Modeling Exponential Growth and Decay

In real-world applications, we need to model the behavior of a function. In mathematical modeling, we choose a familiar general function with properties that suggest that it will model the real-world phenomenon we wish to analyze. In the case of rapid growth, we may choose the exponential growth function:

where [latex]{A}_{0}[/latex] is equal to the value at time zero, [latex]e[/latex] is the natural base (Euler’s constant), and [latex]k[/latex] is a positive constant that determines the rate (percentage) of continuous growth.  We also may use

[latex]y={A}_{0}{b}^{t}[/latex][latex]\\[/latex]

where [latex]{A}_{0}[/latex] is equal to the value at time zero, and [latex]b[/latex] is the growth factor which is greater than 1.  We may use the exponential growth function in applications involving doubling time, the time it takes for a quantity to double. Such phenomena as wildlife populations, financial investments, biological samples, and natural resources may exhibit growth based on a doubling time.

On the other hand, if a quantity is falling rapidly toward zero, without ever reaching zero, then we should probably choose the exponential decay model. Again, we can use the form [latex]y={A}_{0}{e}^{kt}[/latex] where [latex]{A}_{0}[/latex] is the starting value, [latex]e[/latex] is Euler’s constant, and [latex]k[/latex] is the (negative) continuous decay rate or we can use the form [latex]y={A}_{0}{b}^{t}[/latex] where [latex]{A}_{0}[/latex] is the starting value, and [latex]b[/latex] is the decay factor between zero and one.  We may use the exponential decay model when we are calculating half-life, or the time it takes for a substance to exponentially decay to half of its original quantity. We use half-life in applications involving radioactive isotopes.

In our choice of a function to serve as a mathematical model, we often use data points gathered by careful observation and measurement to construct points on a graph and hope we can recognize the shape of the graph. Exponential growth and decay graphs have a distinctive shape, as we can see in the graphs below. It is important to remember that, although parts of each of the two graphs seem to lie on the x-axis, they are really a tiny distance above the x-axis.

Characteristics of the Exponential Function, [latex]y={A}_{0}{e}^{kt}[/latex] and [latex]y={A}_{0}{b}^{t}[/latex]

An exponential function with the form [latex]y={A}_{0}{e}^{kt}[/latex] or [latex]y={A}_{0}{b}^{t}[/latex] has the following characteristics:

• one-to-one function
• horizontal asymptote: [latex]y=0[/latex]
• domain: [latex]\left(–\infty , \infty \right)[/latex]
• range: [latex]\left(0,\infty \right)[/latex]
• x intercept: none
• y-intercept: [latex]\left(0,{A}_{0}\right)[/latex]
• increasing if [latex]k>0[/latex] or [latex]b>1[/latex] and decreasing if [latex]k<0[/latex] or [latex]0<b<1[/latex]

Figure 2. An exponential function models exponential growth when [latex]k>0[/latex] and exponential decay when [latex]k<0.[/latex]

Expressing an Exponential Model in Base e

While powers and logarithms of any base can be used in modeling, the two most common bases are [latex]10[/latex] and [latex]e.[/latex] In science and mathematics, the base [latex]e[/latex] is often preferred. We can use laws of exponents and laws of logarithms to change any base to base [latex]e.[/latex]

How To

Given a model with the form [latex]y=a{b}^{x},[/latex] change it to the form [latex]y={A}_{0}{e}^{kx}.[/latex]

1. Since [latex]{b}^{x}={e}^{\mathrm{ln}\left({b}^{x}\right)}[/latex], rewrite [latex]y=a{b}^{x}[/latex] as [latex]y=a{e}^{\mathrm{ln}\left({b}^{x}\right)}.[/latex]
2. Use the power rule of logarithms to rewrite [latex]y[/latex] as [latex]y=a{e}^{x\mathrm{ln}\left(b\right)}=a{e}^{\mathrm{ln}\left(b\right)x}.[/latex]
3. Note that [latex]a={A}_{0}[/latex] and [latex]k=\mathrm{ln}\left(b\right)[/latex] in the equation [latex]y={A}_{0}{e}^{kx}.[/latex]

Example 1:  Changing to The Natural Base

Change the function [latex]y=2.5{\left(3.1\right)}^{x}[/latex] so that this same function is written in the form [latex]y={A}_{0}{e}^{kx}.[/latex]

