Lesson Explainer: Applications of Exponential Functions | Nagwa Lesson Explainer: Applications of Exponential Functions | Nagwa

Lesson Explainer: Applications of Exponential Functions Mathematics

In this explainer, we will learn how to solve real-world problems involving exponential functions.

Exponential functions appear in many different mathematical models representing real-world problems. In general, exponential models represent scenarios in the real world where the rate of change in a quantity is constant over a period of fixed length. Largely, we can categorize exponential models into two types: exponential growth and exponential decay. We will begin by looking at the growth models. Exponential growth models represent real-world scenarios where a quantity becomes larger with time. The simplest such models are doubling problems, where the quantity doubles in every fixed period.

Let us see how an exponential growth model rises from a real-world doubling problem in our first example.

Example 1: Writing and Solving Exponential Equations in a Real-World Context

A microorganism reproduces by binary fission, where every hour each cell divides into two cells. Given that there were 15‎ ‎141 cells to begin with, determine how many cells there were after 5 hours.

Answer

In this example, we need to find the number of cells after 5 hours. We know that the number of cells at the start was 15‎ ‎141. Since each cell divides into two cells every hour, the number of cells would double after one hour. This means that after 1 hour, we will have 15141×2=30282.cells

Then, after 2 hours, each of these cells would double, resulting in 30282×2=60564.cells

We can continue doubling the number of cells to obtain the number of cells in the next hour until we reach 5 hours: 60564×2=1211283,121128×2=2422564,242256×2=4845125.cellsafterhourscellsafterhourscellsafterhours

Hence, the number of cells in 5 hours is 484‎ ‎512.

In the previous example, we solved a real-world doubling problem by consecutively multiplying 2 by the previous quantity to obtain the next quantity. If we look at the solution closely, we can see a pattern rising. More specifically, we can see that the number of cells after 𝑛 hours is obtained by multiplying the number of cells at the start, that is, 15‎ ‎141, by a power of 2. Let us observe this pattern by listing out these numbers as powers of 2. 15141×20,15141×21,15141×22,15141×23,15141×24,15141×25.cellsafterhourscellsafterhourcellsafterhourscellsafterhourscellsafterhourscellsafterhours

From this list, we can clearly see that numberofcellsinhours𝑛=15141(2).

This is an exponential function where 𝑛 is the independent variable of the function representing the number of hours from the start. In the next example, we will find an exponential model from a real-world doubling problem.

Example 2: Creating Exponential Equations in Two Variables

A start-up company noticed that the number of those who use its product doubles every month. This month, they had 4‎ ‎000 users. Assuming this trend continues, write an equation that can be used to calculate 𝑈(𝑚), the number of users in 𝑚 months’ time.

Answer

In this example, we need to find a mathematical expression for 𝑈(𝑚), which is the number of users in 𝑚 months. Currently, we know that the start-up company has 4‎ ‎000 users. We are given that the number of users doubles each month. This means that the number of users would be 𝑈(1)=4000×2=8000.

Hence, in the next month, the number of users is expected to be 8‎ ‎000. We can continue by using the same pattern to say 𝑈(2)=4000×2=16000,𝑈(3)=4000×2=32000,𝑈(4)=4000×2=64000, and so on. In other words, we can find the number of users in a given month by multiplying the number of users in the previous month by 2. In doing so, we can see a pattern forming, where we multiply the number of users at the start, that is, 4‎ ‎000, by a power of 2 corresponding to the number of months. This leads to the conclusion that the number of users in 𝑚 months’ time is expected to be 𝑈(𝑚)=4000(2).

In our previous example, we found an exponential model for a real-world doubling problem. Since we consecutively multiply the quantity in the previous period by 2 to obtain the new quantity, it makes sense that the resulting expression is a power of 2, hence an exponential function of time. Thus, we can say that, in general, the pattern of doubling a quantity in each time period leads to an exponential function. But it does not have to be “doubling.” Following the same method, we can find exponential models if the quantity changes by any fixed ratio over a given period of time. In such cases, it would be tripling, quadrupling, or more, generally, changing by any positive ratio. Exponential functions capture the pattern of a quantity that changes by a fixed ratio over a given period of time. Allowing for arbitrary ratios enables us to include more sophisticated models such as unrestricted population size and bank savings account balance.

Let us formalize this idea to a greater generality.

