Solving $x - a \log(x)=b$

Question

Let $a>0$ and $b \in \mathbb{R}$: Assume there exists an $x >0 $ s.t. $$x - a\log(x) = b$$ holds. How can it be determined in closed-form?

Answer 1

Hint. You may express $x$ in terms of $a$ and $b$ using a special function.

From $$ x - a\log(x) = b $$ you deduce, by exponentiation, that $$ e^x e^{- a\log(x)} = e^b $$ or, for $a \neq 0$, $$ -\frac{x}a e^{\large-\frac{x}a} = -\frac{1}ae^{\large-\frac{b}a} $$ setting $X=-\dfrac{x}a$, you have to solve $$ Xe^X=-\frac{1}ae^{\large-\frac{b}a} $$ giving

$$ x=-aW\left(-\frac{1}ae^{\large-\frac{b}a}\right) $$

where $W(\cdot)$ is the Lambert function.