Exponential-Growth Word Problems

The exponential-growth formula A = Pe^rt is related to the compound-interest formula, and represents the case of the interest being compounded "continuously".

Note that the particular variables used in the equation may change from one problem to another, or from one context to another, but that the structure of the equation is always the same. For instance, all of the following represent the same relationship:

No matter the particular letters used, the green variable stands for the ending amount, the blue variable stands for the beginning amount, the red variable stands for the growth constant, and the purple variable stands for time. Get comfortable with this formula; you'll be seeing a lot of it.

How do you solve exponential growth problems?

To solve exponential growth word problems, you may be plugging one value into the exponential-growth equation, and solving for the required result. But you may need to solve two problems in one, where you use, say, the doubling-time information to find the growth constant (probably by solving the exponential equation by using logarithms), and then using that value to find whatever the exercise requested.

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What is an example of an exponential growth problem?

• A biologist is researching a newly-discovered species of bacteria. At time t = 0 hours, he puts one hundred bacteria into what he has determined to be a favorable growth medium. Six hours later, he measures 450 bacteria. Assuming exponential growth, what is the growth constant k for the bacteria? (Round k to two decimal places.)

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Many math classes, math books, and math instructors leave off the units for the growth and decay rates. However, if you see this topic again in chemistry or physics, you will probably be expected to use proper units ("growth-decay constant ÷ time"). If I had included this information in my solution above, my answer would have been "k = 0.25/hour". I doubt that your math class will even mention this, let alone require that you include this. Still, it's not a bad idea to get into the habit now of checking and reporting your units.

Note that the constant was positive, because it was a growth constant. If I had come up with a negative value for the growth constant, then I would have known to check my work to find my error(s).

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• A certain type of bacteria, given a favorable growth medium, doubles in population every 6.5 hours. Given that there were approximately 100 bacteria to start with, how many bacteria will there be in a day and a half?

You can do a rough check of this answer, using the fact that exponential processes involve doubling times. The doubling time in this case is 6.5 hours, or between 6 and 7 hours.

If the bacteria doubled every six hours, then there would be 200 in six hours, 400 in twelve hours, 800 in eighteen hours, 1600 in twenty-four hours, 3200 in thirty hours, and 6400 in thirty-six hours.

If the bacteria doubled every seven hours, then there would be 200 in seven hours, 400 in fourteen hours, 800 in twenty-one hours, 1600 in twenty-eight hours, and 3200 in thirty-five hours.

The answer I got above, 4678 bacteria in thirty-six hours, fits nicely between these two estimates.

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Warning: When doing the above simplification of , try to do the calculations completely within your calculator in order to avoid round-off error. It is best to work from the inside out, starting with the exponent, then the exponential, and finally the multiplication, like this:

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Note: When you are given a nice, neat doubling time, another method for solving the exercise is to use a base of 2. First, figure out how many doubling-times that you've been given. In the above case, this would start by noting that "a day and a half" is 36 hours, so we have:

Use this as the power on 2:

Not all algebra classes cover this method. If you're required to use the first method for every exercise of this type, then do so (in order to get the full points). Otherwise, this base-2 trick can be a time-saver. And, yes, you'd use a base of 3 if you'd been given a tripling-time, a base of 4 for a quadrupling-time, etc.