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17Calculus Precalculus - Exponential Growth and Decay

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Exponential growth and decay (also called the Law of Uninhibited Growth) problems are some of the most common applications you will run across. A related concept is half-life, discussed on a separate page.
Since these problems are usually presented as word problems, we recommend that you review the page on how to work word problems.

The equation you will use to describe exponential growth and decay is \(A(t)=A_o e^{kt}\), where A is the amount. Ao and k are constants and t is the variable, usually considered time. (In calculus you get to see where this equation comes from.) You may see different letters used for the constants but the form will be the same. The difference between exponential growth and exponential decay is that k is positive for exponential growth and it is negative for exponential decay. The variable t is usually time. So we think of the domain usually starting at zero and increasing.

Note - On this page we use one equation, \(A(t)=A_o e^{kt}\), and talk about k being positive for exponential growth or negative for decay. Some instructors talk about two equations, \(A(t)=A_o e^{kt}\) for growth and \(A(t)=A_o e^{-kt}\) for decay where k is positive in both cases. The ideas are the same, i.e. since we assume t is always positive, then for growth the exponent is positive, for decay the exponent is negative. Check with your instructor to see what they use and go with that.

The constant Ao, sometimes written A(0), is usually called the initial amount since at time \(t=0\), we have \(y = A_oe^0 = A_o\). This table summarizes the equation.

Equation

\(A(t)=A_o e^{kt}\)

Variable

\(t\)

Constants

\(A_o\) and \(k\)

Notation - You will undoubtedly run into equations that use different names for the function, variable and constants. One example is when the problem is talking about population growth. You may find the equation looks something like \(P(t) = P_o e^{kt}\) or even \(P(x) = P_o e^{kx}\). The same rules apply regardless of the names.
Before we get into the details on how to solve these problems, let's watch this video. This guy explains things very methodically and precisely and is very easy to follow.

Larson Calculus - The Exponential Growth and Decay Model [2min-1sec]

video by Larson Calculus

Here are some specific techniques on how to solve these problems.
1. If the question asks you to find a specific amount at a certain time, they will usually give you enough information to find a value of the constants. To find k and/or Ao, they may give you a specific amount at a different time. For example, one of the practice problems below says something like, after an hour the amount is 420. This means that \(A(1) = 420\). The wording in the problems will be different but you should be able to understand that they are giving a certain amount at a specific time. Here is a simple practice problem demonstrating this.

The population of a town is growing at a rate defined by the equation \(P(t)=P_oe^{0.028t}\). If the population is 100,000 today, what will the population be in 10 years?

Problem Statement

The population of a town is growing at a rate defined by the equation \(P(t)=P_oe^{0.028t}\). If the population is 100,000 today, what will the population be in 10 years?

Solution

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video by Michel vanBiezen

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2. Another thing they might give you is a multiple of the initial amount at a specific time. Use this information to determine k. Half-life is an example of this type of problem. You do not need to know what the initial amount is since the initial amount will cancel out once you set up the equation. Here are a few problems to help you work with this.

If the population of a town is growing at 1% per year, how long will it take for the population to double?

Problem Statement

If the population of a town is growing at 1% per year, how long will it take for the population to double?

Solution

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video by Michel vanBiezen

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At what elevation is the air pressure 1/2 of what it is at sea level? (in miles and in feet) What fraction of the atmospheric pressure at seal level does a climber experience at the top of Mt. Everest? (height of Mt. Everest is 29,028 ft)

Problem Statement

At what elevation is the air pressure 1/2 of what it is at sea level? (in miles and in feet) What fraction of the atmospheric pressure at seal level does a climber experience at the top of Mt. Everest? (height of Mt. Everest is 29,028 ft)

Solution

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3. They may give you growth or decay rate. This is k. In order for the exponent \(kt\) to be dimensionless, k has units 1/time. So sometimes you can pick out whether they have given you k by the units. Here is a practice problem demonstrating this.

A bacteria culture is found to have a population of 8000 after 5 hours. How many were present originally if the growth constant is \(0.25\)/hr?

Problem Statement

A bacteria culture is found to have a population of 8000 after 5 hours. How many were present originally if the growth constant is \(0.25\)/hr?

