6.4: Exponential Growth and Decay
Learning Objectives
A student will be able to:
 Apply the laws of exponential and logarithmic functions to a variety of applications.
 Model situations of growth and decay in a variety of problems.
When the rate of change in a substance or population is proportional to the amount present at any time \begin{align*}t\end{align*}, we say that this substance or population is going through either a decay or a growth, depending on the sign of the constant of proportionality.
This kind of growth is called exponential growth and is characterized by rapid growth or decay. For example, a population of bacteria may increase exponentially with time because the rate of change of its population is proportional to its population at a given instant of time (more bacteria make more bacteria and fewer bacteria make fewer bacteria). The decomposition of a radioactive substance is another example in which the rate of decay is proportional to the amount of the substance at a given time instant. In the business world, the interest added to an investment each day, month, or year is proportional to the amount present, so this is also an example of exponential growth.
Mathematically, the relationship between amount \begin{align*}y\end{align*} and time \begin{align*}t\end{align*} is a differential equation:
\begin{align*} \frac{dy} {dt} = ky.\end{align*}
Separating variables,
\begin{align*} \frac{dy} {y} = k dt,\end{align*}
and integrating both sides,
\begin{align*} \int \frac{dy} {y} = \int k dt,\end{align*}
gives us
\begin{align*}\ln y & = kt + C,\\ y & = e^{kt} e^{C}\\ & = Ce^{kt}.\end{align*}
So the solution to the equation \begin{align*}(dy/dt) = ky\end{align*} has the form \begin{align*}y = C e^{kt}.\end{align*} The box below summarizes the details of this function.
The Law of Exponential Growth and Decay
The function \begin{align*}y = C e^{kt}\end{align*} is a model for exponential growth or decay, depending on the value of \begin{align*}k.\end{align*}
 If \begin{align*}k > 0\end{align*}: The function represents exponential growth (increase).
 If \begin{align*}k < 0\end{align*}: The function represents exponential decay (decrease).
Where \begin{align*}t\end{align*} is the time, \begin{align*}C\end{align*} is the initial population at \begin{align*}t = 0,\end{align*} and \begin{align*}y\end{align*} is the population after time \begin{align*}t.\end{align*}
Applications of Growth and Decay
Radioactive Decay
In physics, radioactive decay is a process in which an unstable atomic nucleus loses energy by emitting radiation in the form of electromagnetic radiation (like gamma rays) or particles (such as beta and alpha particles). During this process, the nucleus will continue to decay, in a chain of decays, until a new stable nucleus is reached (called an isotope). Physicists measure the rate of decay by the time it takes a sample to lose half of its nuclei due to radioactive decay. Initially, as the nuclei begins to decay, the rate starts very fast and furious, but it slows down over time as more and more of the available nuclei have decayed. The figure below shows a typical radioactive decay of a nucleus. As you can see, the graph has the shape of an exponential function with \begin{align*}k < 0.\end{align*}
The equation that is used for radioactive decay is \begin{align*}y = Ce^{kt}.\end{align*} We want to find an expression for the halflife of an isotope. Since halflife is defined as the time it takes for a sample to lose half of its nuclei, then if we starting with an initial mass \begin{align*}C\end{align*} (measured in grams), then after some time \begin{align*}t,\end{align*} \begin{align*}y\end{align*} will become half the amount that we started with, \begin{align*}C/2.\end{align*} Substituting this into the exponential decay model,
\begin{align*}y & = Ce^{kt}\\ \frac{C} {2} & = Ce^{kt}.\end{align*}
Canceling \begin{align*}C\end{align*} from both sides,
\begin{align*} \frac{1} {2} = e^{kt}.\end{align*}
Solving for \begin{align*}t,\end{align*} which is the halflife, by taking the natural logarithm on both sides,
\begin{align*}\ln \frac{1} {2} & = \ln e^{kt}\\ \ln 2 & = kt.\end{align*}
Solving for \begin{align*}t,\end{align*} and denoting it with new notation \begin{align*}t_{1/2}\end{align*} for halflife (a standard notation in physics),
\begin{align*} t_{1/2} = \frac{\ln 2} {k} = \frac{0.693} {k}\end{align*}
This is a famous expression in physics for measuring the halflife of a substance if the decay constant \begin{align*}k\end{align*} is known. It can also be used to compute \begin{align*}k\end{align*} if the halflife \begin{align*}t_{1/2}\end{align*} is known.
