Compound interest (§6.7)

There are several applications of exponential functions. To solve for the input of one of these functions is then an application of logarithms.

If you invest (or borrow) an amount of money at a fixed rate of interest, then the amount of money that you have (or owe) at the end of a period of time is an exponential function of time. There are three basic formulas that you want to use:

Simple interest: A = P (1 + r t);
Intermittent compound interest: A = P (1 + r/n)^nt;
Continuous compound interest: A = P e^rt.

In these formulas, the variables have the following meaning:

P is the original amount of money, called the principal;
A is the amount of money after a period of time;
t is the length of time (in years);
r is the (annual) rate of interest; and
n is the frequency (the number of times per year) that interest is compounded.

With simple interest, the interest is applied once, at the end of the time period; this is effectively compound interest where n = 1/t. With intermittent compound interest, the interest is applied several times and added to the original amount, so that interest can be charged on the interest later. With continuous compound interest, the interest is added to the original amount continuously; this is like compound interest where n is effectively infinite.

Here is an example, which will also illustrate how continuous compound interest comes about:

Or download it: WebM format, Ogg Vorbis format, MPEG-4 format.

Another way to see where the special number e comes into it is to use some properties of exponents to write the formula for intermittent compound interest as A = P (1 + r/n)^nt = P (1 + r/n)^(n/r)(rt) = P ((1 + r/n)^n/r)^rt = P ((1 + x)^1/x)^rt, where I've written x for r/n (so that n/r is its reciprocal, 1/x). With r fixed and positive (as 0.06 for example), as n gets arbitrarily large, x = r/n will get arbitrary close to 0 while remaining positive. So if (1 + x)^1/x gets arbitrarily close to some number e as x gets arbitrarily close to 0 while remaining positive, then A gets arbitrarily close to Pe^rt as n gets arbitrarily large. (In Calculus, this sort of thing is called a limit.) And that's exactly what happens; you can approximate e as closely as you like by using a sufficiently small positive number x in the expression (1 + x)^1/x. (Compare the textbook's definition of e on the top of page 443; they write 1/n in place of x, which amounts to using r = 1.)

If you know the final amount A and want to find the principal P, then solve the equation for P. If you know both A and P and want to find the amount of time t, then you must take a logarithm to solve the equation. It's also possible to solve for r, but not for n (at least not with the operations that we use in this class). Here are the results in the case of intermittent compound interest (which is the most complicated):

P = A (1 + r/n)^−nt;
t = log_1+r/n(P/A)/n;
r = n (^nt√(A/P) − 1).

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