Solutions of $x^ y = y ^ x$, with rational x and irrational y It seems to me that the equation $x^y = y^x$  has no solution in which $x$ is rational and $y$ irrational, or vice versa.
I could not get any counterexample.

The equation is equivelant to $y\log x = x\log y$ or ${y\over\log y}={x\over\log x}$. As $x/\log x$ is convex, for any rational number $x$, there exists a real number $y$ such that $x^y = y^x$. 
Let $y$ be such that $3^y=y^3$. Suppose $y$ is rational, $y=\frac mn$. Then, 
$$n^{3n}3^m=m^{3n}$$
Which implies $n=1$, as $(m,n)=1$. So, $m^3=3^m$. Contradiction, as $(2,4)$ is the only integer solution.
Thus, $(3,y)$ is a rational-irrational pair for the equation.
