Lesson Explainer: Exponential Growth and Decay Mathematics

In this explainer, we will learn how to set up and solve exponential growth and decay equations and how to interpret their solutions.

Exponential growth and decay are both based on exponential functions but exhibit different behavior in their rates of change. Exponential growth is a mathematical change whose magnitude grows without bound over time and the rate of change continues to increase and thus has a divergent limit. Exponential decay is a mathematical change whose magnitude decays over time and the rate of change continues to decrease and thus approaches a finite limit. These changes can occur in either the positive or negative direction, depending on the sign of the starting value.

Exponential growth and decay equations have many real-world applications: for example, in physics with Newton’s law of cooling or the half-life, radiocarbon dating, or decay of radioactive material; in electronics with the discharge of an electric capacitor through a resistance; in computing with Moore’s law describing the growth of the number of transistors in a dense integrated circuit and the processing power of computers; and in finance with compound interest, to name a few.

Examples in biology include modeling the blood concentration of drugs, diseases (such as COVID-19), or cancer growth. Models of tumor growth and treatment have a very long history, modeled by the equation of exponential growth, 𝑁(𝑑)=𝑁(0)⋅𝑒, where πœ†>0 and 𝑁(𝑑) is the number of clonogenic cells in a tumor at time 𝑑. These mathematical models can be used to understand how cancer develops and grows. They can also be used to optimize or personalize treatment, predict the effectiveness of different combinations or new treatments, and give insight into the development of resistance to treatment.

We can see the astounding nature of exponential growth with a story about grains of rice on a chess board. According to legend, grand vizier Sissa Ben Dahir invented the game of chess and gifted the Indian King Shiram with a chessboard. As gratitude for the gift, the Grand Vizier was offered any reward that he requested, as long as it sounded reasonable to the king. The grand vizier made a seemingly humble request with some rice on a chessboard. He requested that one grain of rice be placed on the first square, two on the second, four on the third, and so on, doubling the number of rice each time from the previous square, until the end of the chessboard (which has 64 squares).

The king, impressed by the modest request, obliged and ordered for the rice to be brought in bags. For the first few squares, everything seemed to be going well, but by the 21st square, there were more than a million (1β€Žβ€‰β€Ž048β€Žβ€‰β€Ž576) grains of rice; the bag was emptied and another had to be brought in, which was immediately emptied by the proceeding square. By the 41st square, there were more than a trillion (1β€Žβ€‰β€Ž099β€Žβ€‰β€Ž511β€Žβ€‰β€Ž627β€Žβ€‰β€Ž776) grains of rice, and as this progression continues, doubling each time, there will be more grains of rice in the final squares than exist in the entire world, even without counting all the rice on all the previous squares. Just the grains on the final square would exceed the world production of rice for over 1β€Žβ€‰β€Ž000 years.

It turns out that this was not a modest request at all by the grand vizier, and this highlights how human beings have a difficulty understanding exponential growth, with a tendency to underestimate its progression.

A similar situation arises when we consider the growth in the population of rabbits. If a population of rabbits doubles every month, assuming no death, we will have 2 in the first month, 4 in the second, 8 in the third, and then 16, 32, 64, 128, and so on.

This is a similar situation to the grains of rice on a chessboard, which suggests that the equation describing them will also be similar.

Let’s denote π‘ŽοŠ as the number of grains of rice on the (𝑛+1)th square. Since the number of grains of rice doubles that of the previous square, π‘ŽοŠοŠ±οŠ§, and starting from one grain of rice on the first square, π‘Ž=1, we have the relation π‘Ž=1,π‘Ž=2β‹…π‘Ž,𝑛>0.for

Using this relation, we can express the number of grains of rice on the (𝑛+1)th square as a closed formula in terms of 𝑛, since we have π‘Ž=2β‹…π‘Ž=2,π‘Ž=2β‹…π‘Ž=2β‹…2β‹―π‘Ž=2β‹…2β‹―2β‹…2ο‡Œο†²ο†²ο‡ο†²ο†²ο‡Ž,times which gives the growth as π‘Ž=2.