Definition: Exponential Models

An exponential function can be written in the general form 𝑓(𝑥)=𝐴𝑏, for some constants 𝐴>0 and 𝑏>0, 𝑏1. In exponential models, we need to keep the following in mind:

  • The independent variable 𝑥 usually represents time, in which case we often use 𝑡 rather than 𝑥.
  • Since 𝑓(0)=𝐴𝑏=𝐴, quantity 𝐴 is the initial value of the function, that is, at 𝑥=0.
  • Base 𝑏 tells us about the “rate” at which our quantity is changing with time. Growth corresponds to a base 𝑏>1, while decay occurs when 0<𝑏<1.

The graphs of two particular exponential functions are given below.

In the previous example, we obtained the exponential model 𝑈=4000(2).

This is an exponential model with 𝐴=4000 and 𝑏=2. As expected, 𝐴=4000 agrees with the given number of users at the start. Also, the fact that 𝑏>1 agrees with the fact that this is a growth model. In particular, we can see from our previous two examples that doubling models always lead to exponential functions with 𝑏=2.

If we know that a real-world problem is modeled using an exponential function of the form 𝐴𝑏, we can find the exponential function by identifying the two constants 𝐴 and 𝑏. In the next example, we will find an exponential growth model for a given real-world population problem.

Example 3: Creating Exponential Equations and Using Them to Solve Problems

The US Census is taken every ten years. The population of Texas was 3.05 million in 1900 and 20.9 million in 2000. By modeling the population growth as exponential, answer the following questions.

  1. Write an exponential function in the form 𝑃(𝑑)=𝑃𝑘 to model the population of Texas, in millions, 𝑑 decades after 1900. Round your value of 𝑘 to three decimal places.
  2. According to the model, what was the population of Texas in 1950? Give your answer in millions to two decimal places.
  3. Using the value of 𝑘 from part 1, rewrite your function in the form 𝑃(𝑦)=𝑃(𝑏) where 𝑦 is the time in years after the year 1900. Round your value of 𝑏 to four decimal places.

Answer

Part 1

In this part, we need to find an exponential function of the form

𝑃(𝑑)=𝑃𝑘,(1)

where 𝑑 is the number of decades since 1900 and 𝑃(𝑑) is the population of Texas, in millions, 𝑑 decades since 1900. To find this exponential function, we need to identify the constants 𝑃 and 𝑘. Let us examine the information provided.

We are given that the population of Texas was 3.05 million in 1900. The year 1900 corresponds to 𝑑=0 since 1900 is 0 decades after 1900. This tells us 𝑃(0)=3.05.

If we substitute 𝑑=0 into (1), we obtain 𝑃(0)=𝑃𝑘=𝑃.

Since we know that 𝑃(0)=3.05, this leads to 𝑃=3.05.

Next, we need to find the value of 𝑘. We are also given that the population of Texas was 20.9 million in 2000. The year 2000 is 100 years after year 1900, which means that it is 10 decades since 1900. This tells us 𝑃(10)=20.9.

Substituting 𝑑=10 into (1), we obtain 𝑃(10)=𝑃𝑘=3.05𝑘.

This leads to 3.05𝑘=20.9.

Dividing both sides of this equation by 3.05, 𝑘=6.85245.

Taking the 10th root of both sides of this equation gives us 𝑘=±1.21222. 𝑘 is the base of an exponential function, and we know that the base of an exponential function must be positive. Hence, 𝑘1.212,.roundedtothreedecimalplaces

Substituting 𝑃=3.05 and 𝑘=1.212 into equation (1) gives us 𝑃(𝑑)=3.05(1.212).

Part 2

We need to find the population of Texas in 1950 according to the model we obtained in the previous part. We note that year 1950 is 50 years, or 5 decades, past 1900. So, we can find the population of Texas in 1950 according to our model by substituting 𝑑=5 into the equation above. This gives us 𝑃(5)=3.05(1.212)=7.97651.

Rounding to two decimal places and remembering that the unit of 𝑃(𝑑) is in millions, the population of Texas in 1950 according to our model is 7.98 million.

