exponential decay problem solving

6.8 Exponential Growth and Decay

Learning objectives.

• 6.8.1 Use the exponential growth model in applications, including population growth and compound interest.
• 6.8.2 Explain the concept of doubling time.
• 6.8.3 Use the exponential decay model in applications, including radioactive decay and Newton’s law of cooling.
• 6.8.4 Explain the concept of half-life.

One of the most prevalent applications of exponential functions involves growth and decay models. Exponential growth and decay show up in a host of natural applications. From population growth and continuously compounded interest to radioactive decay and Newton’s law of cooling, exponential functions are ubiquitous in nature. In this section, we examine exponential growth and decay in the context of some of these applications.

Exponential Growth Model

Many systems exhibit exponential growth. These systems follow a model of the form y = y 0 e k t , y = y 0 e k t , where y 0 y 0 represents the initial state of the system and k k is a positive constant, called the growth constant . Notice that in an exponential growth model, we have

That is, the rate of growth is proportional to the current function value. This is a key feature of exponential growth. Equation 6.27 involves derivatives and is called a differential equation. We learn more about differential equations in Introduction to Differential Equations .

Rule: Exponential Growth Model

Systems that exhibit exponential growth increase according to the mathematical model

where y 0 y 0 represents the initial state of the system and k > 0 k > 0 is a constant, called the growth constant .

Population growth is a common example of exponential growth. Consider a population of bacteria, for instance. It seems plausible that the rate of population growth would be proportional to the size of the population. After all, the more bacteria there are to reproduce, the faster the population grows. Figure 6.79 and Table 6.1 represent the growth of a population of bacteria with an initial population of 200 200 bacteria and a growth constant of 0.02 . 0.02 . Notice that after only 2 2 hours ( 120 ( 120 minutes), the population is 10 10 times its original size!

Note that we are using a continuous function to model what is inherently discrete behavior. At any given time, the real-world population contains a whole number of bacteria, although the model takes on noninteger values. When using exponential growth models, we must always be careful to interpret the function values in the context of the phenomenon we are modeling.

Example 6.42

Population growth.

Consider the population of bacteria described earlier. This population grows according to the function f ( t ) = 200 e 0.02 t , f ( t ) = 200 e 0.02 t , where t is measured in minutes. How many bacteria are present in the population after 5 5 hours ( 300 ( 300 minutes)? When does the population reach 100,000 100,000 bacteria?

We have f ( t ) = 200 e 0.02 t . f ( t ) = 200 e 0.02 t . Then

There are 80,686 80,686 bacteria in the population after 5 5 hours.

To find when the population reaches 100,000 100,000 bacteria, we solve the equation

The population reaches 100,000 100,000 bacteria after 310.73 310.73 minutes.

Checkpoint 6.42

Consider a population of bacteria that grows according to the function f ( t ) = 500 e 0.05 t , f ( t ) = 500 e 0.05 t , where t t is measured in minutes. How many bacteria are present in the population after 4 hours? When does the population reach 100 100 million bacteria?

Let’s now turn our attention to a financial application: compound interest . Interest that is not compounded is called simple interest . Simple interest is paid once, at the end of the specified time period (usually 1 1 year). So, if we put $ 1000 $ 1000 in a savings account earning 2 % 2 % simple interest per year, then at the end of the year we have

Compound interest is paid multiple times per year, depending on the compounding period. Therefore, if the bank compounds the interest every 6 6 months, it credits half of the year’s interest to the account after 6 6 months. During the second half of the year, the account earns interest not only on the initial $ 1000 , $ 1000 , but also on the interest earned during the first half of the year. Mathematically speaking, at the end of the year, we have

Similarly, if the interest is compounded every 4 4 months, we have

and if the interest is compounded daily ( 365 ( 365 times per year), we have $ 1020.20 . $ 1020.20 . If we extend this concept, so that the interest is compounded continuously, after t t years we have

Now let’s manipulate this expression so that we have an exponential growth function. Recall that the number e e can be expressed as a limit:

Based on this, we want the expression inside the parentheses to have the form ( 1 + 1 / m ) . ( 1 + 1 / m ) . Let n = 0.02 m . n = 0.02 m . Note that as n → ∞ , n → ∞ , m → ∞ m → ∞ as well. Then we get

We recognize the limit inside the brackets as the number e . e . So, the balance in our bank account after t t years is given by 1000 e 0.02 t . 1000 e 0.02 t . Generalizing this concept, we see that if a bank account with an initial balance of $ P $ P earns interest at a rate of r % , r % , compounded continuously, then the balance of the account after t t years is

Example 6.43

Compound interest.