Similarly, denoting the number of rabbits on the π‘šthΒ month as π‘Žο‰οŠ±οŠ§, starting with 2 rabbits on the first month, π‘Ž=2, we have π‘Ž=2.ο‰ο‰οŠ°οŠ§

We can also use a function to describe their growth as 𝑦=2, where 𝑑=π‘›βˆ’1 for the 𝑛th square or 𝑑=π‘š for each monthΒ π‘š. This function is more general than the formula we obtained for π‘ŽοŠ or π‘Žο‰, since it allows noninteger values of 𝑑 but will otherwise be the same for integer values.

The initial number of grains of rice on the chess board, on the first square (𝑛=1), or the initial population of rabbits, on the first monthΒ (π‘š=1), will be when 𝑑=π‘›βˆ’1=1βˆ’1=0 or 𝑑=π‘š=1, and thus 𝑦=1 and 𝑦=2 respectively. This exponential function describes how the quantity 𝑦 increases after a period of 𝑑 after the initial measurement.

In general, we can describe the exponential growth or decay of a quantity 𝑦 using an exponential function, where the parameters, input, and output of the function can have noninteger values.

Definition: Exponential Growth and Decay

The exponential function describing growth and decay can be represented as 𝑦(𝑑)=π‘Žβ‹…π‘,𝑏>0𝑏≠1,forrealconstantsand where π‘Ž=𝑦(0) is the initial value of the quantity 𝑦 at the start and 𝑏 controls the behavior of the exponential function: exponential growth occurs for 𝑏>1 and exponential decay for 0<𝑏<1.

For example, the exponential equation 𝑦=4(1.28) describes exponential growth, since 𝑏=1.28>1, while the exponential equation 𝑦=3(0.74) describes exponential decay, since 𝑏=0.74<1. Note that we exclude 𝑏=1, since we would have 𝑦=π‘Ž, which is a constant function that does not change.

Let’s consider an example where we have to find the exponential growth equation for a given situation to model the number of bacteria in a laboratory at time 𝑑.

Example 1: Writing an Exponential Expression to Represent an Exponential Growth Model

The number of bacteria in a laboratory quadruples every hour. There were initially 200 bacteria. Write an expression for 𝐡(𝑑), the number of bacteria 𝑑 hours after the initial measurement.

Answer

In this example, we will find an expression for the number of bacteria at 𝑑 hours after the initial measurement.

Micro bacterium and therapeutic bacteria organisms

Recall that exponential growth can be modeled by the function 𝐡(𝑑)=π‘Žβ‹…π‘,𝑏>1,for where π‘Ž=𝐡(0) is the initial value of 𝐡 and 𝑏>1 represents growth.

For the number of bacteria, we are told that there are initially 200 bacteria and the number of bacterial quadruples every hour; hence, π‘Ž=200 and 𝑏=4. An expression for the number of bacteria 𝑑 hours after the initial measurement is 𝐡(𝑑)=200β‹…4.

The graph for the exponential growth function (𝑏>1) has the following shape.

And the graph for the exponential decay function (0<𝑏<1) has the following shape.

The graphs also depict starting values with π‘Ž=𝑦(0)>0 as typically we mostly deal with exponential functions where the starting value for the quantity 𝑦 is positive, particularly for real-world applications and the examples in this explainer, but this is not a requirement for the definition.

Note that the limiting behavior of the exponential function will be different depending on whether there is growth or decay.

For exponential growth (𝑏>1), the limit will depend on the starting value π‘Ž=𝑦(0). In particular, assuming π‘Žβ‰ 0, the absolute or magnitude of the function |𝑦(𝑑)| diverges as 𝑑 approaches infinity. This means that the quantity 𝑦 grows without bound for arbitrarily large 𝑑; that is, its magnitude gets larger after a long time.

For exponential decay (0<𝑏<1), the function 𝑦(𝑑) approaches zero as 𝑑 approaches infinity. This means the quantity 𝑦 decays to zero for arbitrarily large 𝑑; that is, it gets smaller after a long time.

