Algebra and exponential functions

Question

Given $x^n=y$ where $n=y$, I have no problem finding $x$ if $y$ is known. Problem is getting $y$ when only the value of $x$ is known. Is there a way? At the moment I'm working with $x$ and $y$ values less than $1$. Thanks.

Answer 1

If $$x^n=y$$ so we get $$x=\sqrt[n]{y}$$

Answer 2

So you are trying to solve for $y$ in the equation $x^y = y$.

The Lambert W function is useful for this.

$x^y=y \implies x^{-y} = y^{-1} \implies yx^{-y}=1 => ye^{-\ln(x)y} =1$

Multiply by $-\ln x$ on both sides:

$-y\ln xe^{-y\ln x}=-\ln x$

Now use the Lambert W function on both sides.

$-y\ln x=W(-\ln x) \implies y=-\frac {W(-\ln x)}{\ln x}$ For $x \lt 1$, $W(-\ln x)$ is well defined, so this is our solution.