Given that $x^y=y^x$, what could $x$ and $y$ be? It's not too difficult to figure out that $x$ and $y$ can both be 1, and also $x$ can be 2 and $y$ can be 4 (and vice versa). But I can't rule out if there are other solutions. Does it have anything to do with inverse functions? Is there a way to see the solutions by graphing, or algebraically?
Here's a good way to look at this: 
$$
x^y = y^x \implies \\
y \ln x = x \ln y \implies\\
\frac{\ln x}x = \frac{\ln y}y
$$
So one way of solving this is looking at the graph of $y = \frac{\ln x}{x}$ and seeing where it hits a particular value twice.

However, I'm going to go a step further.  I will say that I want my solution $(x,y)$ of $x^y = y^x$ to be of the form $(x,ax)$ so that $y = ax$ for some value $a$.  Furthermore, I'll assume that $x \neq y$ so that $a \neq 1$ (of course, $x = y$ is always a solution if $x^x$ is defined). With that, we have
$$
\frac{\ln x}x = \frac{\ln (ax)}{ax} \implies (\text{assume } x\neq 0)\\
\ln x = \frac{\ln (ax)}{a} \implies\\
\ln x = \frac{\ln (a)}{a} + \frac{\ln x}{a} \implies\\
\frac{a-1}{a}\ln x = \frac {\ln(a)}{a} \implies (\text{we assumed }a \neq 1)\\
\ln x = \frac{\ln(a)}{a-1} \implies\\
x = a^{\frac{1}{a-1}}
$$
So, for any value $a \neq 1$, the pair $x = a^{1/(a-1)},$ $y = a \cdot a^{1/(a-1)} = a^{a/(a-1)}$ will give you a solution to the original equation. 

For example, plugging in $a = 2$ gives you $x=2$ and $y = 4$.  Try some other values.
all numbers of the form $x^{1/(x-1)}$, $x&gt;1$ satisfy this.