# Explicit Representation of $x^{x^y}=y^{y^x}$.

How do you explicitly represent $x^{x^y}=y^{y^x}$ using the Lambert $W$ function?

I started using logarithms to split it up and manipulate it to a form like xe^x. I do this semi-successfully. I go through the steps and get $e^{xln(y)} *ln(y) = e^{yln(x)} *ln(x)$ and basically get stuck. I can multiply $x$ and $y$ to both sides but I still have a problem with having both variables on each side. I may be going about this incorrectly for this type of problem.

• It would be nice to exhibit some work here. For example, what is your understanding of the Lambert $W$ function ? – KonKan Apr 7 '16 at 23:38
• I understand it fairly well, though I've only learned about it on my own (e.g., I know that W(x)*e^W(x) = x). If it is worked out I'll likely understand the process. – Carpenter Apr 7 '16 at 23:43
• when I said "exhibit some work", I meant that it would be a good idea to edit your post and add this info in your post. Try also to explain what problem you are facing when trying to directly applying this to your problem. This might help or even motivate other people to work on it and provide some answer focused in your problem. – KonKan Apr 7 '16 at 23:48

$$xe^x\tag{\leftarrow the x in the exponent is on the second floor.}$$
Now, if I wanted to solve something where a variable was in the first, second, and third floor, I would have to use something beyond the Lambert W function, as the Lambert W function can only solve things (generally, there are some special cases) where the variable is in $2$ consecutive floors. Special cases arrive when we have something like $xe^xe^{xe^x}$, where we see it has the form $f(x)e^{f(x)}$, but if this is not so obvious, it is probably not solvable.
So I'd have to say that it is not possible to solve for $x$ or $y$ here.
If the solution is findable in terms of the Lambert W function, generally make the substitution $x=W(u)$ and use $e^{W(u)}=\frac u{W(u)}$ to reduce the amount of "floors" there are.