Is there any simple ways to compare $x^y$ and $y^x$ without a calculator?

Question

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.

Answer 1

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 $x^{y} < y^{x}$.