Exponential curve fit I want to fit a curve of the form $y = ab^x +c$ where a, b and c are constant
whereby i have a data of points $(x_i, y_i)$
I can reduced my primary equation into a form $log(y - c) = log(a) + xlog(b)$
I need a $w_i$ such that $w_i = log(y_i - c)$ for each value of $i$
But to do this i need to find the constant $c$ first
assuming i have three point $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ on my data points such that $x_2^2 = x_1*x_3$
Is it possible to get a value of $c$ in terms of $x_1, x_2, x_3, y_1, y_2$ and $y_3$
 A: A straightforward method is explained in the paper :
https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales, page 17.
In this paper,  the symbols and notations are different from those used by Toye_Brainz.  It could be confusing. So, the page is rewrited below according to the new notations :

(A typo corrected in the attachment)
I welcome Claude Leibovici and congratulate him for the numerical example. It helps to be more concrete and specific.
With his data, the above method leads to :
$a=20.358422$ ; $b=0.580724$ ; $c=6.249307$ ; Standard deviation$=1.01924$
Computed by Claude Leibovici, the  Yves Doust’s method leads to :
$a=19.9960$ ; $b=0.60656$ ; $c=6.99992$ ; Standard deviation$=1.406842$
Further iterative non-linear regression carried out by Claude Leibovici, leads to :
$a=20.0868$ ; $b=0.603317$ ; $c=6.03258$ ; Standard deviation$=0.999621$
On the practical viewpoint, the deviation is quite the same if we compare $0.999621$ and $1.01924$  . So, there is no major advantage of the above method, except that it avoids the calculus of the preliminary estimate with the  Yves Doust’s method and it avoids further iterative computation. 
A: No you need points such that $x_2=\dfrac{x_1+x_3}2$.
Then take
$$y_i=ab^{x_i}+c.$$
Subtracting for two different $i$, you eliminate $c$:
$$y_i-y_j=a\left(b^{x_i}-b^{x_j}\right).$$
Then rating a ratio you eliminate $a$,
$$\frac{y_i-y_j}{y_i-y_k}=\frac{b^{x_i}-b^{x_j}}{b^{x_i}-b^{x_k}}=\frac{1-b^{x_j-x_i}}{1-b^{x_k-x_i}},$$
and if $x_i-x_j=2(x_k-x_i)$, there is a nice simplification.
A: This is not an answer (already given by Yves Daoust) but just an illustration of the method on a synthetic example.
I generated $n=101$  data points ($x_i=\frac{i}{10}$) and the $y$'s were generated using $$y_i=20\, e^{-i/20}+6+(-1)^i$$ for which the noise is quite significant.
Using the method proposed by Yves Daoust and using the data points corresponding to $x=2$, $x=5$ and $x=8$, we successively obtain $b=0.60656$, $a=19.9960$ and $c=6.99992$.
Starting with these estimates, in a couple of iterations, the nonlinear regression converged to $$y=20.0868 \times0.603317^x+6.03258$$ to which corresponds an adjusted $R^2=0.992$.
