We have three points, which implies we need a logarithmic/exponential function with three constants to be determined.
y=aebx is insufficient and y=aebx+c=aecebx, but aec is just a constant so let's see if we can find a, b, c if y=aebx+c.
For (1,1): 1=aeb+c; for (5,15): 15=ae5b+c; for (10,55): 55=ae10b+c.
From this, 40=ae10b-ae5b, 14=ae5b-aeb. Here, by eliminating c, we now have two variables and two equations.
40=ae5b(e5b-1), 14=aeb(e4b-1),
40/14=20/7=e4b(e5b-1)/(e4b-1)=e4b(eb-1)(e4b+e3b+e2b+eb+1)/[(eb-1)(e3b+e2b+eb+1)],
20/7=e4b(e4b+e3b+e2b+eb+1)/(e3b+e2b+eb+1),
20(e3b+e2b+eb+1)=7e4b(e4b+e3b+e2b+eb+1). From this, b=0.18101 (calculation to be shown later).
So y=ae0.18101x+c.
14=ae0.90505-ae0.18101,
a=14/(e0.90505-e0.18101)=10.9922,
c=1-10.9922e0.18101=-12.1734 (approx).
y=10.9922e0.18101x-12.1734. The three points satisfy this equation which can also be written:
y=10.9922(1.1984x)-12.1734.
How to solve for b: 20/7=e4b(e5b-1)/(e4b-1): rewrite:
20(e4b-1)-7e4b(e5b-1)=0,
20e4b-20-7e9b+7e4b=0. Let f(b)=27e4b-20-7e9b.
Differentiate LHS wrt b:
f'(b)=108e4b-63e9b.
Using Newton's Method (iterative):
bn+1=bn-f(bn)/f'(bn). We need an initial value b0 to start the process. Let's choose b0=1.
b1=0.8904....; b2=0.7820...; b3=0.6756...; b4=0.5723...; b5=0.4743...; etc.
After a few more iterations we get a stable value around b=0.181006155339 (12 decimal places), which was rounded up to 0.18101 in the solution. a=10.9925879274, c=-12.1737653207, and eb=1.19842255597 using this more accurate value for b.
There would have been fewer iterations if b0 had been closer to this computed value. A graph would help to decide on a better approximation than 1 for b0.
Newton's Method could be part of the algorithm to find the logarithmic function.