Part 3

In this part, we want to rewrite our exponential model for the population of Texas as

𝑃(𝑦)=𝑃(𝑏),(2)

where 𝑦 is the number of years, rather than decades, since year 1900 and 𝑃(𝑦) is the population of Texas 𝑦 years since 1900. Rather than starting over from scratch, we can begin with our model from part 1: We know that the population of Texas 𝑑 decades since 1900 is 3.05(1.212).million

Since 1 year equals 110 decades, 𝑦 years equals 𝑦10 decades. In other words, 𝑦 years since 1900 is the same as 𝑦10 decades since 1900. Hence, substituting 𝑑=𝑦10 into the expression above, we obtain 𝑃(𝑦)=3.05(1.212).

Since this is not in the form given in equation (2), let us apply rules of exponents to simplify this expression. Recall that for any base 𝑏 and exponents 𝑥 and 𝑦, 𝑏=(𝑏).

Using this rule and noting 𝑦10=110×𝑦, we can write 3.05(1.212)=3.05(1.212)=3.051.212=3.05(1.01941).×

This tells us 𝑏1.0194 rounded to four decimal places. Hence, 𝑃(𝑦)=3.05(1.0194).

So far, we have considered a few real-world examples of exponential growth. Let us now turn our attention to exponential decay models. As expected from the term “decay,” exponential decay models represent real-world scenarios where a quantity becomes smaller with time. As we motivated exponential growth models with real-world doubling problems, we can understand exponential decay models by considering halving problems. This means that instead of doubling a quantity after each fixed period, the quantity would reduce to half of the current amount after a fixed period.

The most prominent example of a halving problem concerns models for radioactive decay. A radioactive element, such as uranium or carbon-14, is an element that is unstable in nature, which means that the element will split into smaller, more stable elements over time. For each radioactive element, there is an associated half-life. The half-life of a radioactive element is the amount of time it takes for the quantity of the element to reduce to half of the current value. For instance, the half-life of carbon-14 is 5‎ ‎730 years, which means that the quantity of carbon-14 will reduce to half of the current value after this number of years. We can use this information to find an exponential decay model for the amount or concentration of carbon-14 in an object, which then can be used to approximate the age of the object by comparing the present and initial concentrations of carbon-14. This process is known as carbon dating.

In the next question, we will consider a problem concerning radioactive decay.

Example 4: Converting between Different Forms of Exponential Expressions

Radioactive element 𝐽 has a half-life of 1 week. If an experiment starts with 20 g of element 𝐽, the mass 𝑀, in grams, of element 𝐽 remaining after 𝑤 weeks can be found using the equation 𝑀=2012. Write an equation to find the mass of element 𝐽 remaining after 𝑑 days.

Answer

In this example, we are given that the mass 𝑀 of radioactive element 𝐽 is given by an exponential function 𝑀=2012, where 𝑤 is the time from the start of the experiment, measured in weeks. We need to express 𝑀 in terms of 𝑑, which is the time measured in days. Since 1 day equals 17 weeks, 𝑑 days equals 𝑑7 weeks. Hence, substituting 𝑤=𝑑7 into the expression above, we obtain 𝑀=2012.

Recall that an exponential function is generally in the form 𝐴𝑏 for some constants 𝐴 and 𝑏 satisfying 𝐴0, 𝑏>0, and 𝑏1. The expression of 𝑀 we have obtained above is not in the same form since the exponent is 𝑑7, where we want this to be 𝑑. We can apply the power law which states that 𝑎=(𝑎) to write 𝑀 as 𝑀=2012.

In the previous example, we saw a halving problem concerning a radioactive element. Like exponential growth models, decays do not have to be decreasing in halves. Any base 𝑏 of an exponential function with 0<𝑏<1 creates an exponential decay model. In the next example, we will consider an exponential decay model and determine the rate of decrease.

Example 5: Evaluating Functions Involving Exponential Decay

A population of bacteria decreases as a result of a chemical treatment. The population 𝑡 hours after the treatment was applied can be modeled by the function 𝑃(𝑡), where 𝑃(𝑡)=6000×(0.4).

  1. What was the population when the chemical was first applied?
  2. What is the rate of population decrease?

Answer

Part 1

In this part, we need to find the initial population, where the population is given as an exponential function. We recall that in an exponential function of the form 𝑓(𝑥)=𝐴𝑏, the positive constant 𝐴 represents the initial amount. Hence, we can see the population when the chemical was first applied from the given exponential function to be 6‎ ‎000.

We can also obtain this answer directly by substituting 𝑡=0 since this value of 𝑡 corresponds to the time when the chemical was first applied. This leads to 𝑃(0)=6000×(0.4)=6000.

