Solution to an equation with logarithms of type $x\log(x) + ax + b = 0$

Question

I've encountered a transcendental equation with logarithms like $x\log(x) + ax + b = 0$, and I'm wondering if there is a closed-form solution for the equation.

Answer 1

I suspect there is no closed form solution. Transcendental functions such as these usually dont; especially considering the variables $a,b$, there are just too many degrees of freedom and too many unique equations.

Note: I am making the assumption that $\log$ is the natural logarithm. If this is not so, you can easily modify the math.

We can substitute $x=e^u$ and arrive at the equation $ue^u + ae^u + b=0$.

Rewrite it as $e^u(u+a)=-b$

Multiply by $e^a$ to produce $e^{u+a}(u+a)=-be^a$

Take advantage of the Lambert W function: $u+a= W(-be^a)$

Thus: $$\log(x) = W(-be^a) - a$$

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In a comment you asked about the Lambert W having a derivative. It does have an implicit derivative: $$\frac{d W(x)}{dx} = \frac1{x + e^{W(x)}}$$

Answer 2

This kind of equations cannot be solved by rearranging it by applying a finite number of elementary functions.

If Schanuel's conjecture is true and $a,b\in\overline{\mathbb{Q}}$, the equation doesn't have solutions that are elementary numbers.

see Closed-form solvability of elementary transcendental equations?

As CogitoErgoCogitoSum showed in his answer, the non-elementary solution can be represented as

$$x=e^{W(-b{e^a})-a}=-\frac{b}{W(-be^{a})}.$$ $\ $

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$$W'(x)=\frac{W(x)}{x+xW(x)}$$