Question: 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 $\ln(x) = e^{-x}$](https://math.stackexchange.com/questions/993944/solving-lnx-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.

Answer: 
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 $-1$.
Beside Lambert hyperfunction, there is no explicit solution for the zero of function
$$f(x)=e^x\, \log(x)-a$$ and a numerical method would be required and, as usual, a decent initial guess.

$f(x)$ is a very stiff function and it will be much easier to look for the zero of
$$g(x)=x+\log(\log(x))-\log(a)$$ which is quite linear as soon as $x > 2$

Since we can *approximate* $e^x\, \log(x)$ by $x^k\,e^x$, then select $k$ such that
$$\Phi(k)=\int\_1^t \Big[\log (\log (x))-k \log (x)\Big]^2\,dx$$ be minimum. For $t=20$ (this corresponds to $a \sim 1.45\times 10^9$), the optimum value of $k$ is around $0.3494$ which is very close $\frac 7 {20}$. This gives as a first estimate
$$x\_0=\frac{7}{20} W\left(\frac{20 }{7}a^{20/7}\right)$$ With this, start Newton iterations with
$$x\_{n+1}=\frac{x\_n (\log (x\_n) (\log (a)-\log (\log (x\_n)))+1)}{x\_n \log (x\_n)+1}$$

For illustration purposes, let $a=10^p$

$$\left(
\begin{array}{cccc}
 p & x\_0 & x\_1 & \text{solution} \\
 1 & 2.0511456 & 2.3979568 & 2.4241905 \\
 2 & 4.1104351 & 4.2372540 & 4.2377334 \\
 3 & 6.2654856 & 6.2978641 & 6.2978744 \\
 4 & 8.4628503 & 8.4521448 & 8.4521453 \\
 5 & 10.683868 & 10.651835 & 10.651838 \\
 6 & 12.919940 & 12.877274 & 12.877277 \\
 7 & 15.166416 & 15.118952 & 15.118954 \\
 8 & 17.420504 & 17.371663 & 17.371665 \\
 9 & 19.680398 & 19.632289 & 19.632291 \\
 10 & 21.944864 & 21.898834 & 21.898835
\end{array}
\right)$$

So even if not perfectly exact, $x\_1$ (which has a closed form) is *almost* the solution.