How change equation $y=a(1-e^{bx})$ as linear regression to approximate $a$ and $b$? Assume data $(x_i, y_i),i=1,2,...,n $ have a form like $y=a(1-e^{-bx})$.
The question is how to approximate $a$ and $b$ by linear regression.
There is similar question asked here.
Here's what I tried:
$$\ln y = \ln a + \ln(1-e^{-bx})=\ln a + \ln(1-({e^{-x}})^{b})=\ln a + \ln(1-(X)^{b})$$
then
$$Y = A +  \ln(1-(X)^{b})$$
which is not linear relation between $y$ and $x$.
Or as another way:
$$\frac{y}{a}-1=-e^{-bx}\Longrightarrow \ln {(a-y)} - \ln a = \ln(e^{-bx})=-bx$$
which is not linear relationship between $y$ and $x$ again.
I also did'nt find any way to simplify $\ln(1-e^{-bx})$.
$$\ln(1-e^{-bx})=\ln(1-\frac{1}{e^{bx}})=\ln(\frac{e^{bx}-1}{e^{bx}})=\ln(e^{bx}-1)-bx$$
How get rid of $1$? and find linear relationship between $y$ and $x$?
 A: I am afraid that you can't linearize this model.
To solve your problem, two possible approaches:

*

*if you have large enough $x$'s, the ordinate of the horizontal asymptote gives you $a$ (approximately  the $y$ of the largest $x$).


*if you can't estimate the asymptote, hypothesize some $a$ and draw $b$ by
$$b=\frac{\ln\left(1-\dfrac ya\right)}x$$ for every $(x,y)$ pairs. Take the average $b$ and compute the fitting error. Now you have the fitting error as a function if $a$ and you can use a numerical minimizer.

Additional remark:
If you have three equidistant $x$'s, let $x_k=x_0+kh$, you can use the relation
$$\frac{y_2-y_1}{y_1-y_0}=e^{bh}$$ to obtain an approximation of $b$.
A: Of course it is possible to change $\quad y=a(1-e^{bx}) \quad$ for a linear regression. Don't think that is not possible.
The calculus is shown below with a numerical example.


Instead of trying to linearize the original function with elementary transform, we use integral transform. The integral equation is :
$$y(x)=b\int y(x)dx -a\:b\:x +\text{constant}$$
$$y=A\:S+B\:x+C \qquad A=b\quad;\quad B=-a\:b$$
The integral is evaluated by numerical integration (noted $S_k$ for each $x_k$).
Doesn't matter the value of $C$ which depends on the arbitrary lower bound of the integral.
Of course all this can be done with a differential equation instead of an integral equation. But chosing an integral equation is better in practice because numerical integration is more robust than numerical differentiation in case of scattered data.
All is explained with examples in : https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales
Note : Numerical integration is not exact. This introduces some small deviation especially in case of too low number of points.
If one want more accurate fitting according to a particular specified criteria of fitting, one have to use non-linear regression which requires iterative calculus starting from "guessed" values of parameters. The iterative calculus fails if the initial values are not convenient. This is a difficulty which can be overcome in using for initial values the above values first found thanks to the above method.
