Hello,Let $\text{W}$ denote the Lambert W function.

Let $y$ denote the inverse of the function $-x + \log_3(x)$ with domain (0, 1],

then $x = -y + \log_3(y) = -y + \ln(y)/\!\ln(3)$, where $x \leqslant -1$ and $0 < y \leqslant 1$,

so $\ln(3)x = -\ln(3)y + \ln(y)$,

which implies $3^{\large x} = ye^{-\large\ln(3)y}$,

and so $-\ln(3)3^x = -\ln(3)ye^{-\large\ln(3)y}$.

Hence $-\ln(3)y = \text{W}\!\left(-\ln(3)3^{\large x}\right)$,

i.e. $y = -\text{W}\!\left(-\ln(3)3^{\large x})/\!\ln(3\right)$.

If $x \geqslant 1$, the inverse is $-\text{W}_{-1}\!\left(-\ln(3)3^{\large x})/\!\ln(3\right)$.

(I've omitted some justification of the above.)

Do I understand, however, that the function has two inverse functions?Thank you very much!

All the best,

Integrator