TITLE: Inverse of $y=xe^x$
QUESTION [12 upvotes]: I feel like finding the inverse of $y=xe^x$ should have an easy answer but can't find it.

REPLY [2 votes]: As this answer shows, the Lambert W function is not absolutely necessary.  First, notice that $y<\ln(x)$ for large $x$ since $xe^x>e^x$ for $x>1$.
We may then start rewriting as follows:
$$x=ye^y\implies e^y=\frac xy\implies y=\ln\left(\frac xy\right)$$
It then follows that since $y<\ln(x)$, we have
$$y>\ln\left(\frac x{\ln(x)}\right)=\ln(x)-\ln(\ln(x))$$
We've reached Yuval Filmus' short answer at this point.  We may then find an upper bound by noting that
$$y=\ln\left(\frac x{\ln\left(\frac x{y}\right)}\right)<\ln\left(\frac x{\ln\left(\frac x{\ln(x)}\right)}\right)=\ln(x)-\ln(\ln(x)-\ln(\ln(x)))$$
Here is a graph, the dotted line being $y$ and the other two being the bounds:

Note that we can repeat this process indefinitiely, stacking more and more logarithms.  The end result is the following image:

Indeed, this converges for all $x\ge e$.
We may evaluate the inverse for small values of $x$ as well.  Going back to the beginning, we could've had the following:
$$x=ye^y\implies y=xe^{-y}$$
Using $y\le x$ as our initial statement, we can bound small values of $x$:
$$y\ge xe^{-x}$$
And likewise,
$$y=xe^{-xe^{-y}}\le xe^{-xe^{-x}}$$
Here's a graph of the inequalities, the dotted line being the actual inverse again:

Notice the inequalities fail when $x<0$, which is natural.  Again, continuing this process, we end up with the following graph:

which converges to the primary branch for $-e^{-1}\le x\le e$.  To get the pesky other branch is much harder, and I don't know of a neat way to find it off the top of my head :-(