How to solve $e^x \ln(x) = a$? I was wondering if it was possible to solve equations of the form  $e^x \ln(x) = a, \;a > 0$  in terms of the Lambert $W$ function $W(x)$?
I understand that fixed point iteration or the Newton-Rhapson method would work here but am wondering if there is another technique that could give a closed form.
The question Solving $\ln(x) = e^{-x}$ already exists and is equivalent to the case where $a = 1$, but all the answers point to using approximate methods or say that solutions are normally involve the Lambert $W$ function but don’t give any explicit answers.
 A: Let's consider solutions in closed form.
$$e^x\ln(x)=a$$
1.) No algebraic solutions except $1$ for algebraic $a$
According to Schanuel's conjecture, we have $\forall\ 0,1\neq x\in\overline{\mathbb{Q}}\colon$ $\text{trdeg}_\mathbb{Q}\mathbb{Q}\left(x,\ln(x),e^x\right)=2$. That means: $\ln(a_0),e^{a_0}$ are $\overline{\mathbb{Q}}$-algebraically independent for each algebraic $a_0\neq 0,1$. That means, your equation cannot have algebraic solutions except $1$ for algebraic $a$.
2.) No general elementary solutions
According to the theorem of [Ritt 1925], that is proved also in [Risch 1979], the function $f\colon x\mapsto e^x\ln(x)$ seems not to have partial inverses that are elementary functions for non-discrete domains of $f$.
But that doesn't say if solutions in the elementary numbers are possible for single $a$.
[Risch 1979] Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math. 101 (1979) (4) 743-759
[Ritt 1925] Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90
3.) Solutions in terms of Hyper Lambert W
$$e^x\ln(x)=a$$
$$x\to e^z: e^{e^z}(z+2k\pi i)=a\ \ \ \ \ (k\in\mathbb{Z})$$
$$-\pi<\text{Im}(z)\le\pi:$$
$$e^{e^z}z=a$$
This equation can be solved by Hyper Lambert W:
Galidakis, I. N.: On solving the p-th complex auxiliary equation $f^{(p)}(z)=z$. Complex Variables 50 (2005) (13) 977-997
Galidakis, I. N.: On some applications of the generalized hyper-Lambert functions. Complex Variables and Elliptic Equations 52 (2007) (12) 1101-1119
A: If $a=-1$, there is a solution that uses Lambert twice.
$$\begin{align}
e^x\ln(x)&=-1\\
\ln(x)&=-1\cdot e^{-x}\\
x\ln(x)&=-x\cdot e^{-x}\\
W(x\ln(x))&=W(-x\cdot e^{-x})\\
\ln(x) &= -x\\
x &= e^{-x}\\
e^xx &= 1\\
x&=W(1)
\end{align}$$
Maybe you can modify this to handle general values of $a$. To be honest, I have doubts. Somehow you would need to decompose $a$ in such a way that it could be absorbed partly on the left side of the equation and partly on the right side of the equation to get each side in the form $ze^z$.
With this exact solution for $a=-1$, you could also possibly find approximate solutions more quickly for $a$ near $-1$.
