Is the function $y=xe^x$ invertible? I'm wondering if the equation $re^r=se^s$ has any answer. If there is any answer,and $r=-1+v,s=-1-v$ in which $v$ is a positive real number,what can we say about $v$?
Thank you in advance.

Let $r=-1+v$ and $s=-1-v$. You are looking for zeros of the function:
$$
f(v) = (-1+v)e^{-1+v} - (-1-v)e^{-1-v}.
$$
Observe that one solution is $v=0$ which corresponds to $r=s=-1$.
Compute the derivative
$$
f'(v) = ve^{-1-v} (-1+e^{2 v})
$$
and notice that it is never negative and it is zero only for $v=0$. Hence $f$ is strictly increasing. This means $f$ is injective and there cannot be other solutions. So you can say that $v=0$.
