$2^4 = 4^2$ are there any other solutions to $x^y = y^x$? How would I go about solving $x^y = y^x$?
$$\frac{y}{\ln y} = \frac{x}{\ln x} $$
Is trivial to find, but I'm not sure where I'd go from here. 
Thanks in advance
 A: The plot of $\frac{\ln x}{x}$ looks so:

By searching for the solutions of $\frac{x}{\ln x} = \frac{y}{ln y}$ where $x \ne y$, you are looking for the points of this plot with the same $y$-coordinate.
As you can see, $2^4 = 4^2$ is your only integer solution.
A: Plotting (using Geogebra) $\frac{x}{\log_{10}(x)}=\frac{y}{\log_{10}(y)}$ gives

As we can see, the only integral solutions are $(2,4)$ and $(4,2)$ (as well as the trivial $(x,x) \in \Bbb{Z} \times \Bbb{Z}$ solutions). This is because as $y$ (respectively $x$) increases, the curve asymptotes to an $x$ coordinate of $>1$ (respectively $y$ coordinate of $>1$) but never reaches it. 
A: The only integer solution is (2,4).
You are solving
$$y^x=x^y$$
and if you fix $y$, you (almost) always have two solutions for $x$. One is $x=y$, the other is not expressible in closed form with "traditional"  functions, but you can use the transcendental function $W(z)$, the Lambert function, which solves
$$xe^x=z \mapsto x=W(z)$$
To do this, you transform the equation into
$$e^{x\ln y}x^{-y}=1$$
$$x e^{(-x\ln y/y)}=1$$
$$(-x\ln y/y) e^{(-x\ln y/y)}=-\ln y/y$$
The parenthesised expression can be your unknown, arriving at
$$-\frac{x\ln y}{y}=W(-\ln y/y)$$
$$x=-\frac{y}{\ln y}W(-\ln y/y)$$
Provided you have a way to compute $W$, this gives you a solution.
Special case: for $y=e$, the only (double) solution is $x=y=e$. This is the crossover point: for $y<e$, the nontrivial solution is $x>e$ and vice-versa.
