How to solve $x^{y^z}=z$

Question

Initially I isolated the y in $x^y=y$, but I just wanted to expand the infinite power tower to two letters in the tower, but I can't solve for z in the equation $x^{y^z}=z$. I tried to use Lambert W Function but I can't get the form $ze^{z}$ or something like this. I will be thankful if someone help me.

Answer 1

Series solution:

$\def\B{\operatorname B}$

Using $\ln(x)=a,\ln(y)=b$ and Lagrange reversion we get a series for “Hyper Lambert W”

$$x^{y^z}=e^{ae^{bz}}=z\implies z=\sum_{n=1}^\infty \frac1{n!}\left.\frac{d^{n-1}}{dz^{n-1}}e^{ane^{bz}}\right|_{z=0}$$

$e^y$ series:

$$\left.\frac{d^{n-1}}{dz^{n-1}}(e^{ane^{bz}})\right|_{z=0}=\sum_{m=1}^\infty\frac{(an)^m}{m!}\left.\frac{d^{n-1}}{dz^{n-1}}e^{mbz}\right|_{z=0}=\sum_{m=1}^\infty\frac{(an)^m (bm)^{n-1}}{m!}$$

and finally Bell B$_n(x)$ polynomials to get:

$$\boxed{e^{ae^{bz}}=z=1+\frac1b\sum_{n=1}^\infty \frac{(ae^b)^n}{nn!}\B_n(b n)=\frac1b\sum_{n=1}^\infty\frac{(b e^a)^n}{n!}\B_{n-1}(an)}$$

Shown here

Experimental:

A contour integral representation is:

$$\B_n(x)=\frac{n!}{2\pi i}\oint\frac{e^{x(e^t-1)}}{t^{n+1}}dt$$

There are no integral bounds and I have little experience with contour integrals. In case switching the contour integral and sum works:

$$e^{ae^{bz}}=z\mathop=^?1+\frac i{2\pi b}\oint\frac{\ln\left(1-\frac{ae^{be^t}}t\right)}tdt\mathop=^?\frac i{2\pi b}\oint\frac{\ln\left(1-\frac{be^{ae^t}}t\right)}tdt$$

Answer 2

For simplicity, I give an answer for solutions in the reals.

$$x^{y^z}=z$$ $$x^{\left(y^z\right)}=z$$ $$e^{\ln(x)e^{\ln(y)z}}=z$$

Your equation is an equation of elementary functions. It's an algebraic equation in dependence of $e^{\ln(x)e^{\ln(y)z}}$ and $z$. Because the terms $e^{\ln(x)e^{\ln(y)z}},z$ are algebraically independent, we don't know how to rearrange the equation for $z$ by only elementary operations (means elementary functions).
I don't know if the equation has solutions in the elementary numbers.

Your equation cannot be solved in terms of Lambert W but in terms of Hyper Lambert W.

$$\ln(x)e^{\ln(y)z}=\ln(z)$$ $z\to e^t$: $$\ln(x)e^{\ln(y)e^t}=t$$ $$\ln(x)=te^{-\ln(y)e^t}$$ $$te^{-\ln(y)e^t}=\ln(x)$$

This equation can be solved by Hyper Lambert W:

$$G(-\ln(y);t)=\ln(x)$$ $$t=HW(-\ln(y);\ln(x))$$ $$z=e^{HW(-\ln(y);\ln(x))}$$

So we have a closed form for $z$, and the representations of Hyper Lambert W give some hints for calculating $z$.

Galidakis, I. N.: On solving the p-th complex auxiliary equation $f^{(p)}(z)=z$. Complex Variables 50 (2005) (13) 977-997

Galidakis, I. N.: On some applications of the generalized hyper-Lambert functions. Complex Variables and Elliptic Equations 52 (2007) (12) 1101-1119