Suppose you were evaluating a particular site as a possible future location for your new car stereo store, "Rock can Roll". You know that in order to be successful, the store must be located in a town with a population of at least 100,000. You also know that you have saved up enough to carry you through the first two years of becoming established, so the town can start out a little under the minimum population as long as it will hit 100,000 by year three.
The town you like best has a current population of 89,000, and is growing at a rate of 6% per year. Is it a big enough town for your store to be successful?
Watch This
Embedded Video:
 James Sousa: Exponential Growth Models  Part 1 of 2
Guidance
Exponential growth can be a bit surprising, as it can seem to be rather slow at first. At some point though, an exponential function will (sometimes rather suddenly) begin to increase very rapidly.
Population growth can often be modeled with an exponential function (assuming population grows as a percentage of the current population, i.e. 8% per year).
Example A
The population of a small town was 2,000 in the year 1950. The population increased over time, as shown by the values in the table below.
What is the number of people added to the population yearly? Why is this question more complex than it seems?
Year (1950 = 0)  Population 

0  2000 
5  2980 
10  4450 
20  9900 
30  22,000 
40  50,000 
Solution:
If you plot these data points, you will see that the growth pattern is nonlinear:
The population does not continue to increase by the same number of people each year, it rather increases by a percentage of the population at the end of each year, an exponential function.
Example B
Use a graphing calculator to find a function of the form y = a(b^{x}) that fits the data in the table.
Year (1950 = 0)  Population 

0  2000 
5  2980 
10  4450 
20  9900 
30  22,000 
40  50,000 
Solution
 Using a TI83/84 graphing calculator to find an exponential function that best fits a set of data

1. Entering the data
 a. Data must be entered into “lists”. The calculator has six named lists, L1, L2, ... L6. We will enter the x values in L1 and the y values in L2. One way to do this is shown below:
 Press <TI font_2nd> [{] and then enter the numbers separated by commas, and close by pressing the following: <TI font_2nd> [{]<TI font_STO> <TI font_2nd> [L1].
 The top three lines of the figure below show the entry into list L1, followed by the entry of the y values into list L2.
 Now press <TI font_STAT>, and move to the right to the CALC menu. Scroll down to option 10, ExpReg. Press <TI font_ENTER>, and you will return to the home screen. You should see ExpReg on the screen. As long as the numbers are in L1 and L2, the calculator will proceed to find an exponential function to fit the data you listed in List L1 and List L2. You should see on the home screen the values for a and b in the exponential function (See figure below).
Therefore the function y = 1992.7(1.0837)^{x} is an approximate model for the data.
 2. Plotting the data and the equation

To view plots of the data points and the equation on the same screen, do the following.
 a. First, press <TI font_Y=> and clear any equations.

You can type in the equation above, or to get the equation from the calculator, do the following:

b. Enter the above roundedoff equation in Y1, or use the following procedure to get the full equation from the calculator: put the cursor in Y1, press <TI font_VARS>, 5, EQ, and 1. This should place the equation in Y1 (see figure below).

c. Now press <TI font_2nd>[STAT PLOT] and complete the items as shown in the figure below.

d. Now set your window. (Hint: use the range of the data to choose the window – the figure below shows our choices.)

e. Press <TI font GRAPH> and you will to see the function and the data points as shown in the figure below.

b. Enter the above roundedoff equation in Y1, or use the following procedure to get the full equation from the calculator: put the cursor in Y1, press <TI font_VARS>, 5, EQ, and 1. This should place the equation in Y1 (see figure below).
 3. Comparing the real data with the modeled results