Part 2

In this part, we need to find the rate of population decrease. We recall that an exponential function 𝑓(𝑥)=𝐴𝑏 with 0<𝑏<1 represents a quantity that decreases by a fixed ratio over a fixed period. Since 𝑡 is measured in the number of hours after the treatment was applied, the rate of population decrease is given by the percentage of population of bacteria decreased over a one-hour period. Using the given model 𝑃(𝑡)=6000×(0.4), we can see 𝑃(1)=6000×0.4, which tells us that the population of the bacteria is 40% of the initial population. We can also see that 𝑃(2)=6000×(0.4)=𝑃(1)×0.4,𝑃(3)=6000×(0.4)=𝑃(2)×0.4,𝑃(4)=6000×(0.4)=𝑃(3)×0.4, and so on. In each case, the population after 𝑡 hours is given by 40% of the population of the bacteria in the previous hour. In other words, the rate of population decrease is given by 100%40%=60%.perhour

In the previous example, we considered an exponential decay model to identify the rate of decrease. The rate of decrease is the proportion of the quantity that decreases over a fixed period. If we denote this rate by 𝑑, we can obtain the amount in the next period by multiplying the amount in the previous period by 1𝑑.

Definition: Rate of Decay in Exponential Decay Models

If the rate of decrease of a quantity over a given time unit is 𝑑 for 0<𝑑<1 and the initial amount of the quantity is 𝐴, then the exponential decay model is given by 𝑓(𝑡)=𝐴(1𝑑).

It should be noted here that 𝑑 is measured in proportions, not in percentages. In part 2 of the previous example, we concluded that the rate of population decrease is given by 60%. To obtain 𝑑 in this case, we would need to convert the percentage into proportion as follows: 𝑑=60×1100=0.6.

Also, we concluded in part 1 of this example that the initial population is 6‎ ‎000, which means that 𝐴=6000. Applying the definition above leads to the exponential decay model 𝑃(𝑡)=6000×(10.6)=6000×0.4.

We can see that this is the same as the exponential model provided in the problem.

In our final example, we will find an exponential decay model from a real-world problem.

Example 6: Creating Exponential Equations in Two Variables

The number of people visiting a museum is decreasing by 3% a year. This year, there were 50‎ ‎000 visitors. Assuming the decline continues, write an equation that can be used to find 𝑉, the number of visitors there will be in 𝑡 years’ time.

Answer

We are given that the number of visitors is decreasing by 3% a year and also that there were 50‎ ‎000 visitors this year. We need to find the number of visitors after 𝑡 years. Since the number of visitors decrease by 3% per year, the number of visitors next year would be 97% of that of this year. This means 𝑉=50000×0.97𝑡=1.for

The number of visitors in two years will be 97% of this number, so we can find this number by multiplying 0.97 by the number above. Continuing this this pattern, we obtain 𝑉=50000×0.97𝑡=2,𝑉=50000×0.97𝑡=3,𝑉=50000×0.97𝑡=4,forforfor and so on. We can see a pattern rising here where the number of visitors after 𝑡 years is obtained by multiplying 50‎ ‎000 by 0.97. Hence, the number of visitors in 𝑡 years’ time is 𝑉=50000(0.97).

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • Exponential models represent scenarios in the real world where the rate of change in a quantity is constant over a period of fixed length. If the quantity is increasing, this is called an exponential growth model. If the quantity is decreasing, it is called an exponential decay model.
  • An exponential function can be written in the general form 𝑓(𝑥)=𝐴𝑏, for some constants 𝐴>0 and 𝑏>0, 𝑏1. In exponential models, we need to keep in mind the following.
    • The independent variable 𝑥 usually represents time, in which case we often use 𝑡 rather than 𝑥.
    • Since 𝑓(0)=𝐴𝑏=𝐴, quantity 𝐴 is the initial value of the function, that is, at 𝑥=0.
    • Base 𝑏 tells us about the “rate” at which our quantity is changing with time. Growth corresponds to a base 𝑏>1, while decay occurs when 0<𝑏<1.
  • If the rate of decrease of a quantity over a given time unit is 𝑑 for 0<𝑑<1 and the initial amount of the quantity is 𝐴, then the exponential decay model is given by 𝑓(𝑡)=𝐴(1𝑑).

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