how to find the solutions of this equation (with Lambert Function)? I need the solutions $x$, such that $x\ln|x|+\frac{1}{4}=0 $ is true. Wolframalpha gives http://www.wolframalpha.com/input/?i=xln|x|%2B1%2F4%3D0, but I never heard of the Lambert W-Function before. Can you give me a hint how to find the solutions of this equation? Best.
 A: The Lambert-W function is the solution to the inverse of $f(x)=xe^x$, where $W(xe^x)=f^{-1}(xe^x)=x$. This is also true for $W(xlnx)$. So let's get your equation into that form:
$$
xln|x|+\frac{1}{4} = 0
$$
$$
xln|x| = -\frac{1}{4}
$$
Now, in order to take the $W$ of both sides, we need to resolve this absolute value by splitting our equation into two parts, one with $ln(x)$ and one with $ln(-x)$. Here's the positive branch:
$$
W(xlnx) = W(-\frac{1}{4})
$$
$$
lnx = W(-\frac{1}{4})
$$
$$
x = e^{W(-\frac{1}{4})}
$$
This branch gives us two of our answers, $e^{W_0(-1/4)} \approx .699491$ and $e^{W_{-1}(-1/4)} \approx .116101$. Then, our negative counterpart is as follows:
$$
xln(-x) = -\frac{1}{4}
$$
$$
-xln(-x) = \frac{1}{4}
$$
$$
W\biggl(-xln(-x)\biggr) = W\left(\frac{1}{4}\right)
$$
$$
ln(-x) = W\left(\frac{1}{4}\right)
$$
$$
-x = e^{W\left(\frac{1}{4}\right)}
$$
$$
x = -e^{W\left(\frac{1}{4}\right)}
$$
Thus, our final value for $x$ is $-e^{W_0\left(\frac{1}{4}\right)} \approx -1.226161$. There exist, in fact, 3 more solutions in total from the $W_{-1}$ and $W_1$ branches for both $lnx$ and $ln(-x)$, however those solutions are complex answers, not real.