Solution

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video by Michel vanBiezen

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4. Unless otherwise stated, time will start at zero and increase. Initial time means \(t=0\). (Honestly, I have never seen or given my students a problem that starts time at any other value than zero. However, just because I haven't seen one, doesn't mean they don't exist.) Most of the time the problem will not say at t=0. Usually it will use the words initial, initially or currently. Here is a classic problem demonstrating this.

If a town's population, currently at 50,000, experiences 5% growth every 10 years, what will its population be in 25 years?

Problem Statement

If a town's population, currently at 50,000, experiences 5% growth every 10 years, what will its population be in 25 years?

Solution

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video by Michel vanBiezen

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5. Units are extremely important in word problems and they can tell you what is going on in the problem. Also, once the units get very large or very small, the problem may state that the units are in thousandths or billions for example. This can be confusing. The idea is that you work with only the values they give you. For example, this next problem states that the units are in billions of dollars and then they give you a constant of 100 billion dollars. In this case, you just work with 100. Try that on this practice problem before you look at the answer.

An exponential growth model that approximates the amounts (in billions of dollars) of US online advertising spending from 2011 through 2015 is given by \(S=9.30e^{0.1129t}, 11 \leq t \leq 15\), where S is the amount of spending (in billions of dollars) and \(t=11\) represents 2011. According to this model, when will the amount of US online advertising spending ready $100 billion?

Problem Statement

An exponential growth model that approximates the amounts (in billions of dollars) of US online advertising spending from 2011 through 2015 is given by \(S=9.30e^{0.1129t}, 11 \leq t \leq 15\), where S is the amount of spending (in billions of dollars) and \(t=11\) represents 2011. According to this model, when will the amount of US online advertising spending ready $100 billion?

Solution

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6. You need understand how to use the natural log identity \(\ln e^x = x\). Review logarithms on the precalculus logarithm page, if you need to.

7. The best way to learn how to solve these problems is to try them yourself. At first, you will be lost and it will seem confusing but work plenty of problems and you will see how deceptively simple these problems are to solve. The key is to decipher what the word problem is saying.

Practice

The size P of a certain insect population at time t, in days, obeys the function \(P(t) = 500e^{0.02t}\).
a. Determine the initial number of insects.
b. What is the growth rate of the insect population?
c. What is the population after 10 days?
d. When will the insect population reach 800?
e. When will the insect population double?

Problem Statement

The size P of a certain insect population at time t, in days, obeys the function \(P(t) = 500e^{0.02t}\).
a. Determine the initial number of insects.
b. What is the growth rate of the insect population?
c. What is the population after 10 days?
d. When will the insect population reach 800?
e. When will the insect population double?

Final Answer

a. 500 insects
b. \(k = 0.02\)/day
c. \(P(10) = 500e^{0.2} \approx 610\) insects
d. \( \displaystyle{ \frac{\ln(8/5)}{0.02}} \approx 23.50 \) days
e. \( \displaystyle{ \frac{\ln(2)}{0.02}} \approx 34.66 \) days

Problem Statement

The size P of a certain insect population at time t, in days, obeys the function \(P(t) = 500e^{0.02t}\).
a. Determine the initial number of insects.
b. What is the growth rate of the insect population?
c. What is the population after 10 days?
d. When will the insect population reach 800?
e. When will the insect population double?

Solution

Given \(P(t) = 500e^{0.02t}\)
a. You should know that the coefficient 500 is the initial value. However, if you forgot or want to check your answer (highly recommended), you can calculate it. The initial amount is given by \(P(0) = 500 e^0 = 500\). Answer: 500 insects
b. The growth rate is the exponent coefficient of t. In this problem, \(k=0.02\) or 2%. Answer: \(k=0.02\)/days
c. The population after 10 days is calculated by letting \(t=10\) in the equation. So \(P(10) = 500e^{0.02(10)} = 500e^{0.2}\). Answer: \(P(10) = 500e^{0.2}\) insects. This is the exact answer. If your instructor wants an approximate answer, plugging into your calculator gives 610 insects. Note that since we cannot have a partial insect, the calculator gives 610.70138 but our answer is 610 insects.
d. The question asks for the value of t when \(P(t)=800\).
\(\begin{array}{rcl} 800 & = & 500 e^{0.02t} \\ \displaystyle{ \frac{8}{5} } & = & e^{0.02t} \\ \ln(8/5) & = & \ln e^{0.02t} \\ & = & 0.02t \\ t & = & \displaystyle{ \frac{\ln(8/5)}{0.02}} \approx 23.50 \end{array} \)
Answer: \( \displaystyle{ \frac{\ln(8/5)}{0.02}} \approx 23.50 \) days
For this answer, a partial day does make sense.
e. The question asks for the value of t when \(P(t)= 2P(0) = 1000\). We do the same kind of calculation as we did for part d.
\(\begin{array}{rcl} 1000 & = & 500 e^{0.02t} \\ 2 & = & e^{0.02t} \\ \ln(2) & = & \ln e^{0.02t} \\ & = & 0.02t \\ t & = & \displaystyle{ \frac{\ln(2)}{0.02}} \approx 34.66 \end{array} \)
Answer: \( \displaystyle{ \frac{\ln(2)}{0.02}} \approx 34.66 \) days
Thinking about this answer, we can compare it to the answer we got for part d. Since 1000 is larger than 800, our value for part e should be greater than for part d, which is true.

