Given two sets of points from data growing consistently over time, such as subscriptions to a magazine numbering 6.25 million in 1988 and 11.1 million in 1993,

(a) express subscriptions as a natural exponential function of time S = soert and

(b) find the annual rate of growth G in subscriptions.

Setting the values of S0 and r in the proper form S0ert,

(b) The growth rate G of a function is given by the derivative of the natural log of the function:
d/dx[ln f(x)] = f’(x)/f(x). Taking the natural log of S, therefore, and again using (1.13),

Then taking the derivative, recalling that ln 6.25 is a constant,

The growth rate of a natural exponential function will always be the coefficient of the variable in the exponent.

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(a) Letting t=0 for the base year 1988, then t=5 for 1993. Expressing the two sets of data points in terms of S=Se", and recalling that e0 = 1, 6.25 = Soe"(0) = So 11.1 = Soer (5) (1119) (1120) Substituting S, from (11.19) in (11.20) and simplifying algebraically, 11.16.255