Exponential Growth and Decay Tutorial



There are many situations where the increase or decrease of some variable in a fixed time interval will be proportional to the magnitude of the variable at the beginning of that time interval.

For example, let's look at a population of wee beasties which increases by 10% per year. If there were 100 wee beasties now, there would be an increase of 10 wee beasties after a year. We would see an increase of 500 wee beasties in a year when there were 5000 at the beginning.

Likewise, we can look at a population which decreases by 50% (i.e. a decrease to 1/2, or by a factor of 2) every day. A population of 100 would be down to 50 a day later, and a population of 5000 would drop to 2500 after one day.

These are all examples of exponential growth and decay.

 A single equation can be used to solve all problems involving this type of change:

\(N = N_o \cdot e^{k \cdot t}\)

where '\(N\)' is the number in the population after a time '\(t\)', '\(N_o\)' is the initial number, '\(k\)' is the growth constant (if positive) or the decay constant (if negative), and 'e' is the base of natural logarithms (approximately 2.71828).

We can re-write this equation in another convenient form. Dividing the equation by \(N_o\), and then taking the natural logarithms of both sides, we get

\(ln \Bigl(\frac {N}{N_o}\Bigr) = k \cdot t\)

Note that in this form, we do not need to know the absolute values of '\(N\)' or '\(N_o\)'; all we need to know here is the ratio of these two values.

These two problems are used to solve questions for both exponential growth and exponential decay.

Exponential Growth

Be sure you are acquainted with the two forms of the equations for exponential growth and decay. Recall that they are:
\(N = N_o \cdot e^{k \cdot t}\)
\(ln \Bigl(\frac {N}{N_o}\Bigr) = k \cdot t\)
When introducing the equations, we mentioned a case of wee beasties. There were 10% increases in the population. One population began with a population of 100, and after a year, there were 110. The other population had a population of 5000, and one year later, it grew to 5500. Note the ratios of final to initial populations, \(N/N_o\), were both the same:
\(\frac {N}{N_o} = \frac{110}{100} = \frac{5500}{5000}\)
As you can see, for a one year interval, this ratio was 1.1.
You can solve for '\(k\)', the growth constant, for this particular example using the second equation. Since \(N/N_o = 1.1\), and \(t = 1.0\) (year), we have
\(ln (1.1) = k (1.0)\)
Since \(ln (1.1) = 0.09531\), then \(k = 0.0953/ \mathrm year\).
Recall that an exponent must be dimensionless. So '\(k\)' will always have dimensions of reciprocal time. In the case of the wee beasties, \(k\) has units of \(\mathrm year ^{-1}\).
Now that we know the value of the growth constant for our wee beasties, \(k = 0.0953\), we can substitute this into our first equation.
\(N = N_o \cdot e^{0.0953t}\)
Suppose the initial population were 2000 and we wish to calculate what it will be in 5.5 years. We know '\(k\)' (\(= 0.0953 \; year^{-1}\)), '\(t\)' (\(= 5.5 \; year\)), and '\(N_o\)' (\(= 2000\)). Then we can calculate '\(N\)':
\(N = 2000 \cdot e ^{(0.953 \cdot t)}\)
Note that if the growth constant '\(k\)' were larger, then 'kt' would be larger at any given time, and so the increase in population would be greater.

Exponential Decay

Be sure you have reviewed the meanings of factor and percentage change.
Recall the two equations for exponential growth and decay:
\(N = N_o \cdot e^{k \cdot t}\)
\(ln \Bigl(\frac {N}{N_o}\Bigr) = k \cdot t\)
Suppose some environmental stress reduced a population of 1000 wee beasties to 800 in two days. How many will there be 7 days after the intial count of 1000 wee beasties?
This problem must be done in two steps. First, we use the information about the first 2 days to find the decay constant, '\(k\)'. Second, we use '\(k\)' and the time \(t = 7\) days, and the intial population to find the final population.
For the first step, the logarithmic form of the equation is most useful. We know '\(N_o\)' (the initial population was 1000), '\(N\)' (the final population was 800), '\(t\)' (the time period was 2 days). Substituting into the second equation, we get
\(ln \Bigl( \frac {800}{1000}\Bigr) = k \cdot (2)\)
\(- 0,0223 = 2 \cdot k\)

\(k = -0.112 day ^{-1}\)
So our decay constant is \(k = -0.112 \; \mathrm {day}^{-1}\).
Now we can do the second step. This time, the first equation (the exponential form of the equation) will be easier. Substituting \(k = -0.112 \; \mathrm {day}\), \(t = 7\; \mathrm {day}\), and  \(N_o = 1000\), we get
\(N = 1000 \cdot e ^{(-0.112) \cdot(7)}\)
\(= 457\)
You should be able to get \(N = 457\) wee beasties after 7 days.


Before we go on to discuss exponential decay, we should pause and discuss some of the terms that are frequently used in these problems. First let's talk about the term factor.

Suppose you invested \(\$100\), and after a time, your investment was worth \(\$300\). The final value (\(\$300\)) would be three (3) times the initial value. We would say that your investment had increased by a factor of 3.

On the other hand, if you made a poor investment, and the value decreased from \(\$100\) to \(\$25\), then the final value would be a quarter (1/4) of the initial value. We would say the investment had decreased by a factor of 4.

If a population of 500 increased by a factor of 1.5, then there would be a population of (500)(1.5) = 750. If the same population decreased from 500 by a factor of 1.5, then there would be 500/1.5 = 333 remaining.

When there is an increase, the ratio of the final number to the initial number is the factor. That is

\(\frac {N_\mathrm {final}}{N_\mathrm {initial}} = \mathrm {the \; factor}\)

Similarly, if there is a decrease, the ratio of the final number to the initial number is the reciprocal of the factor:

\(\frac {N_\mathrm {final}}{N_\mathrm {initial}} =\frac {1}{ \mathrm {the \; factor}}\)

Percentage Changes

You should be sure you understand percentage changes. The general equation for percentage changes is

\(\frac {N_{\mathrm final} - N_{\mathrm {initial}}}{N_{\mathrm {initial}}} \cdot (100) = \% \; \mathrm {change}\)

Thus, a \(\$100\) investment that increased to \(\$300\) (increased by a factor of 3), had a percentage increase of:

\(\frac {\$300 - \$100}{\$100} \cdot (100) = 200\%\)

The \(\$100\) investment that decreased to \(\$25\) (decreased by a factor of 4), had a percentage decrease of:

\(\frac {\$25 - \$100}{\$100} \cdot (100) = 75\%\)