Final Answer

a. 500 insects
b. \(k = 0.02\)/day
c. \(P(10) = 500e^{0.2} \approx 610\) insects
d. \( \displaystyle{ \frac{\ln(8/5)}{0.02}} \approx 23.50 \) days
e. \( \displaystyle{ \frac{\ln(2)}{0.02}} \approx 34.66 \) days

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The Gross Domestic Product (GDP) of the US (in billions of dollars) t years after the year 2000 can be modeled by \( G(t) = 9743.77 e^{0.0514t} \)
a. Find and interpret \(G(0)\).
b. According to the model, what should have been the GDP in 2007? In 2010? (According to the US Department of Commerce, the 2007 GDP was $14,369.1 billion and the 2010 GDP was $14,657.8 billion.)

Problem Statement

The Gross Domestic Product (GDP) of the US (in billions of dollars) t years after the year 2000 can be modeled by \( G(t) = 9743.77 e^{0.0514t} \)
a. Find and interpret \(G(0)\).
b. According to the model, what should have been the GDP in 2007? In 2010? (According to the US Department of Commerce, the 2007 GDP was $14,369.1 billion and the 2010 GDP was $14,657.8 billion.)

Final Answer

\( G(0) = 9742.77 \) billions of dollars in 2000
\(G(7) \approx 13,963.24 \) billions of dollars in 2007
\(G(10) \approx 16,291.25 \) billions of dollars in 2010

Problem Statement

The Gross Domestic Product (GDP) of the US (in billions of dollars) t years after the year 2000 can be modeled by \( G(t) = 9743.77 e^{0.0514t} \)
a. Find and interpret \(G(0)\).
b. According to the model, what should have been the GDP in 2007? In 2010? (According to the US Department of Commerce, the 2007 GDP was $14,369.1 billion and the 2010 GDP was $14,657.8 billion.)

Solution

a. \( G(0) = 9743.77e^{0.0514(0)} = 9742.77 \) billions of dollars in 2000.
b. In 2007, \(t=7\) so \(G(7) = 9743.77 e^{0.0514(7)} \approx 13,963.24 \) billions of dollars.
in 2010, \(t=10\) so \(G(10) = 9743.77 e^{0.0514(10)} \approx 16,291.25 \) billions of dollars.
The results are summarized below and compared to the actual values.

year

projected

actual

% difference from projected

2000

$9,743.77 billion

$9,743.77 billion

0

2007

$13,963.24 billion

$14,369.1 billion

+2.91

2010

$16,291.25 billion

$14,657.8 billion

-10.03

Final Answer

\( G(0) = 9742.77 \) billions of dollars in 2000
\(G(7) \approx 13,963.24 \) billions of dollars in 2007
\(G(10) \approx 16,291.25 \) billions of dollars in 2010

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Under optimal conditions, the growth of a certain strain of E. Coli is modeled by the Law of Uninhibited Growth \( N(t) = N_o e^{kt} \) where \(N_o\) is the initial number of bacteria and t is the elapsed time, measured in minutes. From numerous experiments, it has been determined that the doubling time of this organisim is 20 minutes. Suppose 1000 bacteria are present initially.
a. Find a function which gives the number of bacteria \(N(t)\) after t minutes.
b. How long until there are 9000 bacteria?

