Question 406573
Suppose our exponential function is in the form


{{{f(x) = ab^x}}}, where a and b are nonzero constants. Since f(0) = 5, then


{{{5 = ab^0}}} --> a = 5.


Also, since f(2) = 20, we can substitute and get {{{20 = 5*(b^2)}}} --> b = 2 (the solution b = -2 is somewhat extraneous because it's not easily defined on real numbers, and not exponential). Therefore we can write {{{f(x) = 5*2^x}}}.