Are there real solutions to $x^y = y^x = 3$ where $y \neq x$? I need to solve the following equation for (x,y) $$x^y = y^x = 3$$
Everytime I run a numerical method for this problem, I get $$ (x,y) = (1.82546...,1.82546..) $$
I expect there to be a solution where $x \neq y$ but I cannot seem to find it by any means.
Does such a solution exist?
 A: I will show that
the smallest value
of $z$
such that
there are an $x$ and $y$
with
$x^y = y^x
= z$
is
$z = e^e
\approx 15.15426224
$,
so there is no solution
to OP's question.
This is a corrected version,
which should ameliorate
 user21820's
 righteous indignation.
Start with the
usual parameterization
of $x^y = y^x$:
Let $y = rx$
where $r > 1$.
Note that this implies
$x < e < y$.
Then
$x^{rx} = (rx)^x$
or
$x^r = rx$
or
$x^{r-1} = r$
or
$x = r^{1/(r-1)}$
and
$y = rx
=r^{1+1/(r-1)}
=r^{r/(r-1)}
$.
Then
$x^y
=(r^{1/(r-1)})^{r^{r/(r-1)}}
=r^{r^{r/(r-1)}/(r-1)}
$.
As a check,
$y^x
=(r^{r/(r-1)})^{ r^{1/(r-1)}}
=r^{r^{1/(r-1)}r/(r-1)}
=r^{r^{r/(r-1)}/(r-1)}
$.
If $r = 1+s$,
this is
$\begin{array}\\
f(s)
&=(1+s)^{(1+s)^{(1+s)/s}/s}\\
&=(1+s)^{(1+s)^{1+1/s}/s}\\
&=(1+s)^{e^{\ln(1+s)(1+1/s)}/s}\\
\text{so}\\
g(s)
&=\ln(f(s))\\
&=\ln(1+s)e^{\ln(1+s)(1+1/s)}/s\\
&=\frac{\ln(1+s)}{s}e^{\ln(1+s)(1+1/s)}\\
&=\frac{\ln(1+s)}{s}e^{\ln(1+s)+\ln(1+s)/s}\\
&=(1+s)\frac{\ln(1+s)}{s}e^{\ln(1+s)/s}\\
&=e+(e s^2)/24-(e s^3)/24+(73 e s^4)/1920+O(s^5)\\
&\qquad\text{according to Wolfy}\\
&=e\left(1+( s^2)/24-( s^3)/24+(73  s^4)/1920+O(s^5)\right)\\
\text{so}\\
f(s)
&=e^e\left(1+s^2/24-s^3/24+(7 s^4)/180+O(s^5)\right)\\
&\qquad\text{again, according to Wolfy}\\
\end{array}
$
I will now show that
$f(s)$ is increasing
by showing that
$g(s)$
is increasing.
Using Wolfy again,
$g'(s)
= \dfrac{((s+1)^{1/s} (s^2-(s+1) \ln^2(s+1)))}{s^3}
$.
To show that
$g'(s) > 0$ for
$s > 0$,
we need to show that
$h(s)
=s^2-(s+1) \ln^2(s+1)
\gt 0
$
for $s > 0$.
$h(0) = 0$
and
$h'(s)
= 2 s-\ln^2(s+1)-2 \ln(s+1)
$.
We need to do
the same with $h'(s)$.
$h'(0) = 0$
and,
for $s > 0$,
$h''(s)
=  2 \frac{(s-\ln(s+1))}{s+1}
\gt 0
$
since
$\ln(1+s) < s$.
Therefore
$f'(s) > 0$
for $s > 0$.
Since
$f(0) = e^e$,
the smallest value
for which
$x^y = y^x$
is $e^e \approx 15.15426224 $,
so this can not be $3$.
This is why
$2^4 = 4^2
=16$
works.
A: Assume $0<x<y$ and $x^y=y^x=3$.
