3.13: Exponential Growth
Andrew and Kaz have designed a website for fans of their favorite online game "EverStar." The membership started off rather slow, but has been growing at an increasing rate over the last few weeks. Their signup records look like this:
Week  Members 

1  10 
2  22 
3  48 
4  106 
5  233 
How could Andrew and Kaz write a function to allow them to calculate the projected membership of the website over the next few weeks?
Watch This
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 James Sousa: Ex: Exponential Growth Function  Bacterial Growth
Guidance
In prior lessons, we have explored the application of given exponential functions to particular questions. In this lesson we will examine the process of deriving a function from observed data, so that we can extrapolate or interpolate data that was not or could not be observed.
In general, if you have enough information about a situation, you can write an exponential function to model growth in the situation.
Example A
A social networking website is started by a group of 10 friends. They advertise their site before they launch, and membership grows fast: the membership doubles every day.
 a) Derive an equation to model the growth of the site.
 b) At this rate, what will the membership be in a week?
 c) When will the membership reach 100,000?
Solution: To model this situation, let’s look at how the membership changes each day:
Time (in days)  Membership 

0  10 
1  2 × 10 = 20 
2  2 × 2 × 10 = 40 
3  2 × 2 × 2 × 10 = 80 
4  2 × 2 × 2 × 2 × 10 = 160 
Notice that the membership on day x is 10(2^{x}).
 a) We can model membership with the function M(x) = 10(2^{x}).
 b) In seven days, the membership will be M(7) = 10(2^{7}) = 1280.
 c) We can solve an exponential equation to find out when the membership will reach 100,000:


\begin{align*}10(2^x) = 100,000\end{align*}
10(2x)=100,000 
\begin{align*}2^x = 10,000\end{align*}
2x=10,000 
\begin{align*}log 2^x = log 10,000\end{align*}
log2x=log10,000 
\begin{align*}x log 2 = 4\end{align*}
xlog2=4 
\begin{align*}x = \frac{4} {log 2} \approx 13.3\end{align*}
x=4log2≈13.3 days.

\begin{align*}10(2^x) = 100,000\end{align*}
Example B
Suppose the membership of the website in "Example A" grew more reasonably. The new website doubles every 7 days.
 a) Derive a function to model this growth.
 b) What would the membership be in 3 months?
Solution:
Time (in days)  Membership 

0  10 
7  2 × 10 = 20 
14  2 × 2 × 10 = 40 
21  2 × 2 × 2 × 10 = 80 
28  2 × 2 × 2 × 2 × 10 = 160 
We can no longer use the function M(x) = 10(2^{x}). However, we can use this function to find another function to model this new situation. Looking at one data point will help. Consider for example the fact that M(21) = 10(2^{3}). This is the case because 21 days results in 3 periods of doubling. In order for x = 21 to produce 2^{3} in the equation, the exponent in the function must be x/7. So we have \begin{align*}M(x) = 10 \left (2^{\frac{x} {7}}\right )\end{align*}


\begin{align*}M(0) = 10 \left (2^{\frac{0} {7}}\right ) = 10(1) = 10\end{align*}
M(0)=10(207)=10(1)=10 
\begin{align*}M(7) = 10 \left (2^{\frac{7} {7}}\right ) = 10(2) = 20\end{align*}
M(7)=10(277)=10(2)=20 
\begin{align*}M(14) = 10 \left (2^{\frac{14} {7}}\right ) = 10(2^2) = 10(4) = 40\end{align*}
M(14)=10(2147)=10(22)=10(4)=40 
\begin{align*}M(21) = 10 \left (2^{\frac{21} {7}}\right ) = 10(2^3) = 10(8) = 80\end{align*}
M(21)=10(2217)=10(23)=10(8)=80 
\begin{align*}M(28) = 10 \left (2^{\frac{28} {7}}\right ) = 10(2^4) = 10(16) = 160\end{align*}
M(28)=10(2287)=10(24)=10(16)=160

\begin{align*}M(0) = 10 \left (2^{\frac{0} {7}}\right ) = 10(1) = 10\end{align*}
Notice that each x value represents one more event of doubling, and in order for the function to have the correct power of 2, the exponent must be (x/7).

