Solve equation $\tfrac 1x (e^x-1) = \alpha$ I have the equation $\tfrac 1x (e^x-1) = \alpha$ for an positive $\alpha \in \mathbb{R}^+$ which I want to solve for $x\in \mathbb R$ (most of all I am interested in the solution $x > 0$ for $\alpha > 1$). How can I do this?
My attempt
I defined $\phi(x) = \tfrac 1x (e^x-1)$ which can be continuously extended to $x=0$ with $\phi(0)=1$ ($\phi$ is the difference quotient $\frac{e^x-e^0}{x-0}$ of the exponential function). Therefore it is an entire function. Its Taylor series is
$$\phi(x) = \frac 1x (e^x-1) = \frac 1x (1+x+\frac{x^2}{2!}+\frac{x^3}{3!} + \ldots -1) = \sum_{n=0}^\infty \frac{x^n}{(n+1)!}$$
Now I can calculate the power series of the inverse function $\phi^{-1}$ with the methods of Lagrange inversion theorem or the Faà di Bruno's formula. Is there a better approach?
Diagram of $\phi(x)=\begin{cases} \tfrac 1x (e^x-1) & ;x\ne 0 \\ 1 & ;x=0\end{cases}$:

 A: The solution can be expressed in terms of Lambert's W (the inverse function of $x \mapsto xe^x$, implemented as ProductLog in Mathematica). The equation 
$$
\frac{e^x - 1}{x} = \alpha
$$
is equivalent to
$$
(-\alpha^{-1} - x) e^{-\alpha^{-1} - x} = - e^{-\alpha^{-1}}\cdot \alpha^{-1}
$$
and therefore
$$
- \alpha^{-1} - x = W(-e^{-\alpha^{-1} }\alpha^{-1})
$$
or
$$
x = - \frac{1}{\alpha} - W\left( -\frac{e^{-\alpha^{-1}}}{\alpha} \right) \, .
$$
A: I just want to complete Hans Engler's answer. He already showed
$$x = -\frac 1\alpha -W\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right)$$
$\alpha > 0$ implies $-\tfrac 1\alpha \in \mathbb{R}^{-}$ and thus $-\tfrac 1\alpha e^{-\tfrac 1\alpha} \in \left[-\tfrac 1e,0\right)$ (The function $z\mapsto ze^z$ maps $\mathbb{R}^-$ to $\left[-\tfrac 1e,0\right)$) The Lambert $W$ function has two branches $W_0$ and $W_{-1}$ on the interval $\left[-\tfrac 1e,0\right)$:

So we have the two solutions
$$x_1 = -\frac 1\alpha -W_0\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right)$$
$$x_2 = -\frac 1\alpha -W_{-1}\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right)$$
One of $W_0\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right)$ and $W_{-1}\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right)$ will always be $-\tfrac 1\alpha$ as $W$ is the inverse of the function $z \mapsto ze^z$.  This solution of $W$ would give $x=0$ which must be canceled out for $\alpha \ne 1$ as $\phi(x)=1$ just for $x=0$.
Case $\alpha=1$: For $\alpha=1$ is $-\tfrac 1\alpha e^{-\tfrac 1\alpha}=-\tfrac 1e$ and thus $W_0\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right)=W_{-1}\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right)=-1$. This gives $\phi^{-1}(1)=0$ as expected.
Case $\alpha > 1$: $\alpha > 1 \Rightarrow 0 < \tfrac 1 \alpha < 1 \Rightarrow -1 < -\tfrac 1 \alpha < 0$.
Because $W_0(y) \ge -1$ it must be $W_0\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right)=-\tfrac 1\alpha$ and so
$$\phi^{-1}(\alpha) = -\frac 1\alpha -W_{-1}\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right)\text{ for } \alpha > 1$$
Case $\alpha < 1$: $0 < \alpha < 1 \Rightarrow \tfrac 1 \alpha > 1 \Rightarrow -\tfrac 1\alpha < -1$
Because $W_{-1}(y) \le -1$ we have $W_{-1}\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right)=-\tfrac 1\alpha$ and thus
$$\phi^{-1}(\alpha) = -\frac 1\alpha -W_{0}\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right)\text{ for } \alpha < 1$$
Solution
The solution is
$$\phi^{-1}(\alpha) = \begin{cases} -\frac 1\alpha -W_{-1}\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right) & ; \alpha > 1 \\ 0 & ; \alpha = 1 \\-\frac 1\alpha -W_{0}\left(-\frac 1\alpha e^{-\tfrac 1\alpha}\right) & ; \alpha < 1 \end{cases}$$ 