In general, we can have 𝑑𝑛 in the exponent 𝑦(𝑑)=π‘Žβ‹…π‘, which will mean that the quantity multiplies by 𝑏 every 𝑛 period of time or, equivalently, the number multiplies by π‘οŽ ο‘ƒ every period 𝑑.

As an example, Moore’s law states that the number of transistors in a dense integrated circuit doubles approximately every two years, which increases the processing power of computers.

Using Moore’s law, we can find an explicit formula for the number of transistors in a single circuit in a year 𝑦. Assuming that in 1971 a circuit had 4β€Žβ€‰β€Ž004 transistors, we have π‘Ž=4004, and since the number doubles every two years, we have 𝑁=4004β‹…2,ο‘‰οŽ‘ where 𝑑 is the number of years after 1971. Note that we have 𝑑2 in the exponent, which tells us that the number of transistors doubles every 2 years or, equivalently, multiplies by 2 every year. We can represent this growth as a table, where the values for the number of transistors are given every two years.

Time (Year)Number of Transistors
1971 4β€Žβ€‰β€Ž004
1973 8β€Žβ€‰β€Ž008
1975 16β€Žβ€‰β€Ž016
1977 32β€Žβ€‰β€Ž032
1979 64β€Žβ€‰β€Ž064
1981 128β€Žβ€‰β€Ž128

A similar equation can be used as a simple model for population growth known as Malthusian Law, where 𝑁 would be replaced by 𝑃(𝑑), the population at time 𝑑, which can represent the growth or spread of bacteria, viruses, plants, animals and people.

Also, if 𝑦(𝑑) describes a quantity which only makes sense as an integer, then we may round the value up or down to the nearest integer at a given time 𝑑, where appropriate for a mathematical model. For example, if we find that at a certain time we have 𝑦(5)=27.48 people in a population, it would make sense to round down to 𝑦=27, since you cannot have 0.48 of a person.

Let’s consider an example where we are given the growth of the population of rabbits as an exponential function, similar to the one discussed above, and we have to determine the number of rabbits after a given time.

Example 2: Evaluating Exponential Function Involving Exponential Growth

Seif has 73 rabbits. He believes that he will have 𝑧=73β‹…(4.23)ο‘ƒοŽ’ rabbits after 𝑛 months. How many rabbits does he expect to have 2 months from now?

Answer

In this example, we will determine the number of rabbits expected after 2 months from the given expression modeling their population.

A plot of this function is given below.

Recall that exponential growth can be modeled by the function 𝑧(𝑑)=π‘Žβ‹…π‘π‘>1,for where π‘Ž=𝑧(0) is the initial value of 𝑧 and 𝑏>0 represents exponential growth.

Since the expression given has 𝑏=4.23, the quantity 𝑧 will indeed grow exponentially and π‘Ž=73, which represents the initial number of rabbits that Seif has.

As 𝑑=𝑛3, this tells us that the population will multiply by 4.23 every 3 months or, equivalently, by 4.23=1.6172β€¦οŽ οŽ’ every month.

Since 𝑛 represents the number of months, the number of rabbits after 2 months can be found by substituting 𝑛=2 into the expression: 𝑧=73β‹…(4.23)=190.9339….

We cannot have 0.9339 rabbits, so we would need to round this answer to the nearest integer below to 190 rabbits.

Therefore, the number of rabbits expected after 2 months would be 190 rabbits.

The exponential growth and decay equations can also be recast in an equivalent form in terms of Euler’s constant 𝑒=2.7182… as 𝑦(𝑑)=π‘Žβ‹…π‘=π‘Žβ‹…π‘’=π‘Žβ‹…π‘’,οοοŒ»ο‡οln where π‘Ž=𝑦(0) is the initial value and π‘˜=𝑏ln is the constant rate of growth when π‘˜>0, equivalent to 𝑏>1, or decay when π‘˜<0, equivalent to 0<𝑏<1.