Then in particular $\frac{\ln x}x=\frac{\ln y}y$. The derivative of $f(t)=\frac{\ln t}t$ is $f'(t)=\frac{\frac1t\cdot t-1\cdot \ln t}{t^2}=\frac{1-\ln t}{t^2}>0$ for $0<t<e$, hence $f$ is injective on $(0,e]$. 
We conclude that $y>e$. 
Also note that for $x,y>0$ we have $x^y\le 1$ if $x\le 1$. Therefore $x>1$.
From $x>1$ and $y^x=3$ it follows that $y<3$. Then $$x=\sqrt[3]{x^3}\ge \sqrt[3]{x^y}=\sqrt[3]{y^x}>\sqrt[3]y>\sqrt[3]e>1+\frac13+\frac1{18}=\frac{25}{18}>\frac54.$$ 
Thus $$y^x>e^{\frac54}>1+\frac54+\frac{25}{32}=\frac{97}{32}>3,$$ contradiction.

Remark: The above does not require any calculator. All "numerical" calculations have been reduced to adding and multiplying fractions with manageably small denominator, and we use that for $t>0$ we have $e^t=\sum_{k=0}^\infty\frac{t^k}{k!}>1+t+\frac{t^2}2$.
Of course, if one trusts in calculator results, $\sqrt[3]e\approx 1.395612425$ and $e^{\sqrt[3]e}\approx4.037446449$ and that is of course also $>3$.
A: Here is an interesting way that works for finding the positive real solutions to $x^y = y^x = c$ for any real $c \in (1,e^e)$.
$\def\rr{\mathbb{R}}$
Take any reals $x,y$.
Firstly, $y^x = c$ implies $y = c^{1/x}$, and since $( t \mapsto c^{1/t} )$ is a strictly decreasing function on $\rr^+$ with range $\rr^+$, it has at least one fixed point $z$, namely $z \in \rr^+$ such that $z = c^{1/z}$. Now clearly $(z,z)$ is a solution.
Secondly, $x^y = y^x = c$ implies $x^{c^{1/x}} = c$, but $( x \mapsto x^{c^{1/x}} )$ is a strictly increasing function on $\rr^+$, and so there is at most one solution for $x$ and hence at most one solution for $(x,y)$.
Therefore we have found the only solution, which must be $(z,z)$.
Technical detail
To prove the strictly increasing nature of the function in the second paragraph above, note that $\frac{d}{dx}( x^{c^{1/x}} ) = \frac{d}{dx}( e^{\ln(x)c^{1/x}} )$ $= e^{\ln(x)c^{1/x}}( \frac{1}{x} c^{1/x} - \ln(x) c^{1/x} \ln(c) \frac{1}{x^2} )$ $= e^{\ln(x)c^{1/x}} c^{1/x} \frac{1}{x^2} ( x - \ln(x) \ln(c) )$ $\ge 0$ because $x - \ln(x) \ln(c) > x - \ln(x) e \ge 0$ which can be proven by noting that $\frac{d}{dx}( x - \ln(x) e ) = 0$ iff $1 - \frac{e}{x} = 0$ iff $x = e$, which is a global minimum because $\frac{d}{dx^2}( x - \ln(x) e ) = \frac{e}{x^2} > 0$.
A: The solution may only be obtained using the Lambert W function for other solutions, which must be complex.
$$x^y=y^x\implies y=e^{W_k(x\ln(x))}$$
Trivially, there is $y=e^{W_0(x\ln(x))}=x$
But there are other branches, as you can see graphically.
I will go about trying to solve:
$$y^x=3$$
$$(e^{W_k(x\ln(x))})^x=e^{xW_k(x\ln(x))}=3$$
$$xW_k(x\ln(x))=3$$
$$W_k(x\ln(x))=\frac3x$$
$$x\ln(x)=\frac3xe^{\frac3x}$$
$$x^x=e^{\frac3xe^{\frac3x}}$$
We have the simple case $x=e^{\frac3x}$, or we don't have that case.
Either way, it is impossible to solve it from here in closed form using the Lambert W function.
I would proceed by attempting to solve the last equality for complex $x$, as it is obvious there are no other real solutions.