Problem Statement

Under optimal conditions, the growth of a certain strain of E. Coli is modeled by the Law of Uninhibited Growth \( N(t) = N_o e^{kt} \) where \(N_o\) is the initial number of bacteria and t is the elapsed time, measured in minutes. From numerous experiments, it has been determined that the doubling time of this organisim is 20 minutes. Suppose 1000 bacteria are present initially.
a. Find a function which gives the number of bacteria \(N(t)\) after t minutes.
b. How long until there are 9000 bacteria?

Final Answer

a. \( N(t) = 1000 e^{(t/20)\ln(2)} \)
b. \( t = 20 \ln(9)/\ln(2) \approx 63.4 \) minutes

Problem Statement

Under optimal conditions, the growth of a certain strain of E. Coli is modeled by the Law of Uninhibited Growth \( N(t) = N_o e^{kt} \) where \(N_o\) is the initial number of bacteria and t is the elapsed time, measured in minutes. From numerous experiments, it has been determined that the doubling time of this organisim is 20 minutes. Suppose 1000 bacteria are present initially.
a. Find a function which gives the number of bacteria \(N(t)\) after t minutes.
b. How long until there are 9000 bacteria?

Solution

a. They give us the initial amount at 1000 bacteria. So \(N_o = 1000\). Now we need to determine k. Since we are told that the doubling time is 20 minutes, we have \(N(20) = 2000\). We can use this to calculate k.

\( 2000 = 1000 e^{20k} \)

\( 2 = e^{20k} \)

\( \ln(2) = \ln( e^{20k} ) \)

\( \ln(2) = 20k \)

\( (1/20) \ln(2) = k \)

\( N(t) = 1000 e^{(t/20)\ln(2)} \)

b. We can determine the time when we have 9000 bacteria by setting the last equation to 9000 and solving for t.

\( 9000 = 1000 e^{(t/20)\ln(2)} \)

\( 9 = e^{(t/20)\ln(2)} \)

\( \ln(9) = \ln(e^{(t/20)\ln(2)}) \)

\( \ln(9) = (t/20)\ln(2) \)

\( 20 \ln(9)/\ln(2) = t \)

\( t = 20 \ln(9)/\ln(2) \approx 63.4 \) minutes

Final Answer

a. \( N(t) = 1000 e^{(t/20)\ln(2)} \)
b. \( t = 20 \ln(9)/\ln(2) \approx 63.4 \) minutes

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Yeast is often used in biological experiments. A research technician estimates that a sample of yeast suspension contains 2.5 million organisms per cubic centimeter (cc). Two hours later, she estimates the population density to be 6 million organisms per cc. Let t be the time elapsed since the first observation, measured in hours. Assume that the yeast growth follows the Law of Uninhibited Growth \(N(t) = N_oe^{kt}\).
a. Find a function which gives the number of yeast (in millions) per cc \(N(t)\) after t hours.
b. What is the doubling time for this strain of yeast?

Problem Statement

Yeast is often used in biological experiments. A research technician estimates that a sample of yeast suspension contains 2.5 million organisms per cubic centimeter (cc). Two hours later, she estimates the population density to be 6 million organisms per cc. Let t be the time elapsed since the first observation, measured in hours. Assume that the yeast growth follows the Law of Uninhibited Growth \(N(t) = N_oe^{kt}\).
a. Find a function which gives the number of yeast (in millions) per cc \(N(t)\) after t hours.
b. What is the doubling time for this strain of yeast?

Final Answer

a. \( N(t) = 2.5 e^{(t/2)\ln(2.4)} \approx 2.5e^{0.4377t} \) million organisms per cc
b. \( t = 2\ln(2)/\ln(2.4) \approx 1.58 \) hours

Problem Statement

Yeast is often used in biological experiments. A research technician estimates that a sample of yeast suspension contains 2.5 million organisms per cubic centimeter (cc). Two hours later, she estimates the population density to be 6 million organisms per cc. Let t be the time elapsed since the first observation, measured in hours. Assume that the yeast growth follows the Law of Uninhibited Growth \(N(t) = N_oe^{kt}\).
a. Find a function which gives the number of yeast (in millions) per cc \(N(t)\) after t hours.
b. What is the doubling time for this strain of yeast?

Solution

a. In order to find an equation, we need to determine k. We are given \(N_o=2.5\) and when \(t=2\), we were given \(N(2) = 6\). So let\'s set up the equation and solve for k.

