This guy solves $x^y=y^x$ by introducing a parameter $t$:
$$x=t^\frac{1}{t-1}$$
$$y=t^\frac{t}{t-1}$$
Fine, but my first instinct was to use the change-of-base formula:
$$\frac{\ln{y}}{\ln{x}} = \frac{\ln{x}}{\ln{y}}$$
$$(\ln{y})^2 = (\ln{x})^2$$
$$\ln{y} = \pm\ln{x}$$
given $y\ne x$,
$$\ln{y} = -\ln{x}$$
Why is this wrong?
$$x^y = y^x$$
does not imply
$${\ln(y)\over \ln(x)} = {\ln(x)\over \ln(y)} $$
but instead
$$y\ln(x) = x\ln(y)$$ or $${\ln(x)\over \ln(y)} = {x\over y}$$