The equation $x^y = y^x$.

Question

The following equation has been of interest for a long time. I'm bringing this to your attention again:

$$x^y = y^x$$

How do you attack to find all real and imaginary solutions for this equation? or

What is the nicest solution that you have seen?

Answer 1

y=-x*W(-(ln(x)/x)/ln(x)

W(x) is the Lambert W function

Answer 2

The standard way I have seen to solve $x^y = y^x$ is to define $\frac{y}{x} = r$.

Then $y = xr$, so $x^{xr} = (xr)^x$ or $x^r = xr$ or $x^{r-1} = r$ or $x = r^{1/(r-1)}$.

From this, $y =xr = r^{1+1/(r-1)} = r^{r/(r-1)} $.

Any value you put in for $r$ (except $r=1$) gives you $x$ and $y$ that satisfy $x^y = y^x$, and conversely.

For example, if you start with $r = i$, $\frac{1}{r-1} =\frac{1}{i-1} =\frac{-i-1}{(-i-1)(i-1)} =\frac{-i-1}{2} $ and $\frac{r}{r-1} =\frac{i}{i-1} =\frac{i(-i-1)}{(-i-1)(i-1)} =\frac{1-i}{2} $. Then $x = i^{(-i-1)/2} $ and $y =i^{(1-i)/2} $. (I'll let you work the standard form for $x$ and $y$.)