Supplemental Videos
The main topics of this section are also presented in the following videos:
SectionComparing Exponential and Linear Growth
SubsectionExponential Growth
Consider the function
\begin{equation*}
P(t) = 100 \cdot 3^t
\end{equation*}
describing the population, \(P\text{,}\) of a bacteria after t minutes. We say a function has exponential growth if during each time interval of a fixed length, the population is multiplied by a certain constant amount call the growth factor. Consider the table:
We can see that the bacteria population grows by a factor of \(3\) each day. For this reason, we say that \(3\) is the growth factor for the function. Functions that describe exponential growth can be expressed in a standard form.
Exponential Growth
The function
\begin{equation*}
P(t) = a (b)^t
\end{equation*}
models exponential growth.
\(a = P(0)\) is the initial value or principal value of \(P\text{;}\)
\(b\) is the growth factor.
For the bacteria population, we have
\begin{equation*}
P(t) = 100 \cdot 3^t
\end{equation*}
so \(a = 100\) and \(b = 3\text{.}\)
Example177
A colony of bacteria starts with \(300\) organisms and doubles every week.
Write a formula for the population of the bacteria colony after \(t\) weeks.
How many bacteria will there be after \(8\) weeks? After \(5\) days?
Solution
-
The initial value of the population was \(a = 300\text{,}\) and its weekly growth factor is \(b = 2\text{.}\) Thus, a formula for the population after \(t\) weeks is
\begin{equation*} P(t) = 300 \cdot 2^t. \end{equation*} -
After \(8\) weeks, the population will be
\begin{equation*} P(8) = 300 \cdot 2^8 = 76,800 \text{ bacteria}. \end{equation*}Because \(5\) days is \(\frac{5}{7}\) of a week, after \(5\) days the population will be
\begin{equation*} P\left(\frac{5}{7}\right)= 300 \cdot 2^{5/7} = 492.2. \end{equation*}We cannot have a fraction of a bacterium, so we round to the nearest whole number, \(492\text{.}\)
Caution178
In Example177a, note that
\begin{equation*}
300 \cdot 2^8 \ne 600^8.
\end{equation*}
According to the order of operations, we compute the power \(2^8\) first, then multiply by \(300\text{.}\)
Example179
A population of \(24\) fruit flies triples every week.
Write a formula for the population of fruit flies after \(t\) weeks.
How many fruit flies will there be after \(6\) months? After \(3\) weeks? (Assume that a month equals \(4\) weeks.)
Solution
The initial value of the population was \(a=24\text{,}\) and its weekly growth factor is \(b=3\text{.}\) Thus \(P(t) = 24\cdot 3^t\)
-
Since \(1\) month is \(4\) weeks, \(6\) months is \(24\) weeks.
\begin{equation*} \begin{aligned} P(24) \amp= 24\cdot24^3=331776 \\ P(3) \amp= 24\cdot3^3=648 \end{aligned} \end{equation*}So after \(6\) months there are \(331776\) flies and after \(3\) weeks there are \(648\) flies.
SubsectionLinear Growth
Recall the following from Chapter 1:
A function \(y = f (x)\) is linear if it can be written in the form
\begin{equation*}
f (x) = (\text{starting value}) + (\text{rate of change}) \cdot x.
\end{equation*}
The starting value, or the value of \(y\) at \(x = 0\text{,}\) is the \(y\)-intercept of the graph, and the rate of change is the slope of the graph. Thus, we can write the equation of a line as
\begin{equation*}
f (x) = b + mx
\end{equation*}
where the constant term, \(b\text{,}\) is the \(y\)-intercept of the line, and \(m\text{,}\) the coefficient of \(x\text{,}\) is the slope of the line. This form for the equation of a line is called the slope-intercept form.
Slope-Intercept Form
If we write the equation of a linear function in the form,
\begin{equation*}
f (x) = b + mx
\end{equation*}
then \(m\) is the slope of the line, and \(b\) is the \(y\)-intercept.
SubsectionComparing Linear Growth and Exponential Growth
It may be helpful to compare linear growth and exponential growth. Consider the two functions
\begin{equation*}
L(t) = 5 + 2t ~\text{ and } ~ E(t) = 5 \cdot 2^t ~~~ (t \ge 0)
\end{equation*}
whose graphs are shown in Figure182.
| \(t\) | \(L(t)\) |
| \(0\) | \(5\) |
| \(1\) | \(7\) |
| \(2\) | \(9\) |
| \(3\) | \(11\) |
| \(4\) | \(13\) |
| \(t\) | \(E(t)\) |
| \(0\) | \(5\) |
| \(1\) | \(10\) |
| \(2\) | \(20\) |
| \(3\) | \(40\) |
| \(4\) | \(80\) |
\(L\) is a linear function with initial value \(5\) and slope \(2\text{;}\) \(E\) is an exponential function with initial value \(5\) and growth factor \(2\text{.}\) In a way, the growth factor of an exponential function is analogous to the slope of a linear function: Each measures how quickly the function is increasing (or decreasing).
However, for each unit increase in \(t\text{,}\) \(2\) units are added to the value of \(L(t)\text{,}\) whereas the value of \(E(t)\) is multiplied by \(2\text{.}\) An exponential function with growth factor \(2\) eventually grows much more rapidly than a linear function with slope \(2\text{,}\) as you can see by comparing the graphs in Figure182 or the function values in Tables180 and 181.
