Solve $x^k \exp(-x)=a$ for $x$ and some inequalities

Question

I am trying to characterize the set i \begin{align} A_{a,k}=\{ x \ge 0: x^k \exp(-x) > a \} \end{align} in terms of inequaliti on $x$ where $a$ and $k$ are some give positive numbers. That is I am trying to find \begin{align} A_{a,k}=\{ x \ge 0: g_1(a) \le x \le g_2(a) \}. \end{align}

The function $x^k \exp(-x)$ has maximum at $x=\sqrt{k}$. So, we have to assume that $a \le k^{k/2}e^{-k}$ for the set to be non-empty.

To characterize $g_1$ and $g_2$ we would need to solve \begin{align} x^k \exp(-x) = a \end{align} This should have to solution if $a$ is strictly less than $k^{k/2}e^{-k}$. However, I cannot find these.

WolframAlpha gives me a solution in terms of $W$ function \begin{align} x= -k W \left(-\frac{a^{1/k}}{k} \right) \end{align}

However, it doesn't give me the second solution.

Finally my question is: What are $g_1$ and $g_2$. ?

I think \begin{align} g_1 = -k W_0 \left(-\frac{a^{1/k}}{k} \right) \end{align} where $W_0 $ is the zero branch

Answer 1

Expression \begin{align} x&= -k \operatorname{W} \left(-\frac{a^{1/k}}{k} \right) \end{align}

means two real solutions,

\begin{align} x_{0}&= -k \operatorname{W_0} \left(-\frac{a^{1/k}}{k} \right) ,\\ x_{-1}&= -k \operatorname{W_{-1}} \left(-\frac{a^{1/k}}{k} \right) \\ \text{for }\quad -\tfrac1k{a^{1/k}}\in(-\tfrac1{\mathrm{e}},0) , \end{align}

where $\operatorname{W_0},\operatorname{W_{-1}}$ are so-called two real branches of the Lambert W function.

In this case $x_0<x_{-1}$, since for the argument $z:\ -\tfrac1{\mathrm{e}}<z<0$,

\begin{align} \operatorname{W_{-1}(z)}&<-1<\operatorname{W_{0}(z)}<0 . \end{align}

Answer 2

The Lambert W function has branches. The (complex) solutions of $x^k e^{-x} = a$ are $x = -k W_i(-a^{1/k}/k)$ where $W_i$ is any branch of the Lambert W function, and $a^{1/k}$ is any branch of the $k$'th root of $a$. The principal ($0$) branch of $W$ gives you a real solution if $-a^{1/k}/k \ge -e^{-1}$, while the $-1$ branch gives you a real solution if $-e^{-1} \le -a^{1/k}/k < 0$.