Example 1:
A radioactive sample contains \begin{align*}2 \;\mathrm{grams}\end{align*} of nobelium. If you know that the halflife of nobelium is \begin{align*}25 \;\mathrm{seconds}\end{align*}, how much will remain after \begin{align*}3 \;\mathrm{minutes}\end{align*}?
Solution:
Before we compute the mass of nobelium after \begin{align*}3 \;\mathrm{minutes}\end{align*}, we need to first know its decay rate \begin{align*}k.\end{align*} Using the halflife formula,
\begin{align*}t_{1/2} & = \frac{\ln 2} {k}\\ k & = \frac{\ln 2} {t_{1/2}}\\ & = \frac{\ln 2} {25}\\ & = 0.028 \ \sec^{1}\end{align*}
So the decay rate is \begin{align*}k = 0.028\mathrm{/sec}.\end{align*} The common unit for the decay rate is the Becquerel \begin{align*}(Bq)\end{align*}: \begin{align*}1 \;\mathrm{Bq}\end{align*} is equivalent to \begin{align*}1\end{align*} decay per sec. Since we found \begin{align*}k\end{align*}, we are now ready to calculate the mass after \begin{align*}3 \;\mathrm{minutes}\end{align*}. We use the radioactive decay formula. Remember, \begin{align*}C\end{align*} represents the initial mass, \begin{align*}C = 2\;\mathrm{grams}\end{align*}, and \begin{align*}t = 3 \;\mathrm{minutes} = 180 \;\mathrm{seconds}\end{align*}. Thus
\begin{align*}y & = Ce^{kt}\\ & = 2e^{(0.028)(180)}\\ & = 0.013 \ \text{grams}.\end{align*}
So after \begin{align*}3 \;\mathrm{minutes}\end{align*}, the mass of the isotope is approximately \begin{align*}0.013\;\mathrm{grams}\end{align*}.
Population Growth
The same formula \begin{align*}y = Ce^{kt}\end{align*} can be used for population growth, except that \begin{align*}k > 0,\end{align*} since it is an increasing function.
Example 2:
A certain population of bacteria increases continuously at a rate that is proportional to its present number. The initial population of the bacterial culture is \begin{align*}140\end{align*} and jumped to \begin{align*}720\end{align*} bacteria in \begin{align*}4 \;\mathrm{hours}\end{align*}.
 How many will be there in \begin{align*}10 \;\mathrm{hours}\end{align*}?
 How long will it take the population to double?
Solution:
From reading the first sentence in the problem, we learn that the bacteria is increasing exponentially. Therefore, the exponential growth formula is the correct model to use.
