Where does the curve $x^y = y^x$ intersect itself? This problem is quite easily solved by using logarithms and derivative and forming the function $f(x,y) = x^y - y^x$, however there are assertions that this problem can be solved without using either of the two. I can not see how one would proceed to solve this problem in the absence of logarithmic manipulation and differentiation. Can anyone shed some light on this?
 A: Here's a method that uses logarithms, but not differentiation.
Of course, one portion of the line is simply that of $y = x$.
Following the method here, we find that the other portion of the curve $x^y = y^x$ for which $x \neq y$ is parameterized by
$$
(a^{1/(a-1)},a^{a/(a-1)}), \quad a \neq 1
$$
It stands to reason (given that the curve intersects) that the point of this intersection is
$$
\lim_{a \to 1}(a^{1/(a-1)},a^{a/(a - 1)})
$$
So, it suffices to verify the limits 
$$
\lim_{a \to 1} a^{1/(a-1)} = \lim_{a \to 1} a^{a/(a-1)} = e
$$
Note that if we rewrite these limits in terms of $b = 1/(a - 1)$ (or conversely $a = 1/b + 1$) and consider only the limit from the right, we're really evaluating
$$
\lim_{b \to \infty}\left(1 + \frac 1b \right)^b \qquad 
\lim_{b \to \infty} \left(1 + \frac 1b\right)^{b + 1}
$$
which both come out to $e$, by whichever definition of $e$ you prefer.
A: Ok the function $f\colon \mathbb R^2 \to \mathbb R$ is continuous. And $f(1,2)=-1$ and $f(2,1)=1$ so it has to take the values between -1 and 1. Thus for some point we have that $f$ is zero.
