I would like to determine the solutions to the inequality $$ \log(1 - x) + 2x > 0. $$
I'm aware of the approach of using the Lambert W function for solving equalities of the form $$e^{x} + ax = b,$$
but I'm unsure how to extend this to an inequality.
Wolframalpha tells me that the range of $x$ is $$0
$$ \log(1-x) + 2x \ge 0 \\ \Longleftrightarrow \log(1-x) + 2(x-1) > -2 \\ \Longleftrightarrow (1-x)e^{2(x-1)} > e^{-2} \\ \Longleftrightarrow 2(x-1)e^{2(x-1)} < -2 e^{-2} $$ Now $y e^y = -2 e^{-2}$ has the two solutions $y_1 = -2$ and $y_2 = W(-2 e^{-2})$, and $y e^y < -2 e^{-2}$ exactly if $y_1 < y < y_2$. Therefore we get $$ -2 < 2(x-1) < W(-2 e^{-2}) \\ \Longleftrightarrow 0 < x < 1 + \frac 12 W(-2 e^{-2}) \approx 0.7968 \, . $$