How to solve $e^x \ln(x) = a$?

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 $\ln(x) = e^{-x}$ 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 1

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

Answer 2

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 $-1$.

Answer 3

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.