# if $x^x=y^y$, and $x>0$ and $y>0$, solve for y in terms of x or prove this impossible

So, I was looking at the graph of $$x^x=y^y$$ and saw that they were equal when $$x=y$$, obviously, but there was also another curve. Can you tell me the equation of this curve or prove to me why it is impossible. If possible, please answer only using algebra, geometry, trigonometry, and calculus (derivitives only)

• What do you mean "there was also another curve"? – sammy gerbil Apr 1 at 23:23

So, in general, there is no closed form of the solution. Solving this involves a function called the Lambert W-function, which is defined as the inverse function of $$f(z)=ze^z$$. We can however solve for the case where $$y\geq 1$$
If $$y \geq 1$$, then $$y^y\geq 1$$. For $$0, $$x^x<1$$, so we must have $$x\geq 1$$.
Now, the derivative of $$x^x$$ is given by $$x^x(\ln x +1)$$ This is positive for $$x\geq 1$$ and so $$x^x$$ is strictly increasing, so, for fixed $$y$$, we can have at most one solution to $$x^x=y^y$$ We have one solution ($$x=y$$), and so that must be the only solution for $$y\geq 1$$.
For $$y<1$$, there can be two solutions: $$x=y$$ and $$x$$ given by an exponential of this W function, or, rather, a thing called an analytic continuation of $$W$$. If you are interested in this, I'd read a bit about the complex logarithm first.