For example, if there are an initial population of π‘Ž=200 bacteria and a growth constant of π‘˜=0.02, this can be described as 𝑃(𝑑)=200⋅𝑒.οŠ¦οŽ–οŠ¦οŠ¨ο

Time (Minutes)Population Size (Number of Bacteria)
0200
10244
20298
30364
40445
50544
60664

In many real-world situations, it is often convenient to replace the parameter 𝑏 with 1+𝑅, where 𝑅 tells us the proportion by which a quantity changes after each time period 𝑑. For example, if a quantity grows exponentially by 25%, then 𝑅=0.25, while a quantity decaying exponentially by 25% would have 𝑅=βˆ’0.25.

Definition: Exponential Constant Rate

The exponential equation 𝑦=π‘Žβ‹…π‘ο can also be expressed in terms of a constant change rate, 𝑅, with 𝑏=1+𝑅 and 𝑦(𝑑)=π‘Žβ‹…(1+𝑅).

The sign of 𝑅 is determined by whether the quantity 𝑦 is growing (𝑅>0) or decaying (𝑅<0) over time and it is most often represented as a decimal.

As an example, consider the decay function 𝑦=272994=27299β‹…ο€Ό14, where π‘Ž=27999 and 𝑏=14 in the general form. We can rewrite this in the form 𝑦=π‘Žβ‹…(1+𝑅) by finding the constant change rate 𝑅 as follows: 𝑅=π‘βˆ’1=14βˆ’1=βˆ’34<0.

Thus, the exponential equation can be written as 𝑦=27299β‹…ο€Ό1βˆ’34.

The constant change rate is 𝑅=βˆ’34=βˆ’0.75<0, which represents a decay rate of 75%, as a percentage, after every period 𝑑.

Let’s consider a real-world example where we will determine the percentage by which a drug concentration decreases every hour by finding the constant change rate 𝑅 from the given expression.

Example 3: Interpreting the Parameters in a Linear or Exponential Function in terms of a Context

The given figure shows the concentration 𝑐, in micrograms per litre, of a certain drug in human blood plasma measured at different times. Considering that the concentration after β„Ž hours can be modeled with the function 𝑐=18β‹…0.75, by what percentage does the drug’s concentration decrease every hour?

Answer

In this example, we will determine the percentage by which the drug’s concentration decreases every hour from the given expression modeling the concentration in a human blood plasma.

Recall that an exponential decrease (or decay) can be modeled by the function 𝑐=π‘Žβ‹…π‘,0<𝑏<1,=π‘Žβ‹…(1+𝑅),𝑅<0,forfor where 𝑏=1+𝑅 and 𝑅 is the constant change rate of the quantity 𝑐. For the drug concentration, we have π‘Ž=18 and 𝑏=0.75 and thus the constant change rate 𝑅=π‘βˆ’1=0.75βˆ’1=βˆ’0.25.

This means the drug’s concentration decreases at a rate of 0.25 per hour, which is equivalent to the percentage 25%.

Most of the time, we will be given or we will need to determine the rate as a percentage, according to which the quantity, 𝑦, grows or decays exponentially. For a quantity that grows exponentially, we can represent the change rate in terms of a growth rate as a percentage. Since 𝑅>0 for exponential growth, we have 𝑅=π‘Ÿ100, where π‘Ÿ% is the growth rate. The exponential growth equation can be expressed as 𝑦(𝑑)=π‘Žβ‹…ο€»1+π‘Ÿ100.

The growth rate π‘Ÿ can be written in terms of 𝑏 as π‘Ÿ=100⋅𝑅=100β‹…(π‘βˆ’1).

Similarly, for a quantity that decreases exponentially, we can represent the change rate in terms of a decay rate as a percentage. Since 𝑅<0 for exponential decay, we have 𝑅=βˆ’π‘Ÿ100, where π‘Ÿ% is the decay rate. The exponential decay equation can be expressed as 𝑦(𝑑)=π‘Žβ‹…ο€»1βˆ’π‘Ÿ100.

The decay rate π‘Ÿ can be written in terms of 𝑏 as π‘Ÿ=βˆ’100⋅𝑅=βˆ’100β‹…(π‘βˆ’1)=100β‹…(1βˆ’π‘).

For example, if the equation 𝑦=π‘Žβ‹…π‘ο represents a π‘Ÿ=15% increase in each period of time 𝑑, the value of 𝑏=1+𝑅=1+0.15=1.15.

