Original problem: Find the real solution of $$e^{-x} = -\ln(x)$$ The first thing I did was divide both sides by $e^{-x}$. With a few more steps I arrived at $$e^x\ln(x) = -1$$ This was great, but still not a solution. I checked Wolfram|Alpha to see what the solution was, and it said that that $x$ was $W(1)$, or alternatively $e^{-W(1)}$.
I'm just wondering, what are the steps you need to make to get to the answer of $W(1)$?
On
let $w=W(1)$. Then, by definition, we have $we^w=1$.
It follows that $e^{-w}=w$ and, taking logs, we also have $w=-\ln(w)$. Combining these we see that $$e^{-w}=-\ln w$$ as desired.
Not sure that this generalizes in any particularly useful way....
On
$-\log(x)$ is the symmetric function of $e^{-x}$ rotated around the symmetry axis $x=y$ (i.e. the 45-degree axis). Therefore they both will intersect at a point $x_0$ along the symmetry axis where $x=y$. Thus both $e^{-x_0}=x_0=-\log(x_0)$. The function $w\cdot e^{w}=z$ is involved in the definition of the Lambert function with $w=W_k(z)$. Therefore for $z=1$ we have $e^{-w}=w$, in which $w=W(1)$.
$$e^{-x}=-\ln(x)$$
We want to solve the equation in terms of Lambert W. We therefore try to bring the equation to a form $te^t=\text{constant}$:
$x\to e^t$: $$e^{-e^t}=-t$$ But we see we cannot achieve this goal.
To be able to apply Lambert W anyway, we try to bring at least one side of the equation to a form $f(x)e^{f(x)}$:
$$-xe^{-x}=x\ln(x)$$ $$-x=W(x\ln(x))$$ $$-x=\ln(x)$$ $x\to e^t$: $$-e^t=t$$ $$-1=te^{-t}$$ $$1=-te^{-t}$$ $$-te^{-t}=1$$ $$-t=W(1)$$ $$t=-W(1)$$ $$x=e^{-W(1)}$$ $$x=W(1)$$