\( N(2) = 2.5 e^{2k} = 6 \)

\( e^{2k} = 6/2.5 = 2.4 \)

\( \ln(e^{2k}) = \ln(2.4) \)

\( 2k = \ln(2.4) \)

\( k = (1/2)\ln(2.4) \)

So our equation is \( N(t) = 2.5 e^{(t/2)\ln(2.4)} \)
b. Now we need to find the doubling time. This means the time when the initial sample is doubled.

\( 2(2.5) = 2.5e^{(t/2)\ln(2.4)} \)

\( 2 = e^{(t/2)\ln(2.4)} \)

\( \ln(2) = \ln(e^{(t/2)\ln(2.4)}) \)

\( \ln(2) = (t/2)\ln(2.4) \)

\( \ln(2)/\ln(2.4) = (t/2) \)

\( 2\ln(2)/\ln(2.4) = t \)

Final Answer

a. \( N(t) = 2.5 e^{(t/2)\ln(2.4)} \approx 2.5e^{0.4377t} \) million organisms per cc
b. \( t = 2\ln(2)/\ln(2.4) \approx 1.58 \) hours

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Suppose a bacteria population starts with 10 bacteria and that they divide every hour.
a. What is the population 7 hours later?
b. When will there be 1,000,000 bacteria?

Problem Statement

Suppose a bacteria population starts with 10 bacteria and that they divide every hour.
a. What is the population 7 hours later?
b. When will there be 1,000,000 bacteria?

Final Answer

a. 1280 bacteria; b. 16.6 hours

Problem Statement

Suppose a bacteria population starts with 10 bacteria and that they divide every hour.
a. What is the population 7 hours later?
b. When will there be 1,000,000 bacteria?

Solution

673 video

video by PatrickJMT

Final Answer

a. 1280 bacteria; b. 16.6 hours

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A radioactive material is know to decay at a yearly rate of 0.2 times the amount at each moment. Suppose there are 1000 grams of the material now. What is the amount after 10 years?

Problem Statement

A radioactive material is know to decay at a yearly rate of 0.2 times the amount at each moment. Suppose there are 1000 grams of the material now. What is the amount after 10 years?

Final Answer

135 grams

Problem Statement

A radioactive material is know to decay at a yearly rate of 0.2 times the amount at each moment. Suppose there are 1000 grams of the material now. What is the amount after 10 years?

Solution

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video by PatrickJMT

Final Answer

135 grams

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Suppose a bacteria population grows at a rate proportional to the population. There were 200 bacteria 3 days ago and 1000 bacteria 1 day ago. How many bacteria will there be tomorrow?

Problem Statement

Suppose a bacteria population grows at a rate proportional to the population. There were 200 bacteria 3 days ago and 1000 bacteria 1 day ago. How many bacteria will there be tomorrow?

Final Answer

5000 bacteria

Problem Statement

Suppose a bacteria population grows at a rate proportional to the population. There were 200 bacteria 3 days ago and 1000 bacteria 1 day ago. How many bacteria will there be tomorrow?

Solution

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Final Answer

5000 bacteria

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A bacteria population increases sixfold in 10 hours. Assuming normal growth, how long did it take for their population to double?

Problem Statement

A bacteria population increases sixfold in 10 hours. Assuming normal growth, how long did it take for their population to double?

Final Answer

3.87 hours

Problem Statement

A bacteria population increases sixfold in 10 hours. Assuming normal growth, how long did it take for their population to double?

Solution

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Final Answer

3.87 hours

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It has been shown that the atmospheric pressure can be calculated with the equation \(P(h) = P_o e^{-0.21h}\) where \(P_o= 14.7\) lbs/in2 and the height above sea level,h, is in miles. What is the air pressure in Denver (h=1 mile)? What is the pressure in Denver as a percentage of the pressure at sea level?

Problem Statement

It has been shown that the atmospheric pressure can be calculated with the equation \(P(h) = P_o e^{-0.21h}\) where \(P_o= 14.7\) lbs/in2 and the height above sea level,h, is in miles. What is the air pressure in Denver (h=1 mile)? What is the pressure in Denver as a percentage of the pressure at sea level?

Solution

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A student measured the number of bacteria in a culture twice and found that the number increased from 100 bacteria after 3 hours to 400 bacteria after 5 hours. Determine the number of bacteria in the culture after 6 hours and the time when we will have a population of 10,000.

