The problems can be generalized as follows: find the exponential curve which passes through the following points (x0,y0) and (x1,y1)
We assume that the exponential form is of the form F(x)=AeBx which depends upon two parameters, A and B. We need to find the values of A and B such that the exponential curve passes through (x0,y0) and (x1,y1).
We impose that the curve passes through (x0,y0) by imposing F(x0)=y0=AeBx0 we also impose that the curve passes through (x1,y1) by imposing F(x1)=y1=AeBx1
We now take the ratio between these two quantities and we obtain
F(x1)/F(x0)=y1/y0=AeBx1/AeBx0 = eB(x1-x0) . By taking the logarithms of both sides of the equality we obtain
log(y1/y0)=B(x1-x0) which implies B=(1/(x1-x0))*log(y1/y0)
If we multiply the above conditions we obtain y1*y0 = A2eB(x1-x0) and hence A=sqrt(y0*y1/eB(x1-x0))
In practice, in your case you have that the two points are given by (x0,y0)=(6,16) and (x1,y1)=(20,58)
and therefore you have that B=1/(20-6)*log(58/16)=(1/14)*log(3.625)=0.091989 and the value of A is given by
The exponential is therefore F(x)=AeBx where A and B are the values evaluated here above.