Example183
A solar energy company sold $\(80,000\) worth of solar collectors last year, its first year of operation. This year its sales rose to $\(88,000\text{,}\) an increase of \(10\)%. The marketing department must estimate its projected sales for the next \(3\) years.
If the marketing department predicts that sales will grow linearly, what should it expect the sales total to be next year? Graph the projected sales figures over the next \(3\) years, assuming that sales will grow linearly.
If the marketing department predicts that sales will grow exponentially, what should it expect the sales total to be next year? Graph the projected sales figures over the next \(3\) years, assuming that sales will grow exponentially.
Solution
-
Let \(L(t)\) represent the company's total sales \(t\) years after starting business, where \(t = 0\) is the first year of operation. If sales grow linearly, then \(L(t)\) has the form \(L(t) = mt + b\text{.}\) Now \(L(0) = 80,000\text{,}\) so the intercept is \((0,80000)\text{.}\) The slope of the graph is
\begin{equation*} \frac{\Delta S}{\Delta t}= \frac{8000 \text{ dollars}}{1\text{ year}}= 8000 \text{ dollars/year} \end{equation*}where \(\Delta S = 8000\) is the increase in sales during the first year. Thus, \(L(t) = 8000t + 80,000\text{,}\) and sales grow by adding $\(8000\) each year. The expected sales total for the next year is
\begin{equation*} L(2) = 8000(2) + 80,000 = 96,000. \end{equation*}The values of \(L(t)\) for \(t=0\) to \(t=4\) are shown in the middle column of Table184. The linear graph of \(L(t)\) is shown in Figure185.
-
Let \(E(t)\) represent the company's sales assuming that sales will grow exponentially. Then \(E(t)\) has the form \(E(t) = E_0b^t\) . The percent increase in sales over the first year was \(r = 0.10\text{,}\) so the growth factor is
\begin{equation*} b = 1 + r = 1.10 \end{equation*}The initial value, \(E_0\text{,}\) is \(80,000\text{.}\) Thus, \(E(t) = 80,000(1.10)^t\text{,}\) and sales grow by being multiplied each year by \(1.10\text{.}\) The expected sales total for the next year is
\begin{equation*} E(2) = 80,000(1.10)^2= 96,800. \end{equation*}The values of \(E(t)\) for \(t=0\) to \(t=4\) are shown in the last column of Table184. The exponential graph of \(E(t)\) is shown in Figure185.
\(t\) \(L(t)\) \(E(t)\) \(0\) \(80,000\) \(80,000\) \(1\) \(88,000\) \(88,000\) \(2\) \(96,000\) \(96,800\) \(3\) \(104,000\) \(106,480\) \(4\) \(112,000\) \(117,128\) 
Table184 Figure185
Example186
A new car begins to depreciate in value as soon as you drive it off the lot. Some models depreciate linearly, and others depreciate exponentially. Suppose you buy a new car for $\(20,000\text{,}\) and \(1\) year later its value has decreased to $\(17,000\text{.}\)
If the value decreased linearly, what was its annual rate of decrease?
If the value decreased exponentially, what was its annual decay factor? What was its annual percent depreciation?
Calculate the value of your car when it is \(5\) years old under each assumption, linear or exponential depreciation.
Solution
-
If the annual rate of decrease is \(m\text{,}\) then
\begin{equation*} 20000-m=17000\text{.} \end{equation*}Thus \(m=3000\) so the rate of decrease is \(\$3000\) per year
-
If the annual decay factor is \(b\text{,}\) then
\begin{equation*} 20000b=17000 \end{equation*}Thus \(b= 0.85\) so the annual decay factor is \(0.85\text{.}\) The annual percent depreciation is the percent change from \(\$20,000\) to \(\$17,000\text{:}\)
\begin{equation*} \text{percent change}=\frac{\text{amount change}}{\text{original amount}}=\frac{\$17,000-\$20,000}{\$20,000}=-0.15 \end{equation*}Therefore the car's value depreciated by \(15\%\text{.}\)
-
Based on the work above, if the car's value decreased linearly then the value of the car after \(t\) years is
\begin{equation*} L(t)=20000-3000t \end{equation*}If the value of the car depreciated exponentially, then the value of the car after \(t\) years is
\begin{equation*} E(t)=20000\cdot0.85^t \end{equation*}After \(5\) years,
\begin{equation*} \begin{aligned} L(5)\amp=20000-3000(5)=5000 \\ E(5)\amp=20000\cdot0.85^5\approx8874 \end{aligned} \end{equation*}After \(5\) years, the car will be worth \(\$5000\) under the linear model and worth approximately \(\$8874\) under the exponential model.
In the last section, we listed several properties of exponential functions; namely, for \(f(x)=a(b)^x\) with \(a>0\text{,}\)
- The domain is all real numbers and the range is all positive numbers.
- If \(b>1\) then the function is increasing, if \(0\lt b\lt 1\) then the function is decreasing.
- The \(y\)-intercept is \((0,a)\text{;}\) there is no \(x\)-\intercept.
Not convinced of the Properties of Exponential Functions listed above? Try varying the \(a\) and \(b\) parameters in the following applet to see many more examples of graphs of exponential functions, and convince yourself that the properties listed above hold true.