How do I express $y = 1 - e^{-ax^b}$ as a linear function in terms of $y$, $x$, $a$, and $b$?

Question

I tried using the log rules and eventually got to $\ln(\ln(1-y)) = \ln(-a) + b\ln(x)$, but $\ln(-a)$ is obviously not defined across all $a$. This assumes that $x\ge 0$ and that $y<1$.

Answer 1

$$1-y = \exp(-ax^b)$$

From this expression, we already conclude that $y < 1$.

Also, in general, fractional power of a negative number is a point of conflict as discussed here. Hence, I think it is reasonable to assume that $x \ge 0$.

Once you get the intercept $c$, you can solve for $a=-\exp(c)$, we just need to be able to solve for $a$, we do not need $\ln(-a)$ to be defined for all $a$.

Answer 2

If $a=0$, it is $y=1$.

Otherwise: $$1-y=e^{-ax^b} \Rightarrow \\ \ln{(1-y)}=-ax^b \Rightarrow \\ \ln{(1-y)^{-1/a}}=x^b \Rightarrow\\ \ln{(\ln{(1-y)^{-1/a}})}=b\ln x, y<1, x>0 \Rightarrow \\ s=bt.$$

Answer 3

You are done up to $\ln (1-y)=-ax^b$ but from now, different cases are considerable. If $a=0$ then $y=1$ which is the trivial case. Otherwise, we multiply the sides of the equation by $-\text{sgn}(a)$ to obtain$$-\text{sgn}(a)\ln(1-y)=|a|\cdot x^b$$since both sides are always non-negative we can take $\ln$ from the sides and write $$\ln\Big(-\text{sgn}(a)\ln(1-y)\Big)=\ln|a|+b\cdot\ln x$$