Problem Statement

A student measured the number of bacteria in a culture twice and found that the number increased from 100 bacteria after 3 hours to 400 bacteria after 5 hours. Determine the number of bacteria in the culture after 6 hours and the time when we will have a population of 10,000.

Solution

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The Law of Uninhibited Growth also applies to situations where an animal is re-introduced into a suitable environment. Such a case is the reintroduction of wolves to Yellowstone National Park. According to the National Park Service, the wolf population in Yellowstone was 52 in 1996 and 119 in 1999. Using these data points, find a function of the form \(N(t) = N_oe^{kt}\) which models the number of wolves t years after 1996. According to the model, how many wolves were in Yellowstone in 2002? (The recorded number is 272.)

Problem Statement

The Law of Uninhibited Growth also applies to situations where an animal is re-introduced into a suitable environment. Such a case is the reintroduction of wolves to Yellowstone National Park. According to the National Park Service, the wolf population in Yellowstone was 52 in 1996 and 119 in 1999. Using these data points, find a function of the form \(N(t) = N_oe^{kt}\) which models the number of wolves t years after 1996. According to the model, how many wolves were in Yellowstone in 2002? (The recorded number is 272.)

Hint

Since they say to start in the year 1996, this will be t=0. So for 1999, t=1999-1996=3. In 2002, t=2002-1996=6.

Problem Statement

The Law of Uninhibited Growth also applies to situations where an animal is re-introduced into a suitable environment. Such a case is the reintroduction of wolves to Yellowstone National Park. According to the National Park Service, the wolf population in Yellowstone was 52 in 1996 and 119 in 1999. Using these data points, find a function of the form \(N(t) = N_oe^{kt}\) which models the number of wolves t years after 1996. According to the model, how many wolves were in Yellowstone in 2002? (The recorded number is 272.)

Final Answer

\(N(t)=52 e^{(t/3)\ln(119/52)}\), 273 wolves

Problem Statement

The Law of Uninhibited Growth also applies to situations where an animal is re-introduced into a suitable environment. Such a case is the reintroduction of wolves to Yellowstone National Park. According to the National Park Service, the wolf population in Yellowstone was 52 in 1996 and 119 in 1999. Using these data points, find a function of the form \(N(t) = N_oe^{kt}\) which models the number of wolves t years after 1996. According to the model, how many wolves were in Yellowstone in 2002? (The recorded number is 272.)

Hint

Since they say to start in the year 1996, this will be t=0. So for 1999, t=1999-1996=3. In 2002, t=2002-1996=6.

Solution

Let's start by laying out all the information they gave us. Our general equation is \(N(t) = N_oe^{kt}\). We were told that initially, in 1996, the population was 52, so when t=0, \( N(0) = N_o = 52 \). So our equation is \(N(t)=52 e^{kt}\). Our next data point was 119 in 1999. This means that when t=1999-1996=3, N(3) = 119. We use this data point to calculate k.

\(N(t)=52 e^{kt}\)

\(N(3)=119=52 e^{3k}\)

\(119/52= e^{3k}\)

\(\ln(119/52)= 3k\)

\(\ln(119/52)= 3k\)

\((1/3)\ln(119/52)= k\)

So now we have the equation they asked for, i.e. \(N(t)=52 e^{(t/3)\ln(119/52)}\).

Now we use this equation to determine the number of wolves in 2002. In 2002, t=2002-1996=6, so the we need to find \(N(6)\).
\(N(6)=52 e^{(6/3)\ln(119/52)} = 52 e^{2\ln(119/52)} \approx 272.33\). Since we can't have partial wolves, we say we have 272 or 273 wolves.

Final Answer

\(N(t)=52 e^{(t/3)\ln(119/52)}\), 273 wolves

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During the early years of a community, it is not uncommon for the population to grow according to the Law of Uninhibited Growth. According to the Painesville Wikipedia entry, in 1860, the Village of Painesville had a population of 2649. In 1920, the population was 7272. Use these two data points to fit a model of the form \(N(t)=N_oe^{kt}\) where \(N(t)\) is the number of Painesville residents t years after 1860. According to this model, what was the population of Painesville in 2010. (The 2010 census gave the population as 19,563.) What could be some causes for such a vast discrepancy?

