I am very new to the Lambert W function, and would gladly have some details about how to use it for solving inequalities when required (and I know it is for the hereafter inequality).
I have seen on the Lambert W function – Wikipedia page that a general equation of the form $x=a+be^{cx}$ has the solution: $x=a-\frac{1}{c}W\left(-bce^{ac}\right)$. But don't understand the steps made in-between, and would love to, instead of using it as a "formula". Moreover, I am rather interested in the (weak) inequality case.
Hence, would someone be able and agree to walk me through the steps of solving algebraically such a general case of the following inequality for $x$:
$$ x \leq a + be^{cx}$$
Where all $x, a, b, c$ are reals. Moreover, in my case (if that would to change something for the steps), I know: $a>0$, $b<0$, $c<0$ and $x\in [0, 1]$.
Thank you,
EDIT: to maybe be clearer, I'm looking to understand how to get the solution – i.e know the steps – which is: $x \leq a- \frac{1}{c}W(-bce^{ac})$
Best Answer
$$x=a+be^{cx}$$ $$x-a=be^{cx}$$ $$x-a=be^{c(x-a)}e^{ac}$$ $$c(x-a)=bce^{ac}e^{c(x-a)}$$ $$c(x-a)e^{-c(x-a)}=bce^{ac}$$ $$-c(x-a)e^{-c(x-a)}=-bce^{ac}$$ $\begin{cases} X=-c(x-a)\\ Y=-bce^{ac} \end{cases}\quad\implies\quad Xe^X=Y$ $$X=W(Y)$$ $$-c(x-a)=W\left(-bce^{ac}\right)$$ $$x=a-\frac{1}{c}W\left(-bce^{ac}\right)$$