Show Solution

The formula is derived as follows

[latex]\begin{align*}y&=2.5{\left(3.1\right)}^{x}&& \hfill \\ &=2.5{e}^{\mathrm{ln}\left({3.1}^{x}\right)}&& \text{Insert exponential and its inverse}\text{.}\hfill \\ &=2.5{e}^{x\mathrm{ln}\left(3.1\right)}&& \text{Laws of logs}\text{.}\hfill \\ &=2.5{e}^{\left(\mathrm{ln}\left(3.1\right)\right)}{}^{x}&& \text{Commutative law of multiplication.}\hfill \end{align*}[/latex]

In the above equation, we see that [latex]{A}_{0}=2.5, [/latex] and [latex]k=\mathrm{ln}\left(3.1\right)\approx 1.1314.[/latex]

Try it #1

Change the function [latex]y=3{\left(0.5\right)}^{x}[/latex] to one having [latex]e[/latex] as the base.

Show Solution

[latex]y=3{e}^{\left(\mathrm{ln}\left(0.5\right)\right)x}\approx3{e}^{-.6931x}[/latex]

Calculating Doubling Time

For growing quantities, we might want to find out how long it takes for a quantity to double. The time it takes for a quantity to double is called the doubling time.  Let [latex]{A}_{0}[/latex] represent the initial quantity at time zero.  We then want to find when the quantity is doubled or equal to [latex]{2A}_{0}.[/latex]  Using the exponential growth equation [latex]f\left(t\right)={A}_{0}{b}^{t} \text{ or } f\left(t\right)={A}_{0}{e}^{kt},[/latex] we solve the equation [latex]{2A}_{0}={A}_{0}{b}^{t}[/latex] or the equation [latex]{2A}_{0}={A}_{0}{e}^{kt}[/latex] for [latex]t[/latex] to find the doubling time [latex]t,[/latex]  if the growth rate or growth factor is known.  Similarly, if we know the doubling time but not the growth rate or factor, we can use [latex]{2A}_{0}={A}_{0}{e}^{kt}[/latex] to solve for the continuous growth rate [latex]k[/latex] or we can use [latex]{2A}_{0}={A}_{0}{b}^{t}[/latex] to solve for the growth factor [latex]b.[/latex]

Example 2:  Graphing Exponential Growth

A population of bacteria doubles every hour. If the culture started with 10 bacteria, graph the population as a function of time.

Show Solution

When an amount grows at a fixed percent per unit time, the growth is exponential. To find [latex]{A}_{0}[/latex] we use the fact that [latex]{A}_{0}[/latex] is the amount at time zero, so [latex]{A}_{0}=10,[/latex] and we can use the model [latex]A=10e^{kt}.[/latex]  To find [latex]k,[/latex] use the fact that after one hour [latex]\left(t=1\right)[/latex] the population doubles from [latex]10[/latex] to [latex]20,[/latex] and plug in the point [latex]\left(1,\text{ } 20\right)[/latex] to the formula. The formula is derived as follows:

[latex]\begin{align*}20&=10{e}^{k\cdot 1}&&\text{ }\\2&={e}^{k}&&\text{Divide by 10.}\\ \mathrm{ln}\left(2\right)&=k&&\text{Take the natural logarithm of both sides.}\end{align*}[/latex][latex]\\[/latex]

Therefore, [latex]k=\mathrm{ln}\left(2\right)\approx0.6931.[/latex] Thus the equation we want to graph is [latex]y=10{e}^{\left(\mathrm{ln}2\right)t}[/latex] or [latex]y\approx10e^{0.6931t}.[/latex]  We can also write this as, [latex]y=10{\left({e}^{\mathrm{ln}2}\right)}^{t}=10{\left(2\right)}^{t}.[/latex]

The graph is shown in Figure 3.

 

Figure 3. The graph of [latex]y=10{e}^{\left(\mathrm{ln}2\right)t}[/latex]

Analysis

The population of bacteria after 10 hours is 10,240.  We could describe this amount as being of the order of magnitude [latex]{10}^4.[/latex]  The population of bacteria after twenty hours is 10,485,760 which is of the order of magnitude [latex]{10}^7,[/latex] so we could say that the population has increased by three order of magnitude in 10 hours.

Half-Life

We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time.

