Is it possible to solve $x\ln(x) + \ln(x) + ax = b$ analytically with the Lambert W function?

Question

The question is as described in the title, with $a$ and $b$ being two positive scalars. After a few attempts, my intuition is that it may not be possible. Is it possible to solve the equation analytically for some special value of $a$? Thanks.

Answer 1

By rearranging the equation, we see that the general equation is not of a form solvable by Lambert W:

$$x\ln(x)+\ln(x)+ax=b$$

$$(x+1)\ln(x)+ax=b$$

$$\ln(x)=\frac{-ax+b}{x+1}$$

$$e^\frac{ax-b}{x+1}=\frac{1}{x}$$

$$xe^\frac{ax-b}{x+1}=1$$

The coefficient of the exponential on the left-hand side of the equation is $x$, whereas the argument of this exponential is a rational expression with $x$ in its numerator and denominator.

Let's rearrange the equation to see if it can be transformed at least to a form that is solvable with generalized Lambert W functions for which a polynomial or a rational coefficient is allowed.

Set

$$t=\frac{ax-b}{x+1},\ \ x\to\frac{b+t}{a-t}.$$

$$\frac{b+t}{a-t}e^t=1$$

We see, the equation cannot be solved in terms of Lambert W. But it can be solved in terms of Generalized Lambert W:

$$t=W\left(^{-b}_{+a};-1\right)=-W(^{-a}_{+b};-1)$$

$-$ see the references below.
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[Mezö 2017] Mezö, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553

[Mezö/Baricz 2017] Mezö, I.; Baricz, A.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934 (On the generalization of the Lambert W function with applications in theoretical physics. 2015)

[Castle 2018] Castle, P.: Taylor series for generalized Lambert W functions. 2018