Exponential Growth and Decay
In July 2002, National Geographic ran an
article about the problems that
When scientists talk about halflife, they are referring to how long it will take for half of a sample to decay. In the case of nuclear waste, it refers to how long it takes for half of the radioactive material to turn into lead.
Exponential growth is very similar, but deals with, you guessed it, growth, instead of decay. The most common example is the growth of bacteria colonies. Bacteria multiply at an alarming rate. If we assume that bacteria can double every hour and if we start with just a single bacteria, then after one day there will be over 16 million bacteria!
Obviously exponential growth, or decay for that matter, cannot continue indefinitely. Eventually there would no longer be any space or nutrients available for the bacteria, or the last atom of plutonium would decay into lead. As a result, exponential growth and decay only refers to the early stages of both processes.
The mathematics behind exponential growth and decay is rather simple. In fact, we use the same formula as for continuous compound interest.
Exponential Growth
If is the initial population and the growth rate is k then the population N at time t is:

The formula for exponential decay is exactly the same, except the k value is negative instead of positive. Below is written out formally.
Exponential Decay
If is the initial population and the decay rate is k then the population N at time t is:

Now, recall the equation for continuous compound interest: . While it does not seem apparent, this equation is the same Eq.(1). The only difference is the variables.
For continuous compound interest, we let the initial amount, the principle, be denoted by the letter P. Think of the principle as the initial population in a savings account. In the interest formula, we let r denote the interest rate. Think of the interest rate as the rate at which the amount of money in a savings account grows. It is analogous to k in the exponential growth/decay formulas. And while A denoted the amount in the account at time t, you can think of it as the population of money in the account at time t.
Example 1:
At the start of an experiment, there are 100 bacteria. If the bacteria follow an exponential growth pattern with rate k = 0.02, (a) what will be the population after 5 hours? (b) how long will it take for the population to double?
Solution:
For (a) we are asked to find the population N at t = 5. To solve this question, we shall use the growth formula, . We are told that , k = 0.02 and that t = 5. Plugging this information in, we have the following:
In (b), we are asked to determine the time that the population doubles, ie reaches 200. Again, we use the same formula, but this time we are solving for t. We have:
So, the bacteria population will double in about 34.7 hours.
You do not need to know the growth rate to begin with to solve problems, as the next example illustrates.
Example 2:
Suppose that the population of a colony of bacteria increases exponentially. At the start of an experiment, there are 6,000 bacteria, and one hour later, the population has increased to 6,400. How long will it take for the population to reach 10,000? Round your answer to the nearest hour.
Solution:
We are given that , and at t = 1, N = 6,400. Plugging this information into the formula for exponential growth, we can solve for k. Then we can use k to find when the population will reach 10,000.
Using this k value, we can determine when the population N will reach 10,000.
Often times when dealing with exponential decay problems, we need to use the halflife. If we know the rate of decay, k, there is a nice formula for halflife. It is:
HalfLife
If k is the rate of decay, then

To show the halflife formula, we merely use formula (2), setting and solve for t. Doing this, we get the following:
While this formula is very helpful and will be used quite a bit, it is better to remember how to derive the formula, not just remember the formula.
Example 3:
The halflife of Plutonium239 is 24,000 years. If 10 grams are present now, how long will it take until only 10% of the original sample remains? Round your answer to the nearest 10,000^{th}.
Solution:
We can use the halflife formula to find the decay rate k. We know that t = 24,000 years. Plugging into the halflife formula, we have:
Now, we need to find when only 10% (1 gram) remains. Plugging into the exponential decay formula, we get the following:
Notice that we rounded our answer. If you used the exact value for k, your answer would be around 79,700 years, but if you used 0.000029, your answer would be around 79,400 years. Because of this variation, only rounding to the nearest ten thousandth will yield the same answer.
Unlike the previous example, the next example does not use the halflife formula. However, it does use a method similar to the derivation of the halflife formula.
Example 4:
Suppose that at the start of an experiment there are 8,000 bacteria. A growth inhibitor and a lethal pathogen are introduced into the colony. After two hours 1,000 bacteria are dead. If the death rates are exponential, (a) how long will it take for the population to drop below 5,000? (b) How long will it take for twothirds of the bacteria to die? Round your answers to the nearest tenth.
Solution:
Part (a): We are starting with 8,000 bacteria, so that is . We know that at t = 2, 1,000 bacteria are dead, so the population is 7,000. Using this information, we can determine the decay rate, k by using the exponential decay formula.
Using k, we can find out the t value when the population will drop below 5,000.
Notice that we rounded up, because we are asked to find when the population drops below 5,000. So, we need to round 7.0391 up to the next tenth. At t = 7.0, the population is still above 5,000.
For part (b), recognize that when twothirds of the population is dead, only onethird of the population remains. To solve, we follow the derivation of the halflife formula, but replace with and use k = 0.06677. Doing this, we get the following:
Again, we rounded up to the next tenth, for the same reasons as in part (a).