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

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 $$x and $$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.

• "... all real values of $x,y$ ..." Presumably just positive? Since $x^y$ might be undefined otherwise. Cool question otherwise (+1) – MSDG Feb 2 '19 at 15:43
• Nice question, have a look at this: mathforum.org/library/drmath/view/70271.html – Anas Khaled Feb 2 '19 at 17:03
• @MisterRiemann Yes you are right – Ma Joad Feb 2 '19 at 22:52
• @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. – Ma Joad Feb 2 '19 at 23:20
• That's why I didn't add the link as an answer. – Anas Khaled Feb 3 '19 at 16:00

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

• But how can I get $y_0$ at the first place? – Ma Joad Mar 4 '19 at 14:04
• 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 $y <2$. – Andreas Mar 5 '19 at 10:24