Lambert Function as a solution

Question

The solution to the equation

$$Xe^X=K$$

is given by

$$X=W(K)$$

where $W$ is the Lambert function.

Is it possible to adapt this such that we can find a solution for

$$\frac{1-e^X}{X}=K?$$

Answer 1

We have $$ \tag+\frac{1-e^X}X = K \iff 1-KX = e^X \iff (1-KX)\cdot e^{-X} = 1$$ We want to have something of the form $Ye^Y$ to apply $W$, hence we write $$ e^{-X} = e^{-\frac{KX}K} = \frac{e^{\frac 1K - \frac{KX}K}}{e^{1/K}} = \frac{e^{\frac{1-KX}K}}{e^{1/K}} $$ So, we divide $(+)$ by $K$ to get $$ \frac{1-KX}{K} \cdot e^{\frac{1-KX}K} = \frac{e^{1/K}}K \iff \frac{1-KX}K = W\left(\frac{e^{1/K}}K\right) \iff X = \frac 1K-W\left(\frac{e^{1/K}}K\right) $$

Answer 2

The solution given by user1337 in the comments is correct. To derive it rewrite your equation

$$ \frac{1-e^{X}}{X} = K$$

to

$$ \frac{1-KX}{K}e^{\frac{1-XK}{K}} = \frac{e^{\frac{1}{K}}}{K}$$

This is now on the form $W(Z) e^{W(Z)} = Z$ with $W(Z) = \frac{1-XK}{K}$ and $Z = \frac{e^{\frac{1}{K}}}{K}$.

Answer 3

The solution using variation of parameters is $$ x=\frac{1}{K}-\operatorname{W}\left( \frac{{{e}^{\frac{1}{K}}}}{K}\right) $$