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)=Ae^{Bx} 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=Ae^{Bx0} we also impose that the curve passes through (x1,y1) by imposing F(x1)=y1=Ae^{Bx1}

We now take the ratio between these two quantities and we obtain

F(x1)/F(x0)=y1/y0=Ae^{Bx1}/Ae^{Bx0 } = e^{B(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 = A^{2}e^{B(x1-x0)} and hence A=sqrt(y0*y1/e^{B(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

A=sqrt(58*16/e^{B(20-6)})=16

The exponential is therefore F(x)=Ae^{Bx} where A and B are the values evaluated here above.

Best,

Davide