1. Just like we did in the previous example, we need to first find \begin{align*}k,\end{align*} the growth rate. Notice that \begin{align*}C = 140, t = 4,\end{align*} and \begin{align*}y = 720.\end{align*} Substituting and solving for \begin{align*}k.\end{align*}
\begin{align*}y & = Ce^{kt}\\ 720 & = 140 e^{k(4)}.\end{align*}
Dividing both sides by \begin{align*}140\end{align*} and then projecting the natural logarithm on both sides,
\begin{align*}\ln \frac{720} {140} & = \ln e^{4k}\\ \ln 5.143 & = 4k\\ k & = 0.409.\end{align*}
Now that we have found \begin{align*}k,\end{align*} we want to know how many will be there after \begin{align*}10 \;\mathrm{hours}\end{align*}. Substituting,
\begin{align*}y & = Ce^{kt}\\ & = 140e^{(0.409)(10)}\\ & = 8364 \ \text{bacteria}.\end{align*}
2. We are looking for the time required for the population to double. This means that we are looking for the time at which \begin{align*}y = 2C.\end{align*} Substituting,
\begin{align*}y & = Ce^{kt}\\ 2C & = C e^{kt}\\ 2 & = e^{kt}.\end{align*}
Solving for \begin{align*}t\end{align*} requires taking the natural logarithm of both sides:
\begin{align*}\ln 2 & = \ln e^{kt}\\ \ln 2 & = kt.\end{align*}
Solving for \begin{align*}t,\end{align*}
\begin{align*}t & = \frac{\ln 2} {k}\\ & = \frac{\ln 2} {0.409}\\ t & = 1.7 \ \text{hours}.\end{align*}
This tells us that after about \begin{align*}1.7 \;\mathrm{hours}\end{align*} (around \begin{align*}100 \;\mathrm{minutes}\end{align*}) the population of the bacteria will double in number.
Compound Interest
Investors and bankers depend on compound interest to increase their investment. Traditionally, banks added interest after certain periods of time, such as a month or a year, and the phrase was “the interest is being compounded monthly or yearly.” With the advent of computers, the compunding could be done daily or even more often. Our exponential model represents continuous, or instantaneous, compounding, and it is a good model of current banking practices. Our model states that
\begin{align*}A = Pe^{rt},\end{align*}
where \begin{align*}P\end{align*} is the initial investment (present value) and \begin{align*}A\end{align*} is the future value of the investment after time \begin{align*}t\end{align*} at an interest rate of \begin{align*}r.\end{align*} The interest rate \begin{align*}r\end{align*} is usually given in percentage per year. The rate must be converted to a decimal number, and \begin{align*}t\end{align*} must be expressed in years. The example below illustrates this model.
Example 3:
An investor invests an amount of \begin{align*}\$10,000\end{align*} and discovers that its value has doubled in 5 years. What is the annual interest rate that this investment is earning?
Solution:
We use the exponential growth model for continuously compounded interest,
\begin{align*}A & = Pe^{rt}\\ 20,000 & = 10,000e^{r(5)}\\ 2 & = e^{5r}\\ \ln 2 & = 5r.\end{align*}
Thus
\begin{align*}r & = \frac{\ln 2} {5}\\ & = 0.139\\ r & = 13.9\%\end{align*}
The investment has grown at a rate of \begin{align*}13.9\%\end{align*} per year.
Example 4:
Going back to the previous example, how long will it take the invested money to triple?
Solution:
\begin{align*}A & = Pe^{rt}\\ 30,000 & = 10,000e^{(0.139)(t)}\\ 3 & = e^{0.139t}\\ \ln 3 & = 0.139t\\ t & = \frac{\ln 3}{0.139}\\ & = 7.9 \ \text{years}.\end{align*}
Other Exponential Models and Examples
Not all exponential growths and decays are modeled in the natural base \begin{align*}e\end{align*} or by \begin{align*}y = Ce^{kt}.\end{align*} Actually, in everyday life most are constructed from empirical data and regression techniques. For example, in the business world the demand function for a product may be described by the formula
\begin{align*}p = 12, 400  \frac{11, 000} {2.2 + e^{0.0003x}},\end{align*}
where \begin{align*}p\end{align*} is the price per unit and \begin{align*}x\end{align*} is the number of units produced. So if the business is interested in basing the price of its unit on the number that it is projecting to sell, this formula becomes very helpful.
