This answer (https://math.stackexchange.com/a/680695/438808) was helpful to me in finding the quadratic equation for three known points.
Is there a similar solution for finding the exponential equation (y = ab^x + c) for three known points?
Thanks in advance!
For certain x-values of points that you're given, you may not be able to solve for the value of $b$ because when you set up a system, a polynomial will be formed with $b$ and polynomials with large degrees aren't necessarily solvable. However, here is a formula for an exponential function passing through any two points $(m,n)$ and $(p,q)$ no matter what the value of $b$ is:
$y=\frac{q-n}{b^p-b^m}(b^x-b^m)+n$
If you can get this through two of your points then you may be able to approximate a value for b.
I've sat down with pen and paper for a bit to try and figure this out and I've come with a solution (not very elegent but it works):
First instead of a variable base, let your equation be y=ae^bx+c. You don't have to do this but I prefer it:
Let m^x = e^bx
e^ln(m)x = e^bx
ln(m) = b
Substituting your 3 points gives 3 equations:
Take equations [ 1) - 2) ] / [ 1) - 3) ] and simplify to get:
(y1-y2)/(y1-y3) = (e^bx1-e^bx2)/(e^bx1-e^bx3)
Now you can solve for b (just shove it into symbolab or whatever). If an answer can't be found, then it is impossible to represent the set of points as an exponential
Now that you have b, sub this back into equations 1) - 2) and solve for a. Now that you have a, solve for c and done
I know this post is pretty old, but posting because it's one of the first links on google when you search how to fit an exponential
You can find the base, $b$ by dividing equally separated differences. if $q - p = n - m$ then $b^{p-m} = (f(q) - f(p)) / ((f(n)-f(m))$
once you have the base, $b$, you can find the added constant.
$c = (b^{m-n}f(n) - f(m)) / (b^{m-n} -1)$
