3.7: Growth and Decay
Learning objectives
 Model situations using exponential and logistic functions.
 Solve problems involving these models, using your knowledge of properties of logarithms, and using a graphing calculator.
Introduction
In lesson 5 you learned about modeling phenomena with exponential and logarithmic functions. In the examples in lesson 5, you used a graphing calculator to find a line that fits a given set of data. Here we will use algebraic techniques to develop models, and you will learn about another kind of function, the logistic function, that can be used to model growth.
Exponential growth
In general, if you have enough information about a situation, you can write an exponential function to model growth in the situation. Let’s start with a straightforward example:
Example 1: 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. At this rate, what will the membership be in a week? 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}). Therefore we can model membership with the function M(x) = 10(2^{x}). In seven days, the membership will be M(7) = 10(2^{7}) = 1280.
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*}
 \begin{align*}2^x = 10,000\end{align*}
 \begin{align*}log 2^x = log 10,000\end{align*}
 \begin{align*}x log 2 = 4\end{align*}
 \begin{align*}x = \frac{4} {log 2} \approx 13.3\end{align*}
At this rate, the membership will reach 100,000 in less than two weeks. This result may seem unreasonable. That’s very fast growth!
So let’s consider a slower rate of doubling. Let’s say that the membership doubles every 7 days.
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*}. Let’s verify that this equation makes sense for the data in the table:

 \begin{align*}M(0) = 10 \left (2^{\frac{0} {7}}\right ) = 10(1) = 10\end{align*}
 \begin{align*}M(7) = 10 \left (2^{\frac{7} {7}}\right ) = 10(2) = 20\end{align*}
 \begin{align*}M(14) = 10 \left (2^{\frac{14} {7}}\right ) = 10(2^2) = 10(4) = 40\end{align*}
 \begin{align*}M(21) = 10 \left (2^{\frac{21} {7}}\right ) = 10(2^3) = 10(8) = 80\end{align*}
 \begin{align*}M(28) = 10 \left (2^{\frac{28} {7}}\right ) = 10(2^4) = 10(16) = 160\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).
With the new function \begin{align*}M(x) = 10 \left (2^{\frac{x} {7}}\right )\end{align*}, the membership doubles to 20 in one week, and reaches 100,000 in about 3 months:

 \begin{align*}10 \left (2^{\frac{x} {7}}\right ) = 100,000\end{align*}
 \begin{align*}2^{\frac{x} {7}} = 100,000\end{align*}
 \begin{align*}log 2^{\frac{x} {7}} = log 100,000\end{align*}
 \begin{align*}\frac{x} {7} log 2 = 4\end{align*}
 \begin{align*}x log 2 = 28\end{align*}
 \begin{align*}x = \frac{28} {log 2} \approx 93\end{align*}
The previous two examples of exponential growth have specifically been about doubling. We can also model a more general growth pattern with a more general growth model. While the graphing calculator produces a function of the form y = a(b^{x}), population growth is often modeled with a function in which e is the base. Let’s look at this kind of example:
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. If the growth continues at this rate, how long will it take for the population to reach 1 million?
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*}
 \begin{align*}P(10) = 35000 = 20000 e^{k \cdot 10}\end{align*}
 \begin{align*}\frac{35,000} {20000} = e^{10k}\end{align*}
 \begin{align*}1.75 = e^{10k}\end{align*}
 \begin{align*}ln 1.75 = ln e^{10k}\end{align*}
 \begin{align*}ln 1.75 = 10k ln e\end{align*}
 \begin{align*}ln 1.75 = 10k(1)\end{align*}
 \begin{align*}ln 1.75 = 10k\end{align*}
 \begin{align*}k = \frac{ln 1.75} {10} \approx 0.056\end{align*}
Therefore we can model the population growth with the function \begin{align*}P(t) = 20000 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.
 Here is a solution using an equation:

 \begin{align*}1000000 = 20000 e^{\frac{ln 1.75} {10}t}\end{align*}
 \begin{align*}50 = e^{\frac{ln 1.75} {10}t}\end{align*}
 \begin{align*}ln 50 = ln \left ( e^{\frac{ln 1.75} {10}t}\right )\end{align*}
 \begin{align*}ln 50 = \frac{ln 1.75} {10} t (ln e)\end{align*}
 \begin{align*}ln 50 = \frac{ln 1.75} {10} t (1)\end{align*}
 \begin{align*}10 ln 50 = ln 1.75 t\end{align*}
 \begin{align*}t = \frac{10 ln 50} {ln 1.75} \approx 70\end{align*}
At this rate, it would take about 70 years for the population to reach 1 million. Like the initial doubling example, the growth rate may seem very fast. In reality, a population that grows exponentially may not sustain its growth rate over time. Next we will look at a different kind of function that can be used to model growth of this kind.
Logistic models
Given that resources are limited, a population may slow down in its growth over time. Consider the last example, the town whose population exploded in the 1990s. If there are no more houses to be bought, or tracts of land to be developed, the population will not continue to grow exponentially. The table below shows the population of this town slowing down, though still growing:
t (1990=0)  Population 