a) A function that models the new growth is: \begin{align*}M(x) = 10 \left (2^{\frac{x} {7}}\right )\end{align*}
M(x)=10(2x7)  b) The membership reaches 100,000 in about 3 months (93 days):


\begin{align*}10 \left (2^{\frac{x} {7}}\right ) = 100,000\end{align*}
10(2x7)=100,000 
\begin{align*}2^{\frac{x} {7}} = 100,000\end{align*}
2x7=100,000 
\begin{align*}log 2^{\frac{x} {7}} = log 100,000\end{align*}
log2x7=log100,000 
\begin{align*}\frac{x} {7} log 2 = 4\end{align*}
x7log2=4 
\begin{align*}x log 2 = 28\end{align*}
xlog2=28 
\begin{align*}x = \frac{28} {log 2} \approx 93\end{align*}
x=28log2≈93

\begin{align*}10 \left (2^{\frac{x} {7}}\right ) = 100,000\end{align*}
Example C
The population of a town was 20,000 in 1990. Because of its proximity to technology companies, the population grew to 35,000 by the year 2000.
a:) Derive an equation to model the described growth rate. b:) If the growth continues at this rate, how long will it take for the population to reach 1 million?
Solution:
The general form of the exponential growth model is much like the continuous compounding function you learned in the previous lesson. We can model exponential growth with a function of the form P(t) = P_{0}e^{kt}. The expression P(t) represents the population after t years, the coefficient P_{0} represents the initial population, and k is a growth constant that depends on the particular situation.
In the situation above, we know that P_{0} = 20,000 and that P(10) = 35,000. We can use this information to find the value of k:


\begin{align*}P(t) = P_0 e^{kt}\end{align*}
P(t)=P0ekt 
\begin{align*}P(10) = 35000 = 20,000 e^{k \cdot 10}\end{align*}
P(10)=35000=20,000ek⋅10 
\begin{align*}\frac{35,000} {20,000} = e^{10k}\end{align*}
35,00020,000=e10k 
\begin{align*}1.75 = e^{10k}\end{align*}
1.75=e10k 
\begin{align*}ln 1.75 = ln e^{10k}\end{align*}
ln1.75=lne10k 
\begin{align*}ln 1.75 = 10k ln e\end{align*}
ln1.75=10klne 
\begin{align*}ln 1.75 = 10k(1)\end{align*}
ln1.75=10k(1) 
\begin{align*}ln 1.75 = 10k\end{align*}
ln1.75=10k 
\begin{align*}k = \frac{ln 1.75} {10} \approx 0.056\end{align*}
k=ln1.7510≈0.056

\begin{align*}P(t) = P_0 e^{kt}\end{align*}
a:) We can model the population growth with the function \begin{align*}P(t) = 20,000 e^{\frac{ln 1.75} {10}t}\end{align*}
We can determine when the population will reach 1,000,000 by solving an equation, or using a graph.
 Using an equation:


\begin{align*}1,000,000 = 20,000 e^{\frac{ln 1.75} {10}t}\end{align*}
1,000,000=20,000eln1.7510t 
\begin{align*}50 = e^{\frac{ln 1.75} {10}t}\end{align*}
50=eln1.7510t 
\begin{align*}ln 50 = ln \left ( e^{\frac{ln 1.75} {10}t}\right )\end{align*}
ln50=ln(eln1.7510t) 
\begin{align*}ln 50 = \frac{ln 1.75} {10} t (ln e)\end{align*}
ln50=ln1.7510t(lne) 
\begin{align*}ln 50 = \frac{ln 1.75} {10} t (1)\end{align*}
ln50=ln1.7510t(1) 
\begin{align*}10 ln 50 = ln 1.75 t\end{align*}
10ln50=ln1.75t 
\begin{align*}t = \frac{10 ln 50} {ln 1.75} \approx 70\end{align*}
t=10ln50ln1.75≈70

\begin{align*}1,000,000 = 20,000 e^{\frac{ln 1.75} {10}t}\end{align*}
At this rate, it would take about 70 years for the population to reach 1 million.
Concept question wrapup Andrew and Kaz can model the growth of their site over the next few weeks by using the data they have collected to estimate a weekly growth rate as a percentage of current membership.
Based on their current data, their membership is growing at a rate of apx 220% per week. Future membership could therefore be estimated with a function like: \begin{align*}M_f = M_i (2.2)^t\end{align*} It is important to note that this is only an educated guess, as there will be many many factors which might alter the continued growth of their club! 

Vocabulary
An exponential model is a representation of growth or decay which varies exponentially.
A logistic model is used to represent a function which grows or decays rapidly for a period of time and then 'levels out'.
Guided Practice
Find an equation for the exponential functions described:
1) You are presented with data collected from a National Forest Campground. According to the ranger who provided it, the number of pieces of garbage found in the park has been increasing exponentially during the year as the summer season approaches. Using the data points on the graph below, derive an exponential equation which describes the data.
2) The number of health points that your avatar "Starbringer" has in your favorite game "DoomEnder II  The End of DoomEnder" increases exponentially with her level. Based on the information in the table below:
 a) Estimate the growth factor of health vs level
 b) Derive an equation to estimate the amount of life she has at level "L"
 c) Calculate how much life she will have at her maximum level of 60.