Problem Statement

During the early years of a community, it is not uncommon for the population to grow according to the Law of Uninhibited Growth. According to the Painesville Wikipedia entry, in 1860, the Village of Painesville had a population of 2649. In 1920, the population was 7272. Use these two data points to fit a model of the form \(N(t)=N_oe^{kt}\) where \(N(t)\) is the number of Painesville residents t years after 1860. According to this model, what was the population of Painesville in 2010. (The 2010 census gave the population as 19,563.) What could be some causes for such a vast discrepancy?

Final Answer

\(N(t)=2649e^{(t/60)\ln(7272/2649)}\), 33076 people

Problem Statement

During the early years of a community, it is not uncommon for the population to grow according to the Law of Uninhibited Growth. According to the Painesville Wikipedia entry, in 1860, the Village of Painesville had a population of 2649. In 1920, the population was 7272. Use these two data points to fit a model of the form \(N(t)=N_oe^{kt}\) where \(N(t)\) is the number of Painesville residents t years after 1860. According to this model, what was the population of Painesville in 2010. (The 2010 census gave the population as 19,563.) What could be some causes for such a vast discrepancy?

Solution

Let's start out by gathering all the data points they gave us. We take 1860 as t=0. This means that for 1920, t=1920-1860=60 years.
t=0, \(N(0)=N_o=2649 \)
t=60, \(N(60) = 7272 \)
From the initial data point, t=0, we have \(N(t)=2649e^{kt}\). To determine the equation, we need to find k.

\(N(t)=2649e^{kt}\)

\(N(60)=7272=2649e^{60k}\)

\(7272/2649=e^{60k}\)

\(\ln(7272/2649)=60k\)

\((1/60)\ln(7272/2649)=k\)

So the equation modeling the population growth is \(N(t)=2649e^{(t/60)\ln(7272/2649)}\).

Using this model, we calculate the population in 2010, with t=2010-1860=150, as \(N(150)=2649e^{(150/60)\ln(7272/2649)} \approx 33075.91 \). Since we can't have part of a person, our answer is 33076.

The discrepancy occurs in that the model is accurate for only a short time, not for a long term. It does not take into account resources like food and space. For a more accurate model, see the page on logistic growth.

Final Answer

\(N(t)=2649e^{(t/60)\ln(7272/2649)}\), 33076 people

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After 1000 years, we have 500 grams of a substance with a decay rate of 0.001. What was the initial amount?

Problem Statement

After 1000 years, we have 500 grams of a substance with a decay rate of 0.001. What was the initial amount?

Final Answer

1355 grams

Problem Statement

After 1000 years, we have 500 grams of a substance with a decay rate of 0.001. What was the initial amount?

Solution

681 video

video by Khan Academy

Final Answer

1355 grams

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Suppose a radioactive substance decays at a rate of 3.5% per hour. What percent of the substance is left after 6 hours?

Problem Statement

Suppose a radioactive substance decays at a rate of 3.5% per hour. What percent of the substance is left after 6 hours?

Final Answer

80.75%

Problem Statement

Suppose a radioactive substance decays at a rate of 3.5% per hour. What percent of the substance is left after 6 hours?

Solution

He works this problem a little differently in the video than you may have seen before.
The way to work this problem using the standard equation \(A(t) = A(0)e^{kt}\) is to determine that \(k = \ln(0.965)\) and then set \(A(6) = aA(0)\) and solve for a.

682 video

video by Khan Academy

Final Answer

80.75%

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A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 420.
a. Find an expression for the number of bacteria after t hours.
b. Find the number of bacteria after 3 hours.
c. Find the rate of growth after 3 hours.
d. When will the population reach 10,000?

Problem Statement

A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 420.
a. Find an expression for the number of bacteria after t hours.
b. Find the number of bacteria after 3 hours.
c. Find the rate of growth after 3 hours.
d. When will the population reach 10,000?

Final Answer

a. \(\displaystyle{ A(t) = 100e^{t \ln(4.20)} }\) bacteria
b. \(7409\) bacteria
c. \(10 632\) bacteria/hr
d. 3.21 hours

Problem Statement

A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 420.
a. Find an expression for the number of bacteria after t hours.
b. Find the number of bacteria after 3 hours.
c. Find the rate of growth after 3 hours.
d. When will the population reach 10,000?

Solution

683 video

video by Khan Academy

Final Answer

a. \(\displaystyle{ A(t) = 100e^{t \ln(4.20)} }\) bacteria
b. \(7409\) bacteria
c. \(10 632\) bacteria/hr
d. 3.21 hours

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