How might one go about solving the inequality:
$x^{a}e^{x+c}\leq b$
where $a,b,c$ are arbitrary constants ($b\geq 0$ and $a\neq0$) for $x$.
My first place would be to try and get all of the variables I am solving for on one side, but this problem evades such tactics.
There is an "algebraic solution" if you allow the Lambert W function which is defined by
$$y=W(x)\iff ye^y=x$$
Let's solve the equality. I'll ignore the multiple-branch issue here which is equivalent to saying there may be two answers or none rather than just one.
$$\begin{align} x^ae^{x+c} &= b \\[1.5ex] x^ae^x &= be^{-c} \\[1.5ex] xe^{x/a} &= b^{1/a}e^{-c/a} \\[1.5ex] \frac xa e^{x/a} &= \frac{b^{1/a}e^{-c/a}}a \\[1.5ex] \frac xa &= W\left(\frac{b^{1/a}e^{-c/a}}a\right) \\[1.5ex] x &= aW\left(\frac{b^{1/a}e^{-c/a}}a\right) \end{align}$$
There may be zero, one, or two solutions to that equality, depending on the value inside the $W$ function. Those solutions will divide the real number line into intervals, some of which may satisfy your inequality.
If that kind of solution interests you, let me know and I'll flesh out the rest of the solution.