A 25-year-old student is offered an opportunity to invest some money in a retirement account that pays 5 % 5 % annual interest compounded continuously. How much does the student need to invest today to have $ 1 $ 1 million when she retires at age 65 ? 65 ? What if she could earn 6 % 6 % annual interest compounded continuously instead?

She must invest $ 135,335.28 $ 135,335.28 at 5 % 5 % interest.

If, instead, she is able to earn 6 % , 6 % , then the equation becomes

In this case, she needs to invest only $ 90,717.95 . $ 90,717.95 . This is roughly two-thirds the amount she needs to invest at 5 % . 5 % . The fact that the interest is compounded continuously greatly magnifies the effect of the 1 % 1 % increase in interest rate.

Checkpoint 6.43

Suppose instead of investing at age 25 25 , the student waits until age 35 . 35 . How much would she have to invest at 5 % ? 5 % ? At 6 % ? 6 % ?

If a quantity grows exponentially, the time it takes for the quantity to double remains constant. In other words, it takes the same amount of time for a population of bacteria to grow from 100 100 to 200 200 bacteria as it does to grow from 10,000 10,000 to 20,000 20,000 bacteria. This time is called the doubling time. To calculate the doubling time, we want to know when the quantity reaches twice its original size. So we have

If a quantity grows exponentially, the doubling time is the amount of time it takes the quantity to double. It is given by

Example 6.44

Using the doubling time.

Assume a population of fish grows exponentially. A pond is stocked initially with 500 500 fish. After 6 6 months, there are 1000 1000 fish in the pond. The owner will allow his friends and neighbors to fish on his pond after the fish population reaches 10,000 . 10,000 . When will the owner’s friends be allowed to fish?

We know it takes the population of fish 6 6 months to double in size. So, if t represents time in months, by the doubling-time formula, we have 6 = ( ln 2 ) / k . 6 = ( ln 2 ) / k . Then, k = ( ln 2 ) / 6 . k = ( ln 2 ) / 6 . Thus, the population is given by y = 500 e ( ( ln 2 ) / 6 ) t . y = 500 e ( ( ln 2 ) / 6 ) t . To figure out when the population reaches 10,000 10,000 fish, we must solve the following equation:

The owner’s friends have to wait 25.93 25.93 months (a little more than 2 2 years) to fish in the pond.

Checkpoint 6.44

Suppose it takes 9 9 months for the fish population in Example 6.44 to reach 1000 1000 fish. Under these circumstances, how long do the owner’s friends have to wait?

Exponential Decay Model

Exponential functions can also be used to model populations that shrink (from disease, for example), or chemical compounds that break down over time. We say that such systems exhibit exponential decay, rather than exponential growth. The model is nearly the same, except there is a negative sign in the exponent. Thus, for some positive constant k , k , we have y = y 0 e − k t . y = y 0 e − k t .

As with exponential growth, there is a differential equation associated with exponential decay. We have

Rule: Exponential Decay Model

Systems that exhibit exponential decay behave according to the model

where y 0 y 0 represents the initial state of the system and k > 0 k > 0 is a constant, called the decay constant .

The following figure shows a graph of a representative exponential decay function.

Let’s look at a physical application of exponential decay. Newton’s law of cooling says that an object cools at a rate proportional to the difference between the temperature of the object and the temperature of the surroundings. In other words, if T T represents the temperature of the object and T a T a represents the ambient temperature in a room, then

Note that this is not quite the right model for exponential decay. We want the derivative to be proportional to the function, and this expression has the additional T a T a term. Fortunately, we can make a change of variables that resolves this issue. Let y ( t ) = T ( t ) − T a . y ( t ) = T ( t ) − T a . Then y ′ ( t ) = T ′ ( t ) − 0 = T ′ ( t ) , y ′ ( t ) = T ′ ( t ) − 0 = T ′ ( t ) , and our equation becomes

From our previous work, we know this relationship between y and its derivative leads to exponential decay. Thus,

and we see that

where T 0 T 0 represents the initial temperature. Let’s apply this formula in the following example.

Example 6.45

Newton’s law of cooling.

