Solve the equation $e^x = yx^n$ for $x$? I've been looking at this equation for some time now:

$$e^x = yx^n$$

where $n \in \mathbb{N}_+$.

I found I can transform it to:

$$e^x = yx^n$$
$$xe^x = yx^{n+1}$$
$$x=W(yx^{n+1})$$

Or to this:

$$e^x = yx^n$$
$$x = \ln(yx^n)$$

If $yx^n$ is positive:

$$x = \ln(x^n)+\ln(y)$$

And if both $x^n$ and $y$ are positive:

$$x = n\ln(x)+\ln(y)$$

But I didn't find anyway to solve these for $x$ from there.
$$
yx^n=e^x
$$
First, divide by $ye^x$
$$
x^n\frac{1}{e^x}=\frac 1y
$$
Then, take the $n$-th root, as suggested by Wojowu in comments:
$$xe^{-\frac xn}=y^{-\frac 1n}$$

Now, multiply by $-\dfrac 1n$ in order to use $W(x)e^{W(x)}=x$:
\begin{align*}-\frac xne^{-\frac xn}&amp;=-\frac{y^{-\frac 1n}}{n} \\
-\frac{x}{n}&amp;=\mathop W\left(-\frac{y^{-\frac 1n}}{n}\right) \\
x&amp;=-n\mathop W\left(-\frac{y^{-\frac 1n}}{n}\right)
\end{align*}
and the equation is solved.
You can not solve the above equation explicit in the form $x=f(y)$