If a motorcycle factory is projecting to sell \begin{align*}7000 \;\mathrm{units}\end{align*} in one month, what price should the factory set on each motorcycle?
\begin{align*}p & = 12,400  \frac{11, 000} {2.2 + e^{0.0003x}}\\ & = 12,400  \frac{11, 000} {2.2 + e^{0.0003(7000)}}\\ & = 12,400  \frac{11, 000} {2.2 + 0.122}\\ & = 7,663.\end{align*}
Thus the factory’s base price for each motorcycle should be set at \begin{align*}\$7663.\end{align*}
As another example, let’s say a medical researcher is studying the spread of the flu virus through a certain campus during the winter months. Let’s assume that the model for the spread is described by
\begin{align*}P = \frac{4500} {1 + 4499e^{0.8x}},\ x \ge 0,\end{align*}
where \begin{align*}P\end{align*} represents the total number of infected students and \begin{align*}x\end{align*} is the time, measured in days. Suppose the researcher is interested in the number of students who will be infected in the next week (\begin{align*}7\end{align*} days). Substituting \begin{align*}x = 7\end{align*} into the model,
\begin{align*}P &= \frac{4500} {1 + 4499e^{0.8x}}\\ & = \frac{4500} {1 + 4499e^{0.8(7)}}\\ & = \frac{4500} {1 + 4499(0.004)}\\ & = 255.\end{align*}
According to the model, \begin{align*}255\end{align*} students will become infected with the flu virus. Assume further that the researcher wants to know how long it will take until \begin{align*}1000\end{align*} students become infected with the flu virus. Solving for \begin{align*}x,\end{align*}
\begin{align*}P = \frac{4500} {1 + 4499e^{0.8x}}.\end{align*}
Crossmultiplying,
\begin{align*}P(1 + 4499e^{0.8x)} & = 4500\\ 1 + 4499e^{0.8x} & = \frac{4500} {P}\\ 4499e^{0.8x} & = \frac{4500} {P}  1\\ & = \frac{4500  P} {P}\\ e^{0.8x} & = \frac{4500  P} {4499P}.\end{align*}
Projecting \begin{align*}\ln\end{align*} on both sides,
\begin{align*}0.8x & = \ln \left [\frac{4500  P} {4499P} \right]\\ x & = \ln \left [\frac{4500  P} {4499P} \right] \div (0.8).\end{align*}
Substituting for \begin{align*}P = 1000\end{align*},
\begin{align*}x = 9\ \text{days}.\end{align*}
So the flu virus will spread to \begin{align*}1000\end{align*} students in \begin{align*}9\end{align*} days.
Other applications are introduced in the exercises.
Multimedia Links
For a video presentation of exponential growth involving bacteria (some calculus in part c) (14.0), see Khan Academy, Exponential Growth and Decay (16:00).
For a video presentation of exponential decay (14.0), see Just Math Tutoring, Exponential Decay, Finding HalfLife (6:08).
For additional problems on exponential growth and decay (14.0), see Khan Academy, Word Problem Solving, Exponential Growth and Decay (7:21).
Review Questions
 In 1990, the population of the USA was \begin{align*}249\;\mathrm{million}\end{align*}. Assume that the annual growth rate is \begin{align*}1.8\%\end{align*}.
 According to this model, what was the population in the year 2000?
 According to this model, in which year will the population reach \begin{align*}1 \;\mathrm{billion}\end{align*}?
 Prove that if a quantity \begin{align*}A\end{align*} is exponentially growing and if \begin{align*}A_1\end{align*} is the value at \begin{align*}t_1\end{align*} and \begin{align*}A_2\end{align*} at time \begin{align*}t_2,\end{align*} then the growth rate will be given by \begin{align*} k = \frac{1} {t_1  t_2} \ln \left (\frac{A_1} {A_2} \right).\end{align*}
 Newton’s Law of Cooling states that the rate of cooling is proportional to the difference in temperature between the object and the surroundings. The law is expressed by the formula \begin{align*}T(t) = (T_0  T_r )e^{kt} + T_r\end{align*} where \begin{align*}T_0\end{align*} is the initial temperature of the object at \begin{align*}t = 0,\end{align*} \begin{align*}T_r\end{align*} is the room temperature (the surroundings), and \begin{align*}k\end{align*} is a constant that is unique for the measuring instrument (the thermometer) called the time constant. Suppose a liter of juice at \begin{align*}23^\circ C\end{align*} is placed in the refrigerator to cool. If the temperature of the refrigerator is kept at \begin{align*}11^\circ C\end{align*} and \begin{align*}k = 0.417,\end{align*} what is the temperature of the juice after \begin{align*}3 \;\mathrm{minutes}\end{align*}?
