Find the two solutions of $\log x=\frac{1}{2}(x-1)$

Question

Question: solve $\log x=\frac{1}{2}(x-1)$

With the following I only get one solution (apparently with $W_0$), but I can't find the solution with $W_{-1}$ this way. What is my mistake?

$$\log x=\frac{1}{2}(x-1)$$ $$\log x-\frac{1}{2}x=-\frac{1}{2}$$ $$e^{\log(x)-\frac{1}{2}x}=e^{-\frac{1}{2}}$$ $$xe^{-\frac{1}{2}x}=e^{-\frac{1}{2}}$$ $$-\frac{1}{2}xe^{-\frac{1}{2}x}=-\frac{1}{2}e^{-\frac{1}{2}}$$ $$-\frac{1}{2}x=W(-\frac{1}{2}e^{-\frac{1}{2}})$$ $$x=-2W(-\frac{1}{2}e^{-\frac{1}{2}})$$

Answer 1

When solving $ye^y=x$ for real $y$ (given real $x$) there are two cases:

• If $x\ge0$ then there is only one solution $W_0(x)$.
• If $-1/e\le x<0$ then thre are two solutions $W_0(x)$ and $W_{-1}(x)$.

In this problem $-\frac{1}{2}e^{-1/2}$ is in the second case, so we find the solutions to be

$$ -2W_0(-\frac{1}{2}e^{-\frac{1}{2}})=1, \qquad -2W_{-1}(-\frac{1}{2}e^{-\frac{1}{2}})\approx 3.51286. $$

For these I typed the following two things into Mathematica:

1. N[-2 ProductLog[0, -0.5 Exp[-0.5]]]
2. N[-2 ProductLog[-1, -0.5 Exp[-0.5]]]

Answer 2

Question: solve

\begin{align}\ln x&=\tfrac12(x-1)\tag{1}\label{1}\end{align}

$\require{begingroup} \begingroup$ $\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$

Equation \eqref{1} has one obvious root $x=1$. For the other roots we can divide \eqref{1} by $(1-x)$ to get

\begin{align} \frac{\ln x}{(1-x)}&=-\tfrac12 \tag{2}\label{2} , \end{align} a recognizable form of parametric representation of the real branches $\Wp,\Wm$ of the Lambert $\W$ function, which has only one solution $x>1$:

\begin{align} x&=\frac{\Wm\left(-\tfrac12\exp(-\tfrac12)\right)} {\Wp\left(-\tfrac12\exp(-\tfrac12)\right)} =\frac{\Wm\left(-\tfrac12\exp(-\tfrac12)\right)} {-\tfrac12} =-2{\Wm\left(-\tfrac12\exp(-\tfrac12)\right)} \approx 3.5128624 \tag{3}\label{3} . \end{align}

$\endgroup$