It looks as if the data points lie on the function. However, using the TRACE function you can determine how close the modeled points are to the real data. Press <TI font_TRACE> to enter the TRACE mode. Then press the right arrow to move from one data point to another. Do this until you land on the point with value Y=22000. To see the corresponding modeled value, press the up or down arrow. See the figure below. The modeled value is approximately 22197, which is quite close to the actual data. You can verify any of the other data points using the same method.
Example C
(from the introduction)
You need to find a town that will have a minimum population of 100,000 by the third year from now. The town you are considering has a population of 89,000, with a yearly growth rate of 6%. Will it work?
Solution
Final population is equal to initial population multiplied by the growth rate once each year.
That indicates that final population is: \begin{align*}{[(P_i \cdot growth) \cdot growth] \cdot growth}\end{align*}… etc. Where \begin{align*}P_i\end{align*} is initial population.
Using r for growth rate, and x for years passed, this simplifies to the exponential function:
\begin{align*}P(f) = P_i \cdot r^{x}\end{align*}
In our town, the population after x years would be: \begin{align*}P(x) = 89,000 \cdot (1.06)^{x}\end{align*}
The beginning of the 3rd year will occur after 2 years have passed, substituting 2 in for x gives: \begin{align*}P(2) = 89,000 \cdot (1.06)^{2}\end{align*}
 \begin{align*}\therefore P(2) = 100,000.4\end{align*}
The projected population is 100,000 (and 4/10... someone is pregnant!) in the 3rd year, just big enough.
Vocabulary
To extrapolate from data is to create new data points, or to predict, outside of the domain of the data set.
To interpolate is to create new data points, or to predict, within the domain of the data set, but for points not in the original data set.
An exponential model is a function reflecting a quantity which grows or decays at a rate proportional to its current value.
Guided Practice
1) Tiny Town, CO, currently (year 2012) has a population of 26 people, but it is growing at a rate of 17% per year.
 a) What is the growth factor for Tiny Town?
 b) What will the population be in 2030?
2) Abbi invests $4000 in a savings account with an APR of 6.5% compounded yearly.
 a) How much will she have after 2 years?
 b) How much will she have after 15 years?
 c) How many years will it take to reach $50,000?
3) Brandon bought a new car for $30,000. It wasn't until he drove away that his friend Kyle mentioned that the car was going to depreciate at a rate of 50% per year!
 a) What is the decay factor of the car?
 b) How much will the car be worth in 5 years?
 c) Using your calculation from "b)", apx how long will it take before the car is only worth $100?
Answers
1) Recall from the lesson that the simplified function for population growth is \begin{align*}P_f = P_i \cdot r^t\end{align*} Where "P_{f}" is final population, "P_{i}" is initial (starting) population, "r" is growth rate, and "t" is time (in years).
 a) The growth factor is 1.17, since the population each year is the entire population from the year before: \begin{align*}1 \cdot P\end{align*} added to the new population: \begin{align*}.17 \cdot P\end{align*}.
 b) The population in 2030 will be apx 375:
 \begin{align*}P_f = 26 \cdot 1.17^17\end{align*}
 \begin{align*}P_f = 26 \cdot 14.426\end{align*}
 \begin{align*}P_f = 375\end{align*}
2) Abbi's money can be calculated with the same type of formula as above: \begin{align*}A = P \cdot r^t\end{align*} Where "A" is final Amount, "'P" is principal (starting money), "r" is growth rate (interest), and "t" is time (in years).
 a) After 2 years, Abbi will have apx $4500
 \begin{align*}A = \$4000 \cdot 1.065^2\end{align*}
 \begin{align*}A = \$4000 \cdot 1.134\end{align*}
 \begin{align*}A = \$4536\end{align*}
 b) After 15 years, Abbi will have apx $10,300
 \begin{align*}A = \$4000 \cdot 1.065^15\end{align*}
 \begin{align*}A = \$4000 \cdot 2.5718\end{align*}
 \begin{align*}A = \$10287.20\end{align*}
 c) To calculate how long it will take to reach $50,000, we use the formula with \begin{align*}A = \$50,000\end{align*} and \begin{align*}x\end{align*} (in the exponent) is the number of years.
 \begin{align*}\$50,000 = \$4000 \cdot 1.065^x\end{align*}