Level Health 1 500 5 787 15 2444 30 14,980 45 73,208

Answers
1) To find the equation of the line of an exponential equation from specific points, we can use the "baseintercept" form: \begin{align*}y = P_0 (a^x)\end{align*}

Start by finding the base a with: \begin{align*}a = (\frac{y_2}{y_1})^{\frac{1}{x_2x_1}}\end{align*}
a=(y2y1)1x2−x1 
\begin{align*}a = (\frac{6}{1.5})^{\frac{1}{60}}\end{align*}
a=(61.5)16−0 : using the data from the first and last points on the graph 
\begin{align*}a = 4^{\frac{1}{6}}\end{align*}
a=416 
\begin{align*}a = 1.259921\end{align*}
a=1.259921

\begin{align*}a = (\frac{6}{1.5})^{\frac{1}{60}}\end{align*}

Substitute the new value for a into the baseintercept form:

\begin{align*}y = 1.5(1.259921^x)\end{align*}
y=1.5(1.259921x)

\begin{align*}y = 1.5(1.259921^x)\end{align*}
\begin{align*}\therefore y = 1.5(1.259921^x)\end{align*} is the equation for estimating the number of pieces of trash (in hundreds) in any given week.
2) As we saw in the lesson, a common form of an exponential growth model is \begin{align*}A_f = A_i (e^{kt})\end{align*} Where A_{f} represents the final amount, A_{i} represents the initial amount, t  the time variable  represents the number of levels gained (we will use L) and k is a growth constant that depends on the particular situation.
In the case of Starbringer's health, we know that A_{i} = 500 and that A_{15} = 2,444. We can use this information to find the value of k:

 \begin{align*}A_f = A_i e^{k (L)}\end{align*}
 \begin{align*}2,444 = 500 e^{k (14)}\end{align*}
 \begin{align*}\frac{2,444}{500} = e^{14k}\end{align*}
 \begin{align*}4.888 = e^{14k}\end{align*}
 \begin{align*}ln 4.888 = ln (e^{14k})\end{align*}
 \begin{align*}ln 4.888 = (14k) ln (e)\end{align*}
 \begin{align*}ln 4.888 = 14k(1)\end{align*}
 \begin{align*}ln 4.888 = 14k\end{align*}
 \begin{align*}k = \frac{ln 4.888}{14} \approx .11334\end{align*}
a) Starbringer's growth factor is \begin{align*}\approx e^{.11334} = 1.12 = 12%\end{align*}
b) We can model increase of Starbringer's health based on level with the function \begin{align*}A_f = A_i \cdot e^{(.11334 L)}\end{align*}.
c) At level 60, Starbringer will have approximately 401,000 health points:
 \begin{align*}A_f = 500 (e^{.11334 (59)})\end{align*} : Substituting 59 for "L", since she will gain 59 levels to reach level 60.
 \begin{align*}A_f = 500 (e^{6.68706})\end{align*} : With a calculator
 \begin{align*}A_f = 500 \cdot 801.9610174\end{align*} : With a calculator
 \begin{align*}A_f \approx 400,980\end{align*}
Practice
Write an exponential function to model the situation.
 Your salary of $35,000 increases 8% each year.
 A population of 300,000 increases by 12% each year.
 A business had a $23,000 profit in 2009. Then the profit increased by 20% per year for the next 5 years.
Answer the question about the given investment.
 What is the balance after 7 years if you deposit $2800 in an account that pays 4% interest compounded yearly?
 What is the balance after 3 years if you deposit $3500 in an account that pays 2% interest compounded yearly?
 What is the balance after 15 years if you deposit $1800 in an account that pays 8% interest compounded quarterly?
The property value on a specific block in a large city grows exponentially every year. In the year 2011, you purchased a piece of property on this block. The table below represents the value of the property you purchased each year since 2006
Year  Value 

2006  $379,786 
2007  $401,259 
2008  $423,946 
2009  $447,917 
2010  $473,242 
2011  $500,000 
Use this data for Qs 710:
 By what percentage is the price of the property in this block increasing per year?
 Derive the function to model the growth of the value of the property.
 What will the property be worth in the year 2020?
 When will the property be worth 750 thousand dollars?
 In 1995, the world’s population was 5.7 billion and was growing at a relative rate of 2% per year. Find the population of the world in the year 2020
 You are working with a bacteria that doubles in cellular quantity every thirty minutes. If the initial population of the bacteria is 95, what is the population of bacteria a) After three hours? b) After one day?
 After losing your job, and needing cash fast, you decide to try your hand at gambling. This turns out to be a terrible decision on your part, as you lose more money and in fact end up owing the casino. At an interest rate of 22% per year, how much will you owe on a loan of $4,000: a)after one year? b)after three years?
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