Let’s consider this with a real-world example. Suppose the property market is going down in a city and the expected properties’ value after π‘š months is given by the expression 𝑃=𝑃⋅0.87,οŠ¦ο‰ where π‘ƒοŠ¦ is the initial value. The percentage reduction in the property value per month would be π‘Ÿ=100β‹…(1βˆ’π‘)=100β‹…(1βˆ’0.87)=100β‹…0.13=13%.

Now, let’s look at at a few examples to practice and deepen our understanding of exponential growth and decay equations and interpret their solutions.

In the first example, we will interpret the parameters in a given exponential function and determine the decay rate π‘Ÿ as a percentage.

Example 4: Interpreting the Parameters in a Linear or an Exponential Function in terms of a Given Context

The Asian elephant population 𝑑 years after the yearΒ 1900 is given by 𝑃=100000β‹…0.25ο‘‰οŽ οŽŸοŽŸ.

  1. What was the Asian elephant population in 1900?
  2. According to this model, by what percentage has the Asian elephant population decreased over a century?

Answer

Part 1

In this example, we will interpret the parameters in an exponential function and determine the Asian elephant population at a particular time.

An elephant herd, led by a Magnificent bull in the National Park

Recall that an exponential decay can be represented by the function 𝑃(𝑑)=π‘Žβ‹…π‘0<𝑏<1,for where π‘Ž=𝑃(0) is the initial value of 𝑃 and 0<𝑏<1 represents exponential decay.

Since the expression given has 𝑏=0.25, the quantity 𝑃, the Asian elephant population, will indeed decrease exponentially and π‘Ž=100000 is the initial number of elephants at the start, which is during the yearΒ 1900. We can also substitute 𝑑=0 into the expression given to obtain the same value.

Thus, the Asian elephant population in 1900 was 100β€Žβ€‰β€Ž000.

Part 2

Now, we will determine the constant change rate and then the decay rate as a percentage.

The population of the elephants can also be expressed as 𝑃(𝑑)=100000β‹…(1+𝑅), where 𝑅<0 is the decay rate of the quantity 𝑦, which we can determine from 1+𝑅=0.25 as 𝑅=0.25βˆ’1=βˆ’0.75.

This means that the elephant population decreases at a rate of 0.75 every year (after 1900), which is equivalent to a 75% reduction every year. We can also see this by using a decay rate π‘Ÿ given as a percentage.

Since the constant change rate 𝑅<0, as expected for exponential decay, and the quantity 𝑃 decreases, we can represent this constant change rate as a reduction rate in terms of a percentage by 𝑅=βˆ’π‘Ÿ100, where π‘Ÿ% is the reduction rate. In our case, the percentage is given by π‘Ÿ=βˆ’100⋅𝑅=βˆ’100β‹…βˆ’0.75=75%.

Thus, as a percentage, the elephant population has decreased by 75%.

Now, let’s look at an example where we have to create an exponential decay equation for a cereal manufacturer reducing the amount of sugar in their products. We will also determine an equation that can be used to determine the rate π‘Ÿ.

Example 5: Creating Exponential Equations with One Variable to Solve Problems

A cereal manufacturer decides to make their products healthier by reducing the amount of sugar in them. Their target is to reduce the amount of sugar in their product range by 20%. They plan to achieve their target in 4 years.

Write an equation they could use to find π‘Ÿ, the annual sugar reduction rate required to achieve their target.

Answer

In this example, we want to determine the exponential equation for a cereal manufacturer reducing the amount of sugar in their products with a particular target. We will determine an equation they could use to find π‘Ÿ, the annual sugar reduction rate required to achieve their target.

Recall that an exponential decrease (or decay) can be represented by the function 𝑦(𝑑)=π‘Žβ‹…π‘0<𝑏<1,=π‘Žβ‹…(1+𝑅)𝑅<0,forfor where π‘Ž=𝑦(0) is the initial amount of the quantity 𝑦, 𝑏=1+𝑅, and 𝑅<0 is the constant change rate of the quantity 𝑦. Since we know that the quantity decreases, we can represent this decay rate as a percentage: 𝑅=βˆ’π‘Ÿ100, where π‘Ÿ% is the reduction rate and the function can be written as 𝑦(𝑑)=π‘Žβ‹…ο€»1βˆ’π‘Ÿ100.