Choosing an Appropriate Model for Data

Now that we have discussed various mathematical models, we need to learn how to choose the appropriate model for the raw data we have. Many factors influence the choice of a mathematical model, among which are experience, scientific laws, and patterns in the data itself. Not all data can be described by elementary functions. Sometimes, a function is chosen that approximates the data over a given interval. For instance, suppose data were gathered on the number of homes bought in the United States from the years 1960 to 2013. After plotting these data in a scatter plot, we notice that the shape of the data from the years 2000 to 2013 follow a logarithmic curve. We could restrict the interval from 2000 to 2010, apply regression analysis using a logarithmic model, and use it to predict the number of home buyers for the year 2015.

Three kinds of functions that are often useful in mathematical models are linear functions, exponential functions, and logarithmic functions. If the data lies on a straight line, or seems to lie approximately along a straight line, a linear model may be best. If the data is non-linear, we often consider an exponential or logarithmic model, though other models, such as quadratic models, may also be considered.

In choosing between an exponential model and a logarithmic model, we look at the way the data curves. This is called the concavity. If we draw a line between two data points, and all (or most) of the data between those two points lies above that line, we say the curve is concave down. We can think of it as a bowl that bends downward and therefore cannot hold water. If all (or most) of the data between those two points lies below the line, we say the curve is concave up. In this case, we can think of a bowl that bends upward and can therefore hold water. An exponential curve, whether rising or falling, whether representing growth or decay, is always concave up away from its horizontal asymptote. A logarithmic curve is always concave away from its vertical asymptote. In the case of positive data, which is the most common case, an exponential curve is always concave up, and a logarithmic curve always concave down.

After using the graph to help us choose a type of function to use as a model, we substitute points, and solve to find the parameters. We reduce round-off error by choosing points as far apart as possible.

Example 6:  Choosing a Mathematical Model

Does a linear, exponential, logarithmic, or quadratic model best fit the values listed in Table 2? Find the model, and use a graph to check your choice.

Table 2

[latex]x[/latex]	1	2	3	4	5	6	7	8	9
[latex]y[/latex]	0	1.386	2.197	2.773	3.219	3.584	3.892	4.159	4.394

Show Solution

First, plot the data on a graph as in Figure 4. For the purpose of graphing, round the data to two significant digits.

Figure 4.

 

Clearly, the points do not lie on a straight line, so we reject a linear model. If we draw a line between any two of the points, most or all of the points between those two points lie above the line, so the graph is concave down, suggesting a logarithmic model. We can try [latex]y=a\mathrm{ln}\left(bx\right).[/latex] Plugging in the first point, [latex]\left(\text{1,0}\right)\text{,}[/latex] gives [latex]0=a\mathrm{ln}b.[/latex] We reject the case that [latex]a=0[/latex] (if it were, all outputs would be 0), so we know [latex]\mathrm{ln}\left(b\right)=0.[/latex] Thus, [latex]b=1[/latex] and [latex]y=a\mathrm{ln}\left(\text{x}\right).[/latex] Next, we can use the point [latex]\left(\text{9,4}\text{.394}\right)[/latex] to solve for [latex]a:[/latex]

 

[latex]\begin{align*}y&=a\mathrm{ln}\left(x\right)\hfill \\ 4.394&=a\mathrm{ln}\left(9\right)\hfill \\a&=\frac{4.394}{\mathrm{ln}\left(9\right)}\hfill \end{align*}[/latex][latex]\\[/latex]

Because [latex]a=\frac{4.394}{\mathrm{ln}\left(9\right)}\approx 2,[/latex] an appropriate model for the data is [latex]y=2\mathrm{ln}\left(x\right).[/latex]

To check the accuracy of the model, we graph the function together with the given points as in Figure 5.

Figure 5. The graph of [latex]y=2\mathrm{ln}\left(x\right).[/latex]

We can conclude that the model is a good fit to the data.

Compare Figure 5 to the graph of [latex]y=\mathrm{ln}\left({x}^{2}\right)[/latex] shown in Figure 6.