According to experienced baristas, the optimal temperature to serve coffee is between 155 ° F 155 ° F and 175 ° F . 175 ° F . Suppose coffee is poured at a temperature of 200 ° F , 200 ° F , and after 2 2 minutes in a 70 ° F 70 ° F room it has cooled to 180 ° F . 180 ° F . When is the coffee first cool enough to serve? When is the coffee too cold to serve? Round answers to the nearest half minute.

Then, the model is

The coffee reaches 175 ° F 175 ° F when

The coffee can be served about 2.5 2.5 minutes after it is poured. The coffee reaches 155 ° F 155 ° F at

The coffee is too cold to be served about 5 5 minutes after it is poured.

Checkpoint 6.45

Suppose the room is warmer ( 75 ° F ) ( 75 ° F ) and, after 2 2 minutes, the coffee has cooled only to 185 ° F . 185 ° F . When is the coffee first cool enough to serve? When is the coffee be too cold to serve? Round answers to the nearest half minute.

Just as systems exhibiting exponential growth have a constant doubling time, systems exhibiting exponential decay have a constant half-life. To calculate the half-life, we want to know when the quantity reaches half its original size. Therefore, we have

Note : This is the same expression we came up with for doubling time.

If a quantity decays exponentially, the half-life is the amount of time it takes the quantity to be reduced by half. It is given by

Example 6.46

Radiocarbon dating.

One of the most common applications of an exponential decay model is carbon dating . Carbon- 14 Carbon- 14 decays (emits a radioactive particle) at a regular and consistent exponential rate. Therefore, if we know how much carbon was originally present in an object and how much carbon remains, we can determine the age of the object. The half-life of carbon- 14 carbon- 14 is approximately 5730 5730 years—meaning, after that many years, half the material has converted from the original carbon- 14 carbon- 14 to the new nonradioactive nitrogen- 14 . nitrogen- 14 . If we have 100 100 g carbon- 14 carbon- 14 today, how much is left in 50 50 years? If an artifact that originally contained 100 100 g of carbon now contains 10 10 g of carbon, how old is it? Round the answer to the nearest hundred years.

So, the model says

In 50 50 years, we have

Therefore, in 50 50 years, 99.40 99.40 g of carbon- 14 carbon- 14 remains.

To determine the age of the artifact, we must solve

The artifact is about 19,000 19,000 years old.

Checkpoint 6.46

If we have 100 100 g of carbon- 14 , carbon- 14 , how much is left after 500 500 years? If an artifact that originally contained 100 100 g of carbon now contains 20 g 20 g of carbon, how old is it? Round the answer to the nearest hundred years.

Section 6.8 Exercises

True or False ? If true, prove it. If false, find the true answer.

The doubling time for y = e c t y = e c t is ( ln ( 2 ) ) / ( ln ( c ) ) . ( ln ( 2 ) ) / ( ln ( c ) ) .

If you invest $ 500 , $ 500 , an annual rate of interest of 3 % 3 % yields more money in the first year than a 2.5 % 2.5 % continuous rate of interest.

If you leave a 100 ° C 100 ° C pot of tea at room temperature ( 25 ° C ) ( 25 ° C ) and an identical pot in the refrigerator ( 5 ° C ) , ( 5 ° C ) , with k = 0.02 , k = 0.02 , the tea in the refrigerator reaches a drinkable temperature ( 70 ° C ) ( 70 ° C ) more than 5 5 minutes before the tea at room temperature.

If given a half-life of t years, the constant k k for y = e k t y = e k t is calculated by k = ln ( 1 / 2 ) / t . k = ln ( 1 / 2 ) / t .

For the following exercises, use y = y 0 e k t . y = y 0 e k t .

If a culture of bacteria doubles in 3 3 hours, how many hours does it take to multiply by 10 ? 10 ?

If bacteria increase by a factor of 10 10 in 10 10 hours, how many hours does it take to increase by 100 ? 100 ?

How old is a skull that contains one-fifth as much radiocarbon as a modern skull? Note that the half-life of radiocarbon is 5730 5730 years.

If a relic contains 90 % 90 % as much radiocarbon as new material, can it have come from the time of Christ (approximately 2000 2000 years ago)? Note that the half-life of radiocarbon is 5730 5730 years.

The population of Cairo grew from 5 5 million to 10 10 million in 20 20 years. Use an exponential model to find when the population was 8 8 million.

The populations of New York and Los Angeles are growing at 1 % 1 % and 1.4 % 1.4 % a year, respectively. Starting from 8 8 million (New York) and 6 6 million (Los Angeles), when are the populations equal?