We are told that the cereal manufacturer wants to reduce the amount of sugar in their product range by 20% in 4 years; in other words, the amount of sugar will be 80% of the initial amount π‘Ž. This means we have the condition 𝑦(4)=0.8β‹…π‘Ž.

We can also substitute 𝑑=4 into the function above to obtain 𝑦(4)=π‘Žβ‹…ο€»1βˆ’π‘Ÿ100.οŠͺ

Thus, in order to determine the reduction rate π‘Ÿ, we have to solve π‘Žβ‹…ο€»1βˆ’π‘Ÿ100=0.8β‹…π‘Ž,ο€Ό100βˆ’π‘Ÿ100=0.8.οŠͺοŠͺ

In the next example, we will derive an exponential equation to model the reduction in the value of a car as a function of time. We will also determine the decay rate π‘Ÿ to the nearest whole number, by using the information that the car’s value will be halved in a given period.

Example 6: Writing and Evaluating Exponential Functions to Model Exponential Decay in a Real-World Context

A car’s value depreciates by π‘Ÿ% every year. A new car costs 𝑃dollars.

  1. Write a function that can be used to calculate 𝑉, the car’s value in dollars, after 𝑑 years.
  2. What is the value of π‘Ÿ for which the car’s value will be halved in 3 years? Give your answer to the nearest whole number.

Answer

Part 1

In this example, we will determine the exponential function to describe a value of a car after 𝑑 years, whose value depreciates every year.

Recall that an exponential decrease (or decay) can be represented by the function 𝑉(𝑑)=π‘Žβ‹…π‘0<𝑏<1=π‘Žβ‹…(1+𝑅)𝑅<0,forfor where π‘Ž=𝑉(0) is the initial amount of the quantity 𝑉, 𝑏=1+𝑅, and 𝑅<0 is the constant change rate of the quantity 𝑉.

Since the car’s value 𝑉 depreciates π‘Ÿ% every year, we can represent the decay rate 𝑅 as a percentage in terms of π‘Ÿ: 𝑅=βˆ’π‘Ÿ100.

A new car costs 𝑃dollars, which is the initial value of 𝑉, and hence π‘Ž=𝑃. Thus, the function that can be used to calculate the car’s value in dollars, after 𝑑 years, is given by 𝑉(𝑑)=𝑃1βˆ’π‘Ÿ100.

Part 2

Now, let’s determine the constant rate change π‘Ÿ%, by using the fact that the value of the car will be halved in 3 years.

If the value of the car is halved after 3 years, we have 𝑉(3)=12𝑃 and substituting 𝑑=3 into the function, we find 𝑉(3)=𝑃1βˆ’π‘Ÿ100.

In order to determine the percentage π‘Ÿ, we have to solve 𝑃1βˆ’π‘Ÿ100=12𝑃,ο€»1βˆ’π‘Ÿ100=12,1βˆ’π‘Ÿ100=ο„ž12,π‘Ÿ100=1βˆ’ο„ž12,π‘Ÿ=100β‹…ο€Ώ1βˆ’ο„ž12.

Thus, we have π‘Ÿ=20.6299…, and to the nearest whole number the value of π‘Ÿ is 21%.

Now, let’s consider an example where we have to determine the exponential equation for a ball which loses kinetic energy every time it bounces. By using the information given, we will also determine the height, to the nearest centimetre, that the ball must be dropped from in order to rebound to a given height on a particular bounce.

Example 7: Creating Exponential Equations with One Variable to Solve Problems

A child’s ball loses 15% of its energy every time it rebounds. By considering that the ball’s kinetic energy is proportional to the height from which it was dropped, determine the height, to the nearest centimetre, that the ball must be dropped from so that it rebounds to 20 cm on the fifth bounce.

Answer

In this example, we will determine the height that a ball must be dropped from in order to rebound to another height of 20 cm on the fifth bounce by using the information given in the question.

