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There are plenty of discussion on MSE about how to compare $x^y$ and $y^x$. For $x,y>e$, it is sufficient to just compare $x$ and $y$ to reach a conclusion. But I wonder if there are some general steps that can be used in any situations where $x<e$ and $y>e$. I tried to use logarithm, but the inequities produced often includes multiplication of negative numbers, so I often make mistakes along the way.

Note that I am interested in all REAL values of $x,y$, NOT just integers. For example, $x=2,y=\sqrt{5}$. I wish anyone to give a solution that is applicable in other values of $x,y$ as well.

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  • $\begingroup$ "... all real values of $x,y$ ..." Presumably just positive? Since $x^y$ might be undefined otherwise. Cool question otherwise (+1) $\endgroup$ – MisterRiemann Feb 2 at 15:43
  • $\begingroup$ Nice question, have a look at this: mathforum.org/library/drmath/view/70271.html $\endgroup$ – anas pcpro Feb 2 at 17:03
  • $\begingroup$ @MisterRiemann Yes you are right $\endgroup$ – Jethro Feb 2 at 22:52
  • $\begingroup$ @anaspcpro That link just discusses the case where x, y are integers. It have not considered the case where x, y are generally any real number. $\endgroup$ – Jethro Feb 2 at 23:20
  • $\begingroup$ That's why I didn't add the link as an answer. $\endgroup$ – anas pcpro Feb 3 at 16:00
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I can offer a "comparison test".

W.l.o.g., suppose you are interested to establish whether or not $x^y > y^x$. Now set $a=y/x$. Then it is easy to show that $x^y > y^x$ is equivalent to
$$ x > f(a) = a^{\frac{1}{a-1}} $$ Now you can exploit the fact that $f(a)$ is strictly decreasing with $a$. That allows to come up with the following comparison test:

If you have a pair $(x_0,y_0)$ for which $x_0^{y_0} > y_0^{x_0}$ holds, then $x_0^{y} > y^{x_0}$ will also hold for all $y > y_0$.

And: If you have a pair $(x_0,y_0)$ for which $x_0^{y_0} < y_0^{x_0}$ holds, then $x_0^{y} < y^{x_0}$ will also hold for all $y < y_0$.

A particular "test case" is $x_0=y_0$ which gives $f(1) = e$. Then the above statements are:

If $y>x>e$, then $x^{y} > y^{x}$.

And: If $x<y<e$, then $x^{y} < y^{x}$.

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  • $\begingroup$ But how can I get $y_0$ at the first place? $\endgroup$ – Jethro Mar 4 at 14:04
  • $\begingroup$ Well, since we have two special cases where the comparison is easy, cases remain to be investigated only where one of the two variables is less than $e$ and the other is higher than $e$. For these cases, you need initial values to compare. A good start would be to find all value pairs $(x_0,y_0)$ where $x_0^{y_0} = y_0^{x_0}$. Take as an example $(x_0,y_0) = (2,4)$. Then we know that $ 2^{y} > y^2$ for $y >4$. Reversing the pair, we have that $ 4^{y} > y^4$ for $y <2$. $\endgroup$ – Andreas Mar 5 at 10:24

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