Suppose the value of $ 1 $ 1 in Japanese yen decreases at 2 % 2 % per year. Starting from $ 1 = ¥ 250 , $ 1 = ¥ 250 , when will $ 1 = ¥ 1 ? $ 1 = ¥ 1 ?

The effect of advertising decays exponentially. If 40 % 40 % of the population remembers a new product after 3 3 days, how long will 20 % 20 % remember it?

If y = 1000 y = 1000 at t = 3 t = 3 and y = 3000 y = 3000 at t = 4 , t = 4 , what was y 0 y 0 at t = 0 ? t = 0 ?

If y = 100 y = 100 at t = 4 t = 4 and y = 10 y = 10 at t = 8 , t = 8 , when does y = 1 ? y = 1 ?

If a bank offers annual interest of 7.5 % 7.5 % or continuous interest of 7.25 % , 7.25 % , which has a better annual yield?

What continuous interest rate has the same yield as an annual rate of 9 % ? 9 % ?

If you deposit $ 5000 $ 5000 at 8 % 8 % annual interest, how many years can you withdraw $ 500 $ 500 (starting after the first year) without running out of money?

You are trying to save $ 50,000 $ 50,000 in 20 20 years for college tuition for your child. If interest is a continuous 10 % , 10 % , how much do you need to invest initially?

You are cooling a turkey that was taken out of the oven with an internal temperature of 165 ° F . 165 ° F . After 10 10 minutes of resting the turkey in a 70 ° F 70 ° F apartment, the temperature has reached 155 ° F . 155 ° F . What is the temperature of the turkey 20 20 minutes after taking it out of the oven?

You are trying to thaw some vegetables that are at a temperature of 1 ° F . 1 ° F . To thaw vegetables safely, you must put them in the refrigerator, which has an ambient temperature of 44 ° F . 44 ° F . You check on your vegetables 2 2 hours after putting them in the refrigerator to find that they are now 12 ° F . 12 ° F . Plot the resulting temperature curve and use it to determine when the vegetables reach 33 ° F . 33 ° F .

You are an archaeologist and are given a bone that is claimed to be from a Tyrannosaurus Rex. You know these dinosaurs lived during the Cretaceous Era ( 146 ( 146 million years to 65 65 million years ago), and you find by radiocarbon dating that there is 0.000001 % 0.000001 % the amount of radiocarbon. Is this bone from the Cretaceous?

The spent fuel of a nuclear reactor contains plutonium-239, which has a half-life of 24,000 24,000 years. If 1 1 barrel containing 10 kg 10 kg of plutonium-239 is sealed, how many years must pass until only 10 g 10 g of plutonium-239 is left?

For the next set of exercises, use the following table, which features the world population by decade.

[T] The best-fit exponential curve to the data of the form P ( t ) = a e b t P ( t ) = a e b t is given by P ( t ) = 2686 e 0.01604 t . P ( t ) = 2686 e 0.01604 t . Use a graphing calculator to graph the data and the exponential curve together.

[T] Find and graph the derivative y ′ y ′ of your equation. Where is it increasing and what is the meaning of this increase?

[T] Find and graph the second derivative of your equation. Where is it increasing and what is the meaning of this increase?

[T] Find the predicted date when the population reaches 10 10 billion. Using your previous answers about the first and second derivatives, explain why exponential growth is unsuccessful in predicting the future.

For the next set of exercises, use the following table, which shows the population of San Francisco during the 19th century.

[T] The best-fit exponential curve to the data of the form P ( t ) = a e b t P ( t ) = a e b t is given by P ( t ) = 35.26 e 0.06407 t . P ( t ) = 35.26 e 0.06407 t . Use a graphing calculator to graph the data and the exponential curve together.

[T] Find and graph the derivative y ′ y ′ of your equation. Where is it increasing? What is the meaning of this increase? Is there a value where the increase is maximal?

[T] Find and graph the second derivative of your equation. Where is it increasing? What is the meaning of this increase?

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Exponential Growth and Decay

Exponential growth can be amazing.

The idea: something always grows in relation to its current value, such as always doubling.

Example: If a population of rabbits doubles every month, we would have 2, then 4, then 8, 16, 32, 64, 128, 256, etc!

Amazing tree.

Let us say we have this special tree.

It grows exponentially , following this formula:

Height (in mm) = e x

e is Euler's number , about 2.718

No tree could ever grow that tall. So when people say "it grows exponentially" ... just think what that means.

