# Applications of logarithms

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.

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.

## Growth and decay

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.)

More generally, anything that follows an exponential law of growth or decay looks like

• A = Pbkt.
Here, b can be any base you like (2, e, 10, whatever); once you pick that, k is determined by the problem at hand. If you choose e as your base, then k is the instantaneous rate of growth, which can be explained using calculus.
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This web page was written in 2011 by Toby Bartels, last edited on 2011 May 19. Toby reserves no legal rights to it.

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