Figure 6. The graph of [latex]y=\mathrm{ln}\left({x}^{2}\right).[/latex]

The graphs appear to be identical when [latex]x>0.[/latex] A quick check confirms this conclusion:  [latex]y=\mathrm{ln}\left({x}^{2}\right)=2\mathrm{ln}\left(x\right)[/latex] for [latex]x>0.[/latex]

However, if [latex]x<0,[/latex] the graph of [latex]y=\mathrm{ln}\left({x}^{2}\right)[/latex] includes a “extra” branch, as shown in Figure 9. This occurs because, while [latex]y=2\mathrm{ln}\left(x\right)[/latex] cannot have negative values in the domain (as such values would force the argument to be negative), the function [latex]y=\mathrm{ln}\left({x}^{2}\right)[/latex] can have negative domain values.

Figure 7.

Try it #5

Does a linear, exponential, or logarithmic model best fit the data in Table 3? Find the model.

Table 3

[latex]x[/latex]	1	2	3	4	5	6	7	8	9
[latex]y[/latex]	3.297	5.437	8.963	14.778	24.365	40.172	66.231	109.196	180.034

Show Solution

Exponential.  [latex]y=2{e}^{0.5x}.[/latex]

Using Newton’s Law of Cooling (Optional)

Exponential decay can also be applied to temperature. When a hot object is left in surrounding air that is at a lower temperature, the object’s temperature will decrease exponentially, leveling off as it approaches the surrounding air temperature. On a graph of the temperature function, the leveling off will correspond to a horizontal asymptote at the temperature of the surrounding air. Unless the room temperature is zero, this will correspond to a vertical shift of the generic exponential decay function. This translation leads to Newton’s Law of Cooling, the scientific formula for temperature as a function of time as an object’s temperature is equalized with the ambient temperature

Using Logistic Growth Models (Optional)

Exponential growth cannot continue forever. Exponential models, while they may be useful in the short term, tend to fall apart the longer they continue. Consider an aspiring writer who writes a single line on day one and plans to double the number of lines she writes each day for a month. By the end of the month, she must write over 17 billion lines, or one-half-billion pages. It is impractical, if not impossible, for anyone to write that much in such a short period of time. Eventually, an exponential model must begin to approach some limiting value, and then the growth is forced to slow. For this reason, it is often better to use a model with an upper bound instead of an exponential growth model, though the exponential growth model is still useful over a short term, before approaching the limiting value.

The logistic growth model is approximately exponential at first, but it has a reduced rate of growth as the output approaches the model’s upper bound, called the carrying capacity. For constants [latex]a, \text{ }b,[/latex] and [latex]c,[/latex] the logistic growth of a population over time [latex]x[/latex] is represented by the model

The graph in Figure 8 shows how the growth rate changes over time. The graph increases from left to right, but the growth rate only increases until it reaches its point of maximum growth rate, at which point the rate of increase decreases.

Figure 8.

Example 8:  Using the Logistic-Growth Model

An influenza epidemic spreads through a population rapidly, at a rate that depends on two factors: The more people who have the flu, the more rapidly it spreads, and also the more uninfected people there are, the more rapidly it spreads. These two factors make the logistic model a good one to study the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population.

For example, at time [latex]t=0[/latex] there is one person in a community of 1,000 people who has the flu. So, in that community, at most 1,000 people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is [latex]b=0.6030.[/latex] Estimate the number of people in this community who will have had this flu after ten days. Predict how many people in this community will have had this flu after a long period of time has passed.

Show Solution

We substitute the given data into the logistic growth model

[latex]f\left(x\right)=\frac{c}{1+a{e}^{-bx}}.[/latex][latex]\\[/latex]

Because at most 1,000 people, the entire population of the community, can get the flu, we know the limiting value is [latex]c=1000.[/latex] To find [latex]a,[/latex] we use the formula that the number of cases at time [latex]t=0[/latex] is [latex]\frac{c}{1+a}=1,[/latex] from which it follows that [latex]a=999.[/latex] This model predicts that, after ten days, the number of people who have had the flu is [latex]f\left(10\right)=\frac{1000}{1+999{e}^{-0.6030\times 10}}\approx 293.8.[/latex] Because the actual number must be a whole number (a person has either had the flu or not) we round to 294. In the long term, the number of people who will contract the flu is the limiting value, [latex]c=1000.[/latex]

Analysis

Remember that, because we are dealing with a virus, we cannot predict with certainty the number of people infected. The model only approximates the number of people infected and will not give us exact or actual values.

The graph in Figure 9 gives a good picture of how this model fits the data.

Figure 9. The graph of [latex]f\left(x\right)=\frac{1000}{1+999{e}^{-0.6030x}}[/latex]