 \begin{align*}12.5 = 1.065^x\end{align*} : Divide both sides by $4000
 \begin{align*}log 12.5 = log 1.065^x\end{align*} : Take the log of both sides
 \begin{align*}log 12.5 = xlog 1.065\end{align*} : Using \begin{align*}log x^y = ylog x\end{align*} from prior lesson
 \begin{align*}\frac{log 12.5}{log 1.065} = x\end{align*} : Divide both sides by \begin{align*}log 1.065\end{align*}
 \begin{align*}\frac{1.096}{.0273} = x\end{align*} : With a calculator
 \begin{align*}40.14 = x\end{align*} : With a calculator
It will take just over 40 years for Abbi's initial $4000 to become $50,000 at 6.5% interest compounded yearly.
3) The formula for calculating decay is again very similar: \begin{align*}V_f = V_i \cdot r^t\end{align*} Where "V_{f}" is final value, "V_{i}" is initial value, "r" is decay factor (depreciation rate), and "t" is time (in years).
 a) The decay rate is simply \begin{align*}1  .5 = .5\end{align*} since the car's value decays at a rate of 50% per year.
 b) In 5 years, the car will be worth apx
 \begin{align*}V_f = \$30,000 \cdot .5^5\end{align*}
 \begin{align*}V_f = \$30,000 \cdot .03125\end{align*}
 \begin{align*}V_f = \$937.50\end{align*} OUCH!
 c) If the car loses 1/2 of its value each year, and it is worth apx $1000 after 5 years:
 Year 6 = \begin{align*}\$1000 \cdot .5 = \$500\end{align*}
 Year 7 = \begin{align*}\$500 \cdot .5 = \$250\end{align*}
 Year 8 = \begin{align*}\$250 \cdot .5 = \$125\end{align*}
It will take only about 8 years before the car is only worth $100. Brandon may have made a questionable purchase!
Practice
Calculate the following values using: \begin{align*}A = P \cdot r^t\end{align*}
Assume all rates are x% per year, compounded yearly unless specified otherwise
 What is the value of a $5000 investment after 5 years at a rate of 5% ?
 What is the value of a $15000 investment after 3 years at a rate of 8% ?
 What is the value of a $3500 investment after 12 years at a rate of 2% ?
 What is the value of a $7550 investment after 7 years at a rate of 4.3% ?
 What is the value of a $42,340 investment after 13 years at a rate of 5.034% ?
For problems 610, calculate:
a) The growth factor
b) The final population
 If a population starts at 5,000 people in 1995, and increases at a rate of 7% per year, what is the population in 2032?
 If a population starts at 15,000 people in 2000, and increases at a rate of 3% per year, what is the population in 2027?
 If a population starts at 25,500 people in 1900, and increases at a rate of 2% per year, what is the population in 2008?
 If a population starts at 87,432 people in 1940, and increases at a rate of 4.3% per year, what is the population in 2040?
 If a population starts at 126,352 people in 1776, and increases at a rate of 1.067% per year, what is the population in 2012?
For problems 1115, calculate:
a) The decay factor (recall that decay factor = 1  %decay as a decimal)
b) The final value, using \begin{align*}V_f = V_i \cdot r^t\end{align*} from the lesson.
 A car is worth $4000, and loses value at a rate of 12% per year, what will it be worth in 5 years?
 A boat is purchased for $14,000, and loses value at a rate of 16% per year, what will it be worth in 7 years?
 A car is purchased for $40,500, and loses value at a rate of 21% per year, what will it be worth in 4 years?
 A motorcycle is worth $9350, and loses value at a rate of 6.5% per year, what will it be worth in 3.5 years?
 A plane is purchased for $342,137, and loses value at a rate of 4.67% per year, what will it be worth in 13 years?
For problems 1620, calculate the number of years required before the value reaches the specified total, using \begin{align*}A_f = A_i \cdot r^t\end{align*} and beginning with \begin{align*}A_f =\end{align*} final amount, and \begin{align*}x\end{align*} (in the exponent) as the number of years.
 How many years before a population of 5,000 reaches at least 8,000 at a growth rate of 6%?
 How many years before a value of $4,000 reaches at least $7,000 at a growth rate of 4%?
 How many years before a value of $12,000 reaches at least $25,000 at a growth rate of 12%?
 How many years before a population of 15,500 reaches at least 46,000 at a growth rate of 8.5%?
 How many years before the value of a car currently worth $52,138 depreciates to at least $8,000 at a depreciation rate of 14.7% ?