Applications of logarithms (§5.8)

The formula A = Pe^rt 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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