# 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 ert.
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.

To illustrate how continuous compound interest comes about, suppose that P = \$1000.00, r = 6% = 6/100 = 3/50 = 0.06 (per year), and t = 1 year. With simple interest, this gives A = P(1 + rt) = (\$1000.00)(1 + (0.06)(1)) = \$1060.00. With n = 1/t = 1/(1) = 1, that is compounded annually, the result is the same (although calculated differently): A = P(1 + r/n)nt = (\$1000.00)(1 + (0.06)/(1))(1)(1) = \$1060.00. If we compound n = 4 times per year (quarterly), then there is some interest on the interest, so the result is larger: A = P(1 + r/n)nt = (\$1000.00)(1 + (0.06)/(4))(4)(1) = \$1061.363550624 ≈ \$1061.36. Here I've rounded off to the nearest cent, which I'll do from now on. If we compound n = 12 times per year (monthly), then there is more interest on the interest, so the result is even larger: A = P(1 + r/n)nt = (\$1000.00)(1 + (0.06)/(12))(12)(1) ≈ \$1061.68. If we compound 30 times more often than that, that is n = 360 times per year (daily by bankers' traditional counting methods), then the amount is even larger, although not very much larger: A = P(1 + r/n)nt = (\$1000.00)(1 + (0.06)/(360))(360)(1) ≈ \$1061.83. If you increase n still more, then in principle A will continue to increase, although in fact you won't notice it for a while due to rounding. Even if you use n = 1235, you still get A ≈ \$1061.83. If you use n = 1236, then you finally get A ≈ \$1061.84; after that … you'll always get A ≈ \$1061.84, no matter how large you make n. And continuous compound interest gives you this limiting result directly: A = Pert = (\$1000)e(0.06)(1) = \$1000e0.06 ≈ \$1061.84. No matter how often you compound the interest, the final result will never be larger than that!

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 (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, then A gets arbitrarily close to Pert as n gets arbitrarily large. 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 book's definition of e on the top of page 431; the n in their definition is not my n but rather my 1/x.)

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).

## Growth and decay

Anything that follows an exponential law of growth or decay looks like
• A = P ekt,
where k is the relative growth rate, a term that can be explained using calculus. However, you can replace e with any other valid base (2, 10, whatever), so long as you change k to match. 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 = P 2t/H,
where P is the original size. If instead the quantity goes to half its size every h years, then its size after t years is
• A = P 2t/h.
(In both of these formulas, you can use different units of time than years, as long as you do so for both t and H or h.) In these formulas, H is called the doubling time, and h is called the halflife.

## Variations

If an object is placed in an environment at constant temperature, 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 will undergo exponential decay:
• A − T = (P − T)ekt,
where T is the temperature of the environment. You may prefer to solve for A:
• A = T + (P − T)ekt.
The book's version of this formula uses different variable names, but it is essentially the same.

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:

• A/(C − A) = P/(C − P) ⋅ ekt,
where C is the carrying capacity. This looks rather different if we solve for A:
• A = CPekt ÷ (C − P + Pekt).
The book has a slightly different formula, using not only different variable names but also constants with a different meaning, and dividing both sides of the fraction by ekt (which makes it easier to solve for t).
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