Consider that a) for $$x > e$$ and $$y > e$$ , the function $$f$$ is decreasing, which means

• if $$e < x < y => f(x) > f(y)$$. [both x and y are greater than e, the greater exponent wins.]

and b) for $$x < e$$ and $$y < e$$ the function $$f$$ is increasing, which means

• if $$0 < x < y < e => f(x) < f(y)$$. [both x and y are less than e, the greater base wins.]

• But if $$x < e < y$$, the comparison is not decided through the function’s behaviour. [e is between both.] We have to calculate both powers to compare the outcomes.

So there are pairs of $$x,y$$ existing for which $$x^y = y^x$$, with $$x < e < y$$.

For example, $$x = √(√5)$$ and y = √(√(5^5)) = √(√3125)
both x^y and y^x come out on ≈ 20,25372598…

Many pairs are findable, e.g. by setting y = a·x and resolving the equation (x^y) = (y^x), or

f(x) = f(y).

The equation $$x^y=y^x$$ has solution along the bisector $$y=x$$ and along a curve symmetric with respect to the bisector. In the following image the gray region is where $$x^y > y^x$$

A parametric representation of the curve is $$(e^{f(t)},e^{f(-t)})$$ where $$f(t) = \frac{t}{e^t-1}$$

A curve very similar to that is the equilateral hyperbola $$(x-1)(y-1) = (e-1)^2$$ and here there is a graphical comparison between the two curves

So a possible test with very low error is, assuming $$x < y$$ $$(x-1)(y-1) > (e-1)^2 \quad \Leftrightarrow \quad x^y > y^x$$

As an empirical measure of the error, over a million random samples in the range $$0 the rate of errors is below 0.3%.