Recall that an exponential decrease (or decay) can be represented by the function 𝐾(𝑛)=π‘Žβ‹…π‘0<𝑏<1=π‘Žβ‹…(1+𝑅)𝑅<0,forfor where π‘Ž=𝐾(0) is the initial amount of the quantity 𝐾, 𝑏=1+𝑅, and 𝑅<0 is the constant change rate of the quantity 𝐾. Since we know that the quantity decreases, we can represent this decay rate as a percentage: 𝑅=βˆ’π‘Ÿ100, where π‘Ÿ% is the reduction rate, and the function can be written as 𝐾(𝑛)=π‘Žβ‹…ο€»1βˆ’π‘Ÿ100.

Since the child’s ball loses π‘Ÿ=15% of its energy every time it rebounds, 𝐾(𝑛)=π‘Žβ‹…ο€Ό1βˆ’15100=π‘Žβ‹…(1βˆ’0.15)=π‘Žβ‹…0.85.

Since we are told the kinetic energy π‘Ž=𝐾(0) of the ball is proportional to the height β„Ž from which it was dropped, we have 𝐾(𝑛)=πœ†β„Ž(𝑛), for some constant (of proportionality) πœ†βˆˆβ„οŠ°, since the height and kinetic energy are always positive, and β„Ž(𝑛) is the height to which the ball rebounds after the 𝑛th bounce. The initial kinetic energy can also be written as π‘Ž=𝐾(0)=πœ†β„Ž(0), where β„Ž(0) is the initial height from which it was dropped.

The energy after 𝑛 bounces can be written in terms of the initial height β„Ž(0) as 𝐾(𝑛)=πœ†β„Ž(0)β‹…0.85.

By using the expression for 𝐾(𝑛) in terms of β„Ž(𝑛), this can be written as πœ†β„Ž(𝑛)=πœ†β„Ž(0)β‹…0.85β„Ž(𝑛)=β„Ž(0)β‹…0.85.

Now, we are told that the ball rebounds to 20 cm on the fifth bounce; hence we have β„Ž(5)=20. We can substitute 𝑛=5 to obtain the expression β„Ž(0)β‹…0.85=20,β„Ž(0)=200.85.

Thus, the initial height is β„Ž(0)=45.0749 which to the nearest centimetre is 45 cm.

So far, we have considered exponential functions of the form 𝑦(𝑑)=π‘Žβ‹…π‘=π‘Žβ‹…π‘’, where π‘˜=𝑏ln. We can also transform the graphs of these functions by translating them by a constant 𝑐 as 𝑦(𝑑)=𝑐+π‘Žβ‹…π‘’.

For exponential decay (π‘˜<0), as 𝑑 becomes arbitrary large, the function 𝑦(𝑑) will approach the constant 𝑐, since the second term will decay to zero. This means after a long period of time, the quantity 𝑦 increases to a fixed limit 𝑐.

Finally, let’s consider an example where we look at this transformed exponential function and determine the function for given initial and limiting population values.

Example 8: Transforming Graphs of Exponential Functions

A population that grows over time 𝑑 to a fixed positive limit may be modeled by a transformed exponential function 𝑃(𝑑)=𝑐+π‘Žβ‹…π‘’οŒ»ο for suitable π‘Ž, 𝑏, and 𝑐. The following is a graph of this.

If the initial and limiting populations are 𝐴 and 𝐿, respectively, which would be a suitable function?

  1. 𝐿+(𝐿+𝐴)𝑒()
  2. πΏβˆ’(πΏβˆ’π΄)𝑒()
  3. 𝐿+(πΏβˆ’π΄)𝑒()
  4. πΏβˆ’(πΏβˆ’π΄)𝑒()
  5. πΏβˆ’(𝐿+𝐴)𝑒()

Answer

In this example, we will consider the transformed exponential function, describing a population that grows over time to a fixed positive limit, and determine suitable π‘Ž, 𝑏, and 𝑐, by using information about the initial and limiting population.