Growth and Decay

But sometimes things can grow (or the opposite: decay) exponentially, at least for a while .

So we have a generally useful formula:

y(t) = a × e kt

Where y(t) = value at time "t" a = value at the start k = rate of growth (when >0) or decay (when <0) t = time

Example: 2 months ago you had 3 mice, you now have 18.

Start with the formula:

We know a=3 mice, t=2 months, and right now y(2)=18 mice:

18 = 3 × e 2k

Now some algebra to solve for k :

• The step where we used ln(e x )=x is explained at Exponents and Logarithms .
• we could calculate k ≈ 0.896 , but it is best to keep it as k = ln(6)/2 until we do our final calculations.

We can now put k = ln(6)/2 into our formula from before:

y(t) = 3 e (ln(6)/2)t

Now let's calculate the population in 2 more months (at t=4 months):

y( 4 ) = 3 e (ln(6)/2)× 4 = 108

And in 1 year from now ( t=14 months):

y( 14 ) = 3 e (ln(6)/2)× 14 = 839,808

That's a lot of mice! I hope you will be feeding them properly.

Exponential Decay

Some things "decay" (get smaller) exponentially.

Example: Atmospheric pressure (the pressure of air around you) decreases as you go higher.

It decreases about 12% for every 1000 m: an exponential decay .

The pressure at sea level is about 1013 hPa (depending on weather).

• Write the formula (with its "k" value),
• Find the pressure on the roof of the Empire State Building (381 m),
• and at the top of Mount Everest (8848 m)
• a (the pressure at sea level) = 1013 hPa
• t is in meters (distance, not time, but the formula still works)
• y(1000) is a 12% reduction on 1013 hPa = 891.44 hPa

891.44 = 1013 e k×1000

Now we know "k" we can write:

y(t) = 1013 e (ln(0.88)/1000)×t

And finally we can calculate the pressure at 381 m , and at 8848 m :

y( 381 ) = 1013 e (ln(0.88)/1000)× 381 = 965 hPa

y( 8848 ) = 1013 e (ln(0.88)/1000)× 8848 = 327 hPa

In fact pressures at the top of Mount Everest are around 337 hPa ... good calculations!

The "half life" is how long it takes for a value to halve with exponential decay.

Commonly used with radioactive decay, but it has many other applications!

Example: The half-life of caffeine in your body is about 6 hours. If you had 1 cup of coffee 9 hours ago how much is left in your system?

• a (the starting dose) = 1 cup of coffee!
• t is in hours
• at y(6) we have a 50% reduction (because 6 is the half life)

0.5 = 1 cup × e 6 k

Now we can write:

y(t) = 1 e (ln(0.5)/6)×t

In 6 hours:

y( 6 ) = 1 e (ln(0.5)/6)× 6 = 0.5

Which is correct as 6 hours is the half life

And in 9 hours:

y( 9 ) = 1 e (ln(0.5)/6)× 9 = 0.35

After 9 hours the amount left in your system is about 0.35 of the original amount. Have a nice sleep :)

Have a play with the Half Life of Medicine Tool to get a good understanding of this.

How to Solve Exponential Decay Functions

Algebra Solutions: Answers and Explanations

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Exponential Decay

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Exponential functions tell the stories of explosive change. The two types of exponential functions are exponential growth and exponential decay. Four variables (percent change, time, the amount at the beginning of the time period, and the amount at the end of the time period) play roles in exponential functions. Use an exponential decay function to find the amount at the beginning of the time period.

Exponential decay is the change that occurs when an original amount is reduced by a consistent rate over a period of time.

Here's an exponential decay function:

y = a( 1 -b) x

• y : Final amount remaining after the decay over a period of time
• a : The original amount
• The decay factor is (1- b )
• The variable b is the percent of the decrease in decimal form.

Purpose of Finding the Original Amount

If you are reading this article, then you are probably ambitious. Six years from now, perhaps you want to pursue an undergraduate degree at Dream University. With a $120,000 price tag, Dream University evokes financial night terrors. After sleepless nights, you, Mom, and Dad meet with a financial planner. Your parents' bloodshot eyes clear up when the planner reveals that an investment with an eight percent growth rate can help your family reach the $120,000 target. Study hard. If you and your parents invest $75,620.36 today, then Dream University will become your reality thanks to exponential decay.

