Implicit function equation $f(x) + \log(f(x)) = x$

Question

Is there a function $f \colon \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ such that $$ f(x) + \log(f(x)) = x $$ for all $x \in \mathbb{R}_{>0}$?

I have tried rewriting it as a differential equation by introducing $g(x) := f(x) + \log(f(x)) - x$, which is constant $=0$, so $$ 0 = g'(x) = f'(x) + \frac{f'(x)}{f(x)} - 1 $$ which is equivalent to $$ f'(x) \left( \frac{f(x)+1}{f(x)}\right) = 1, $$ but when I solve that, I get back the original implicit function.

Answer 1

You want the inverse function of $f^{-1}(x)=x+\log x$.

Rewrite the function as $x+\log x=\log e^x + \log x = \log xe^x\implies f(x) = W(e^x)$, where $W(x)$ is the Lambert-W or Product Log function.

Answer 2

Without using the Lambert W function, all we need to do is show that given any $x >0$, we can find a unique $y>0$ such that $y+ \log y = x$.

The answer is yes: For $y>0$ and $y \to 0$, the expression $y + \log y \to -\infty$ and if $y \to \infty$ then $y + \log y \to \infty$. Since $y \mapsto y + \log y$ is strictly increasing and continuous on the positive numbers, by the intermediate value theorem there indeed exists a unique $y > 0$ such that $y + \log y = x$ for the given number $x$.