0  20,000 
10  35,000 
15  38,000 
20  40,000 
As the population growth slows down, the population may approach what is called a carrying capacity, or an upper bound of the population. We can model this kind of growth using a logistic function, which is a function of the form \begin{align*}f(x) = \frac{c} {1 + a (e^{bx})}\end{align*}.
The graph below shows an example of a logistic function. This kind of graph is often called an “s curve” because of its shape.
Notice that the graph shows slow growth, then fast growth, and then slow growth again, as the population or quantity in question approaches the carrying capacity. Logistics functions are used to model population growth, as well as other situations, such as the amount of medicine in a person’s system
Given the population data above, we can use a graphing calculator to find a logistic function to model this situation. The details of this process are explained in the Technology Note in Lesson 3.5. As shown there, enter the data into L_{1} and L_{2}. Then run a logistic regression. (Press <TI font_STAT>, scroll right to CALC, and scroll down to B. Logistic.) An approximation of the logistic model for this data is: \begin{align*}f(x) = \frac{41042.38} {1 + 1.050 e^{.178x}}\end{align*} . A graph of this function and the data is shown here.
Notice that the graph has a horizontal asymptote around 40,000. Looking at the equation, you should notice that the numerator is about 41,042. This value is in fact the horizontal asymptote, which represents the carrying capacity. We can understand why this is the carrying capacity if we consider the limit of the function as x approaches infinity. As x gets larger and larger, e^{.178}^{x} will get smaller and smaller. So 1.05 e^{.178}^{x} will get smaller. This means that the denominator of the function will get closer and closer to 1:
 \begin{align*}\lim_{x \to \infty} (1 + 1.050 e^{.178x}) = 1\end{align*}.
Therefore the limit of the function is (approximately) (41042/1) = 41042. This means that given the current growth, the model predicts that the population will not go beyond 41,042. This kind of growth is seen in population, as well as other situations in which some quantity grows very fast and then slows down, or when a quantity steeply decreases, and then levels off. You will see work with more examples of logistic functions in the review questions.
Exponential decay
Just as a quantity can grow, or increase exponentially, we can model a decreasing quantity with an exponential function. This kind of situation is referred to as exponential decay. Perhaps the most common example of exponential decay is that of radioactive decay, which refers to the transformation of an atom of one type into an atom of a different type, when the nucleus of the atom loses energy. The rate of radioactive decay is usually measured in terms of “halflife,” or the time it takes for half of the atoms in a sample to decay. For example Carbon14 is a radioactive isotope that is used in “carbon dating,” a method of determining the age of organic materials. The halflife of Carbon14 is 5730 years. This means that if we have a sample of Carbon14, it will take 5730 years for half of the sample to decay. Then it will take another 5730 years for half of the remaining sample to decay, and so on.
We can model decay using the same form of equation we use to model growth, except that the exponent in the equation is negative: A(t) = A_{0} e^{}^{kt}. For example, say we have a sample of Carbon14. How much time will pass before 75% of the original sample remains?
We can use the halflife of 5730 years to determine the value of k:

\begin{align*}A(t) = A_0e^{kt}\end{align*} \begin{align*}\frac{1} {2} = 1e^{k\cdot 5730}\end{align*} We do not know the value of A_{0}, so we use “1” as 100%. (1/2) of the sample remains when t = 5730 years \begin{align*}ln \frac{1} {2} = ln e^{k\cdot 5730}\end{align*} Take the ln of both sides \begin{align*}ln \frac{1} {2} = 5730 klne\end{align*} Use the power property of logs \begin{align*}ln \frac{1} {2} = 5730k\end{align*} \begin{align*}ln(e) = 1\end{align*} \begin{align*}ln 2 = 5730k\end{align*} \begin{align*}ln(1/2) = ln(2^{1})=ln2\end{align*} \begin{align*}ln 2 = 5730k\end{align*} \begin{align*}k = \frac{ln2} {5730}\end{align*} Isolate k
Now we can determine when the amount of Carbon14 remaining is 75% of the original:

 \begin{align*}0.75 = 1 e^{\frac{ln 2} {5730}t}\end{align*}
 \begin{align*}0.75 = 1 e^{\frac{ln 2} {5730}t}\end{align*}
 \begin{align*}ln (0.75) = ln e^{\frac{ln 2} {5730}t}\end{align*}
 \begin{align*}ln (0.75) = \frac{ln 2} {5730}t\end{align*}
 \begin{align*}t = \frac{5730 ln (0.75)} {ln 2} \approx 2378\end{align*}
Therefore it would take about 2,378 years for 75% of the original sample to be remaining. In practice, scientists can approximate the age of an artifact using a process that relies on their knowledge of the halflife of Carbon14, as well as the ratio of Carbon14 to Carbon12 (the most abundant, stable form of carbon) in an object. While the concept of halflife often is used in the context of radioactive decay, it is also used in other situations. In the review questions, you will see another common example, that of medicine in a person’s system.
Related to exponential decay is Newton’s Law of Cooling. The Law of Cooling allows us to determine the temperature of a cooling (or warming) object, based on the temperature of the surroundings and the time since the object entered the surroundings. The general form of the cooling function is T(x) = T_{5} + (T_{0}  T_{5}) e^{}^{kx}, where T_{5}, is the surrounding temperature, T_{0} is the initial temperature, and x represents the time since the object began cooling or warming.
The first graph shows a situation in which an object is cooling. The graph has a horizontal asymptote at y = 70. This tells us that the object is cooling to \begin{align*}70^{\circ}\end{align*}F. The second graph has a horizontal asymptote at y = 70 as well, but in this situation, the object is warming up to \begin{align*}70^{\circ}\end{align*}F.
We can use the general form of the function to answer questions about cooling (or warming) situations. Consider the following example: you are baking a casserole in a dish, and the oven is set to \begin{align*}325^{\circ}\end{align*}F. You take the pan out of the oven and put it on a cooling rack in your kitchen which is \begin{align*}70^{\circ}\end{align*}F, and after 10 minutes the pan has cooled to \begin{align*}300^{\circ}\end{align*}F. How long will it take for the pan to cool to 200^{0}F?
We can use the general form of the equation and the information given in the problem to find the value of k:

 \begin{align*}T(x) = T_s + (T_0 T_s) e^{kx}\end{align*}
 \begin{align*}T(x) = 70 + (325  70)e^{kx}\end{align*}
 \begin{align*}T(x) = 70 + (255)e^{kx}\end{align*}
 \begin{align*}T(10) = 70 + 255e^{10k} = 300\end{align*}
 \begin{align*}255e^{10k} = 230\end{align*}
 \begin{align*}e^{10k} = \frac{230} {255}\end{align*}
 \begin{align*}ln e^{10k} = ln \left (\frac{230} {255}\right )\end{align*}
 \begin{align*}10k = ln \left (\frac{230} {255}\right )\end{align*}
 \begin{align*}k = \frac{ln \left (\frac{230} {255}\right )} {10} \approx 0.0103\end{align*}
Now we can determine the amount of time it takes for the pan to cool to 200 degrees:

