I am trying to linearize the following equation: $Y=a+be^{cx}$ where $Y$ and $x$ are given data points and $a$,$b$ and $c$ are constants that needs to be found. And by rearranging the equation, I got to this point:
$$\begin{align*} \ln Y &= \ln a+\ln(be^{cx}) \\ \ln Y &= cx + \ln a + \ln b \\ \ln Y &= cx + \ln(ab) \end{align*}$$
Using a graphical software, I linearized the formula ($y=mx+\text{intercept})$ and got my constant $c$. However I am stuck at finding exact values of constants $a$ and $b$. This is where I have reached:
$$\ln(ab)=p$$
$$a=\frac{e^p}b$$
where $p$ is the $y$-intercept which is known. Any leads/approach to finding the exact values for constants $a$ and $b$?
One cannot linearize $y=a+be^{cx}$ in the common sense of "linnearize". All common approaches can only be more or less approximates on limited ranges.
That is why the usual way to fit this function to a given set of points requires some iterative calulus (non-linear regression).
Taken in broader sense "linearize" can be exactly achieved involving integal and/or differential. In the present case one get $$y=a+be^{cx}\quad\implies\quad \int y\:dx=ax+\frac{b}{c}e^{cx}+constant$$ $$\int y\:dx=ax+\frac{1}{c}y+constant$$ $$\boxed{y=c\:S+C_1x+C_2 \quad \begin{cases} S=\int y\:dx \\ C_1=-a\:c\\ C_2=constant \end{cases}}$$ From the data $(x,y)$ the integral $S(x)$ is computed thanks to numerical integration : See below.
$y$ is a linear function of $S$ and $x$. The parameters $c$ , $C_1$ , $C_2$ are computed thanks to linear regression. This is the unusual sens of "linearisation" in this method.
Since a value of $c$ is obtained $y=a+bX$ is linear wrt the new variable $X=e^{cx}$.
Of course the numerical integration introduces some deviation all the less important than the number of the given points $(x,y)$ is large.
Reference : https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales . Numerical examples can be found in this paper.