Recall that an exponential decay can be represented by the function 𝑃(𝑑)=π‘Žβ‹…π‘‘=π‘Žβ‹…π‘’=π‘Žβ‹…π‘’,ln where π‘Ž=𝑃(0) is the initial value of 𝑃 and 𝑑<1 represents decay. The modified exponential function 𝑃(𝑑)=𝑐+π‘Žβ‹…π‘’οŒ»ο has the same shape as the usual exponential function, translated by a constant 𝑐.

Since the initial population is given as 𝐴, we have 𝑃(0)=𝐴 and we can substitute 𝑑=0 into the modified function to obtain 𝑃(0)=𝑐+π‘Žβ‹…π‘’=𝑐+π‘Ž.

Thus, we have 𝐴=𝑐+π‘Ž. The limiting population is what happens after an arbitrarily large amount of time, or as π‘‘β†’βˆž, which is only well defined for the transformed exponential function for 𝑏<0, as if 𝑏>0, the function would diverge to infinity.

Taking this limit for 𝑏<0 for the modified function, we have 𝑐=𝐿 since the second term π‘Žβ‹…π‘’οŒ»ο will approach zero as 𝑑 becomes arbitrarily large (i.e., approaches infinity). Thus, with π‘Ž=π΄βˆ’π‘=π΄βˆ’πΏ, the modified function can be written as 𝑃(𝑑)=𝐿+(π΄βˆ’πΏ)𝑒=πΏβˆ’(πΏβˆ’π΄)𝑒.

This would work for any suitable 𝑏<0 and hence one particular choice can be 𝑏=βˆ’5 with πΏβˆ’(πΏβˆ’π΄)𝑒.()

This is option D.

So far, we have looked at problems where we determine the value of a quantity at a particular time, but we can also do the reverse. Using the equations of exponential growth and decay, we can determine the time at which a particular quantity will reach a specific value, which is useful for many real-world applications such as calculating the half-life in physics, in particular, for radioactive material, which can be described as 𝑁(𝑑)=𝑁⋅𝑒, where πœ† is a positive number called the decay constant of the decaying radioactive quantity.

The half-life, π‘‘οŽ οŽ‘, is the time required for exactly half of the initial quantity 𝑁≠0 to decay and thus satisfies 𝑁𝑑=12π‘οŽ οŽ‘οŠ¦.

We can determine the half-life by substituting 𝑑=π‘‘οŽ οŽ‘ into the equation for radioactive decay as follows: 𝑁𝑑=𝑁𝑒=12𝑁.

Since 𝑁≠0, we can divide by π‘οŠ¦ to obtain the expression 𝑒=12.

In order to solve this equation for π‘‘οŽ οŽ‘, we would need to apply the logarithm. However, this is beyond the scope of this explainer and will be covered elsewhere.

Key Points

  • The exponential function describing growth and decay can be represented as 𝑦(𝑑)=π‘Žβ‹…π‘,𝑏>0𝑏≠1,forrealconstantsand where 𝑏>1 represents exponential growth and 0<𝑏<1 represents exponential decay, or, equivalently, as 𝑦(𝑑)=π‘Žβ‹…π‘’,π‘˜π‘˜β‰ 0,forrealconstantsand with π‘˜=𝑏ln, where π‘˜>0 represents exponential growth and π‘˜<0 represents exponential decay.
  • The exponential function can also be expressed in terms of a constant change rate, 𝑅, with 𝑏=1+𝑅 and 𝑦(𝑑)=π‘Žβ‹…(1+𝑅). The sign of 𝑅 is determined by whether the quantity 𝑦 is growing (𝑅>0) or decaying (𝑅<0) over time and it is represented as a decimal.
  • We can also write the exponential growth function in terms of a constant growth rate π‘Ÿ% (as a percentage): 𝑦(𝑑)=π‘Žβ‹…ο€»1+π‘Ÿ100, where π‘Ÿ=100⋅𝑅=100β‹…(π‘βˆ’1). Similarly for an exponential decay function in terms of a constant decay rate π‘Ÿ%, 𝑦(𝑑)=π‘Žβ‹…ο€»1βˆ’π‘Ÿ100, where π‘Ÿ=βˆ’100⋅𝑅=100β‹…(1βˆ’π‘).

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