Understanding the graph for $x^y = y^x$ I graphed $x^y=y^x$ and it is a union of the line y=x, with some other curve.  So my first question is, how do I derive that other curve? My next question is, why don't I get the same graph when I plot $x^{1/x}=y^{1/y}$?

Here is the way I first saw it,
many years ago:
If $x^y = y^x$
with $x > 0$ and $y > 0$,
let $r = y/x$,
so $y = rx$.
We will derive a parameterization
for $x$ and $y$
in terms of $r$.
Then
$x^{rx} = (rx)^x$
or,
taking the $x^{th}$ root,
$x^r = rx$
or
$x^{r-1} = r$
or,
if $r \ne 1$,
$x = r^{1/(r-1)}$
and
$y = rx = r^{1+1/(r-1)}
= r^{r/(r-1)}$.
Letting $r$ go through
the reals $> 1$
gives point $(x, y)$ with
$x^y = y^x$.
Another way to look at this
is to consider the curve
$v = u^{1/u}$.
For each $v$ such that
$1 < v < e$,
there are two values of $u$,
$u_1 < e < u_2$,
such that
$u_1^{1/u_1} = v = u_2^{1/u_2}$.
Note that
if $1/(r-1)$ is an integer,
say $n$,
then
$r = 1+1/n$
and
$x = (1+1/n)^n$
and
$y = (1+1/n)^{n+1}$
are two rational numbers
such that
$x^y = y^x$.
$n=1$ gives $x=2$ and $y=4$ (the well known solution).
$n=2$ goves
$x = (3/2)^2 = 9/4$
and $y = (3/2)^3 = 27/8$
(it is definitely less well known that
$(9/4)^{27/8} = (27/8)^{9/4}$.
I believe that
these are the only rational solutions to
$x^y = y^x$,
but I do not have a proof.
What the heck, I'll make it a problem and,
I hope, harvest some points.
