# The inverse of a function

#### Integrator

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.)
Hello,

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

All the best,

Integrator

#### skipjack

Forum Staff
The original function isn't invertible. If its domain is changed (without restricting it more than necessary) to produce an invertible function (there are two ways of doing that), it will have an inverse that's unique, but depends on the domain chosen.

1 person

#### Integrator

The original function isn't invertible. If its domain is changed (without restricting it more than necessary) to produce an invertible function (there are two ways of doing that), it will have an inverse that's unique, but depends on the domain chosen.
Hello,

Now , I understand! Thank you very much!

All the best,

Integrator