Inverse of the function $- \log(1-[1-e^{-x^\alpha}]^\beta)$

Question

I have a function as follows, I would like to get the inverse of this function. What is the inverse of $f(x)$?

$$ y = f(x) = - \log(1-[1-e^{-x^\alpha}]^\beta)$$

Is my answer correct?

$$ f^{-1}(x) = (-\log(1-(1-\exp^{-x})^{1/\beta}))^{1/\alpha}$$

Answer 1

Multiply both sides by $-1$, exponentiate both sides, subtract 1 from both sides, multiply both sides by $-1$, raise both sides to the $\frac{1}{\beta}$ power, subtract 1 from both sides, multiply both sides by $-1$, take natural logarithm of both sides, multiply both sides by $-1$, raise both sides to $\frac{1}{\alpha}$ power.

This solves for $x$ in terms of $y$, which gives you the inverse function in terms of $y$.

(Note that you will need the $\frac{1}{\beta}$ and $\frac{1}{\alpha}$ powers to be defined)

Answer 2

Since you have already done the first step ..$$y = f(x) = - \log(1-[1-\exp(-x^\alpha)]^\beta)$$

$$\implies-{y}=\log(1-[1-e^{-x^{\alpha}}]^{\beta})$$ $$\implies e^{-y}=1-[1-e^{-x^{\alpha}}]^{\beta})$$ $$\implies -e^{-y}+1=[1-e^{-x^{\alpha}}]^{\beta}$$ $$\implies [1-e^{-y}]^{\frac{1}{\beta}}-1=-e^{-x^{\alpha}}$$ which turns out to be $$e^{-x^{\alpha}}=1-[1-e^{-y}]^{\frac{1}{\beta}}$$. Taking $\log$ on both sides we get $$-x^{\alpha}=\log (1-[1-e^{-y}]^{\frac{1}{\beta}})\implies log \frac{1}{(1-[1-e^{-y}]^{\frac{1}{\beta}})}=x^\alpha$$

Hence $$f^{-1}(x)=\left(log \frac{1}{(1-[1-e^{-x}]^{\frac{1}{\beta}})}\right)^{\frac{1}{\alpha}}$$