Applications of logarithms (§§6.7&6.8)

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

Compound interest

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 + rt);
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).

Exponential growth and decay

The formula for continuous compound interest applies to anything that follows an exponential law of growth or decay. Outside of finance, it's more common to write A₀ in place of P; more generally, you can put a subscript zero on any variable to indicate its value when t = 0 (so t₀ = 0, for example). In place of using r for the interest rate, we use k for the relative growth rate, a term that can be explained using calculus. (Especially if k is negative, you can also refer to −k as the relative decay rate.) This gives this formula:

A = A₀e^kt.

(The textbook also writes A(t) sometimes, treating A as a function rather than a quantity, but they are inconsistent about that.) You may wish to solve this equation for k or t:

k = ln(A/A₀)/t;
t = ln(A/A₀)/k.

(These formulas can also be used for continuous compound interest, with r in place of k and P in place of A₀.)

You can replace e with any other valid base (2, 10, whatever), so long as you change k to match (but then k is no longer the relative growth rate). A different choice of the base can make the correct value of k either more or less obvious. For example, if a quantity doubles in size every H years, then its size after t years is

A = A₀ 2^t/H.

If instead the quantity goes to half its size every h years, then its size after t years is

A = A₀ 2^−t/h.

(In both of these formulas, you can use different units of time than years, as long as you do so both for t and for H or h.) In these formulas, H is called the doubling time, and h is called the halflife. (These formulas are not in the textbook, so it has to do more work when solving a problem involving a doubling time or halflife.)

If an object is placed in an environment at constant temperature, then it will cool down or heat up to reach the environment's temperature. This temperature will neither grow nor decay exponentially; but according to Isaac Newton's law of cooling and heating, the difference in temperature between the object and its environment will undergo exponential decay. If u is the temperature of the object and T is the temperature of its environment, then u − T is the quantity A in the general formula for exponential growth and decay, with u₀ − T in place of A₀:

u − T = (u₀ − T)e^kt.

Although the meaning is less obvious, you may prefer to solve for u:

u = T + (u₀ − T)e^kt.

(This is essentially the textbook's version of the formula.) As with doubling times and halflives, it may be easier to set up a problem using a different base.

Exponential decay is one thing, but exponential growth forever is unrealistic. In the model of logistic growth, there is a carrying capacity beyond which a population cannot grow. In this case, there is still an exponential growth, but it is the ratio of the population to the remaining capacity that grows exponentially. If P is the population and c is its carrying capacity, then P/(c − P) is the A in the general formula for exponential growth and decay, with P₀/(c − P₀) in place of A₀:

P/(c − P) = P₀/(c − P₀) e^kt.

Again, you may prefer to solve for P:

P = cP₀e^kt/(P₀e^kt + c − P₀).

This looks a little simpler (and is also easier to solve for t) if you divide both sides by P₀e^kt and write a for (c − P₀)/P₀ (which is the reciprocal of the initial value of the ratio of the population to the remaining capacity):

P = c/(1 + ae^−kt).

(This is essentially the formula in the textbook, but they don't explain where it comes from.)

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