How can I find the solution to $x(e^x)+\ln(x)+c = 0$, where $c$ is constant?
I believe the solution is solved using Lambert $W$-function and derivatives, but I can't figure it out. I can't find any ideas. Can someone tell me the idea behind this problem?
Consider the two functions $$f(x)=x e^x+\log(x)+c \qquad \text{and} \qquad g(x)=x+\log(x)+c$$ Since $f(x)>g(x)$, the solution of $f(x)=0$ is smaller than $$a=W\left(e^{-c}\right)\qquad \text{and} \qquad f(a)=\log \left(W\left(e^{-c}\right)\right)+c+e^{-c}$$ is always positive.
Expanding $f(x)$ as an infinite series around $x=a$ gives $$f(x)=\left(e^a a+\log (a)+c\right)+\sum_{n=1}^\infty \Bigg[\frac{e^a (a+n)}{\Gamma (n+1)}-(-1)^n\frac{ a^{-n}}{n} \Bigg](x-a)^n$$ Truncating to some order and using series reversion with (for easier notations) $$k=\left(e^a a+\log (a)+c\right)\qquad \text{and} \qquad b_n=\frac{e^a (a+n)}{\Gamma (n+1)}-(-1)^n\frac{ a^{-n}}{n}$$
$$\color{blue}{x=a-\frac k{b_1}-\frac{b_2 k^2}{b_1^3}+\frac{\left(b_1 b_3-2 b_2^2\right) k^3}{b_1^5}-\frac{\left(5 b_2^3-5 b_1 b_3 b_2+b_1^2 b_4\right) k^4}{b_1^7}+\cdots}$$
Trying for a few values of $c$, the results
$$\left( \begin{array}{ccc} c & \text{approximation} & \text{solution}\\ 0.00 & \color{red}{0.47067}41211 & 0.4706753748 \\ 0.25 & \color{red}{0.41522}37046 & 0.4152225064 \\ 0.50 & \color{red}{0.361196}8377 & 0.3611964501 \\ 0.75 & \color{red}{0.3097050}650 & 0.3097050036 \\ 1.00 & \color{red}{0.2618048}436 & 0.2618048383 \\ 1.25 & \color{red}{0.218353889}6 & 0.2183538895 \\ 1.50 & \color{red}{0.1798994651} & 0.1798994651 \end{array} \right)$$ which seem to be decent.
For sure, we could take more terms in the expansion in terms of $t$ and obtain better results.