 Referring back to problem 3, if it takes an object \begin{align*}320 \;\mathrm{seconds}\end{align*} to cool from \begin{align*}40^\circ C\end{align*} above room temperature to \begin{align*}22^\circ C\end{align*} above room temperature, how long will it take to cool another \begin{align*}10^\circ C\end{align*} after it reaches \begin{align*}22^\circ C\end{align*} above room temperature?
 Polonium\begin{align*}210\end{align*} is a radioactive isotope with halflife of \begin{align*}140\end{align*} days. If a sample has a mass of \begin{align*}10\;\mathrm{grams}\end{align*}, how much will remain after \begin{align*}10\end{align*} weeks?
 In the physics of acoustics, there is a relationship between the subjective sensation of loudness and the physically measured intensity of sound. This relationship is called the sound level \begin{align*}\beta.\end{align*} It is specified on a logarithmic scale and measured with units of decibels \begin{align*}(dB)\end{align*}. The sound level \begin{align*}\beta\end{align*} of any sound is defined in terms of its intensity \begin{align*}I\end{align*} (in the SImks unit system, it is measured in watts per meter squared, \begin{align*}\mathrm{W/m^2}\end{align*}) as \begin{align*} \beta = 10 \log \frac{I} {10^{12}}.\end{align*} For example, the average decibel level of a busy street traffic is \begin{align*}70\;\mathrm{dB},\end{align*} normal conversation at a dinner table is \begin{align*}55\;\mathrm{dB}\end{align*}, the sound of leaves rustling is \begin{align*}10\;\mathrm{dB}\end{align*}, the siren of a fire truck at \begin{align*}30\;\mathrm{meters}\end{align*} is \begin{align*}100\;\mathrm{dB},\end{align*} and a loud rock concert is \begin{align*}120\;\mathrm{dB.}\end{align*} The sound level \begin{align*}120\;\mathrm{dB}\end{align*} is considered the threshold of pain for the human ear and \begin{align*}0\;\mathrm{dB}\end{align*} is the threshold of hearing (the minimum sound that can be heard by humans.)
 If at a heavy metal rock concert a \begin{align*}dB\end{align*} meter registered \begin{align*}130\;\mathrm{dB},\end{align*} what is the intensity \begin{align*}I\end{align*} of this sound level?
 What is the sound level (in \begin{align*}dB\end{align*}) of a sound whose intensity is \begin{align*}2.0 \times 10^{6} \;\mathrm{W/m^2}\end{align*} ?
 Referring to problem #6, a single mosquito \begin{align*}10\;\mathrm{meters}\end{align*} away from a person makes a sound that is barely heard by the person (threshold \begin{align*}0\;\mathrm{dB}\end{align*}). What will be the sound level of \begin{align*}1000\end{align*} mosquitoes at the same distance?
 Referring back to problem #6, a noisy machine at a factory produces a sound level of \begin{align*}90\;\mathrm{dB.}\end{align*} If an identical machine is placed beside it, what is the combined sound level of the two machines?
Review Answers

 \begin{align*}298.1 \;\mathrm{million}\end{align*}
 \begin{align*}2067\end{align*}
 Hint: use \begin{align*}A_1 = Ce^{kt 1}\end{align*}
 \begin{align*}14.4^\circ \mathrm{C}\end{align*}
 \begin{align*}324 \;\mathrm{seconds}\end{align*}, or about five and a half more minutes.

\begin{align*}7\;\mathrm{grams}\end{align*}
 \begin{align*}10 \;\mathrm{W/m^2}\end{align*}
 \begin{align*}63\;\mathrm{dB}\end{align*}
 \begin{align*}30\;\mathrm{dB}\end{align*}
 \begin{align*}93\;\mathrm{dB}\end{align*}
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