 \begin{align*}T(x) = 70 + (255)e^{.0103}x\end{align*}
 \begin{align*}T(x) = 70 + (255)e^{.0103}x\end{align*}
 \begin{align*}200 = 70 + (255)e^{.0103}x\end{align*}
 \begin{align*}130 = (255)e^{.0103}x\end{align*}
 \begin{align*}\frac{130} {255} = e^{.0103}x\end{align*}
 \begin{align*}ln \left (\frac{130} {255}\right ) = ln e^{.0103}x\end{align*}
 \begin{align*}ln \left (\frac{130} {255}\right ) = .0103 x\end{align*}
 \begin{align*}x = \frac{ln \left (\frac{130} {255}\right )} {.0103} \approx 65\end{align*}
Therefore, in the given surroundings, it would take about an hour for the pan to cool to 200 degrees.
Lesson Summary
In this lesson we have developed exponential and logistic models to represent different phenomena. We have considered exponential growth, logistic growth, and exponential decay. After reading the examples in this lesson, you should be able to write a function to represent a given situation, to evaluate the function for a given value of x, and to solve exponential equations in order to find values of x, given values of the function. For example, in a situation of exponential population growth as a function of time, you should be able to determine the population at a particular time, and to determine the time it takes for the population to reach a given amount. You should be able to solve these kinds of problems by solving exponential equations, and by using graphing utilities, as we have done throughout the chapter.
Points to Consider
 How can we use the same equation for exponential growth and decay?
 What are the restrictions on domain and range for the examples in this lesson?
 How can we use different equations to model the same situations?
Review Questions
 The population of a town was 50,000 in 1980, and it grew to 70,000 by 1995. a. Write an exponential function to model the growth of the population. b. Use the function to estimate the population in 2010. c. What if the population growth was linear? Write a linear equation to model the population growth, and use it to estimate the population in 2010.
 A telecommunications company began providing wireless service in 1994, and during that year the company had 1000 subscribers. By 2004, the company had 12,000 subscribers. a. Write an exponential function to model the situation b. Use the model to determine how long it will take for the company to reach 50,000 subscribers.
 The population of a particular strain of bacteria triples every 8 hours. a. Write a general exponential function to model the bacteria growth. b. Use the model to determine how long it will take for a sample of bacteria to be 100 times its original population. c. Use a graph to verify your solution to part b.
 The halflife of acetaminophen is about 2 hours. a. If you take 650 mg of acetaminophen, how much will be left in your system after 7 hours? b. How long before there is less than 25 mg in your system?
 The population of a city was 200,000 in 1991, and it decreased to 170,000 by 2001. a. Write an exponential function to model the decreasing population, and use the model to predict the population in 2008. b. Under what circumstances might the function cease to model the situation after a certain point in time?
 Consider the following situation: you buy a large box of pens for the start of the school year, and after six weeks, (1/3) of the pens remain. After another six weeks, (1/3) of the remaining pens were remaining. If you continue this pattern, when will you only have 5% of the pens left?
 Use Newton’s law of cooling to answer the question: you pour hot water into a mug to make tea. The temperature of the water is about 200 degrees. The surrounding temperature is about 75 degrees. You let the water cool for 5 minutes, and the temperature decreases to 160 degrees. What will the temperature be after 15 minutes?
 The spread of a particular virus can be modeled with the logistic function \begin{align*}f(x) = \frac{2000} {1 + 600e^{.75x}}\end{align*}, where x is the number of days the virus has been spreading, and f(x) represents the number of people who have the virus. a. How many people will be affected after 7 days? b. How many days will it take for the spread to be within one person of carrying capacity?
 Consider again the situation in problem #2: A telecommunications company began providing wireless service in 1994, and during that year the company had 1000 subscribers. By 2004, the company had 12,000 subscribers. If the company has 15,000 subscribers in 2005, and 16,000 in 2007, what type of model do you think should be used to model the situation? Use a graphing calculator to find a regression equation, and use the equation to predict the number of subscribers in 2010.
 Compare exponential and logistic functions as tools for modeling growth. What do they have in common, and how do they differ?
Review Answers
 a. \begin{align*}A(T) = 50,000 e^{\frac{ln(\frac{7}{5})} {15}t}\end{align*} b. \begin{align*}98,000\end{align*} c. \begin{align*}f(t) = \frac{4000} {3} t + 50000\end{align*}. The population would be 90,000, which is different by about 9%.
 a. \begin{align*}S(t) = 1000 e^{\frac{ln(12)} {10}t}\end{align*} b. \begin{align*}t = \frac{10 ln 50} {ln 12} \approx 15.74\end{align*}
 a. A(t) = A_{0} (3^{t/8}) b. \begin{align*}t = \frac{16} {log 3} \approx 33.53\end{align*} c. The graph below shows y = 100 and y = 3^{x/8}, which intersect at approximately x = 33.53
 a. About 57.45 mg b. About 9.4 hours
 a. \begin{align*}P(t) = 200000 e^{\frac{ln.85} {10}t} , P(17) \approx 151720\end{align*} b. If the economy or other factors change, the population might begin to increase, or the rate of decrease could change as well.
 \begin{align*}t = \frac{6 log (0.05)} {log(\frac{1}{3})} \approx 16\end{align*} weeks
 About 114 degrees.
 a. About 482 people. b. After 19 days, over 1999 people have the virus.
 The graph indicates a logistic model. \begin{align*}f(x) \approx \frac{18872} {1 + 21.45 e^{377x}}\end{align*} gives 17952 subscribers in 2010.
 Both types of functions model fast increase in growth, but the logistic model shows the growth slowing down after some point, with some upper bound on the quantity in question. (Many people argue that logistic growth is more realistic.)
Vocabulary
 Carrying capacity
 The supportable population of an organism, given the food, habitat, water and other necessities available within an ecosystem is known as the ecosystem's carrying capacity for that organism.
 Radioactive decay
 Radioactive decay is the process in which an unstable atomic nucleus loses energy by emitting radiation in the form of particles or electromagnetic waves. This decay, or loss of energy, results in an atom of one type transforming to an atom of a different type. For example, Carbon14 transforms into Nitrogen14
 Halflife
 The amount of time it takes for half of a given amount of a substance to decay. The halflife remains the same, no matter how much of the substance there is.
 Isotope
 Isotopes are any of two or more forms of a chemical element, having the same number of protons in the nucleus, or the same atomic number, but having different numbers of neutrons in the nucleus, or different atomic weights.