TITLE: Algebraic solution to natural logarithm equations like $1-x+x\ln(-x)=0$
QUESTION [5 upvotes]: Can you please show me the method to solving natural logarithm ($\ln$) equations of this type (this is just an example): 
$$1-x+x\ln(-x)=0$$
(such that $x<0$?) 
I mean by this type "natural logarithm with polynomials". I can't seem to solve these no matter what I do. I thought there has to be an "algebraic" method rather than a numerical one.

REPLY [3 votes]: If you cannot use Lambert function and do not want numerical methods, you can use function approximation to get a "reasonable" estimate of the solution.
Since $x=-e$ seems to be interesting because of the logarithm, let us use Padé approximants built around this value. To make life simple, let us use degree $1$ for the numerator and degree $n$ for the denominator. This means that, by the end, we just need to solve $$a^{(n)}_0+a^{(n)}_1 (x_{(n)}-e)=0$$
I produced below the expressions for a few values of $n$ and their numerical approximations
$$\left(
\begin{array}{ccc}
n & x_{(n)} &  x_{(n)}\approx \\
 1 & \frac{-3 e-2 e^2}{1+2 e} & -3.56292 \\
 2 & \frac{-4 e-24 e^2-12 e^3}{-2+12 e+12 e^2} & -3.59832 \\
 3 & \frac{6 e-66 e^2-180 e^3-72 e^4}{6-6 e+108 e^2+72 e^3} & -3.58922 \\
 4 & \frac{-24 e+60 e^2-1260 e^3-2160 e^4-720 e^5}{-36+180 e^2+1440 e^3+720 e^4} &
   -3.59164 \\
 5 & \frac{360 e-240 e^2-2700 e^3-61200 e^4-75600 e^5-21600 e^6}{720+240 e-2700
   e^2+18000 e^3+54000 e^4+21600 e^5} & -3.59098
\end{array}
\right)$$