How to Solve

This function describes the exponential growth of the investment:

120,000 = a (1 +.08) 6

• 120,000: Final amount remaining after 6 years
• .08: Yearly growth rate
• 6: The number of years for the investment to grow
• a : The initial amount that your family invested

Thanks to the symmetric property of equality, 120,000 = a (1 +.08) 6 is the same as a (1 +.08) 6 = 120,000. Symmetric property of equality states that if 10 + 5 = 15, then 15 = 10 + 5.

If you prefer to rewrite the equation with the constant (120,000) on the right of the equation, then do so.

a (1 +.08) 6 = 120,000

Granted, the equation doesn't look like a linear equation (6 a = $120,000), but it's solvable. Stick with it!

Do not solve this exponential equation by dividing 120,000 by 6. It's a tempting math no-no.

1. Use order of operations to simplify

a (1 +.08) 6 = 120,000 a (1.08) 6 = 120,000 (Parenthesis) a (1.586874323) = 120,000 (Exponent)

2. Solve by dividing

a (1.586874323) = 120,000 a (1.586874323) / (1.586874323) = 120,000 / (1.586874323) 1 a = 75,620.35523 a = 75,620.35523

The original amount to invest is approximately $75,620.36.

3. Freeze: You're not done yet; use order of operations to check your answer

120,000 = a (1 +.08) 6 120,000 = 75,620.35523(1 +.08) 6 120,000 = 75,620.35523(1.08) 6 (Parenthesis) 120,000 = 75,620.35523(1.586874323) (Exponent) 120,000 = 120,000 (Multiplication)

Answers and Explanations to the Questions

Woodforest, Texas, a suburb of Houston, is determined to close the digital divide in its community. A few years ago, community leaders discovered that their citizens were computer illiterate. They did not have access to the internet and were shut out of the information superhighway. The leaders established the World Wide Web on Wheels, a set of mobile computer stations.

World Wide Web on Wheels has achieved its goal of only 100 computer illiterate citizens in Woodforest. Community leaders studied the monthly progress of the World Wide Web on Wheels. According to the data, the decline of computer illiterate citizens can be described by the following function:

100 = a (1 - .12) 10

1. How many people are computer illiterate 10 months after the inception of the World Wide Web on Wheels?

Compare this function to the original exponential growth function:

100 = a (1 - .12) 10 y = a( 1 + b) x

The variable y represents the number of computer illiterate people at the end of 10 months, so 100 people are still computer illiterate after the World Wide Web on Wheels began to work in the community.

2. Does this function represent exponential decay or exponential growth?

• This function represents exponential decay because a negative sign sits in front of the percent change (.12).

3. What is the monthly rate of change?

4. How many people were computer illiterate 10 months ago, at the inception of the World Wide Web on Wheels?

Use ​ order of operations to simplify.

100 = a (.88) 10 (Parenthesis)

100 = a (.278500976) (Exponent)

Divide to solve.

100(.278500976) = a (.278500976) / (.278500976)

359.0651689 = 1 a

359.0651689 = a

Use the order of operations to check your answer.

100 = 359.0651689(1 - .12) 10

100 = 359.0651689(.88) 10 (Parenthesis)

100 = 359.0651689(.278500976) (Exponent)

100 = 100 (Multiply)

5. If these trends continue, how many people will be computer illiterate 15 months after the inception of the World Wide Web on Wheels?

Add in what you know about the function.

y = 359.0651689(1 - .12) x

y = 359.0651689(1 - .12) 15

Use Order of Operations to find y .

y = 359.0651689(.88) 15 (Parenthesis)

y = 359.0651689 (.146973854) (Exponent)

y = 52.77319167 (Multiply).

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Exponential-Decay Word Problems

Log Probs Expo Growth Expo Decay

What is the formula for exponential decay?

Exponential growth word problems work off the exponential-decay formula, A  =  Pe kt , where A is the ending amount of whatever you're dealing with (for example, carbon-14 in a biological sample), P is the beginning amount of that same whatever, k is the decay constant, and t is time.

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What is the difference between the exponential-growth formula and the exponential-decay formula?

The only difference between exponential-growth equations and exponential-decay equations is that the constant for decay situations is negative. The equation itself is just the same as for exponential growth, but you should expect a negative value for the constant. If you get a positive value, you should probably go back and check your work.

Note that the particular variables used in the equation may change from one problem to another, or from one context to another, but that the structure of the equation is always the same. For instance, all of the following represent the same relationship:

A = P e r t A = P e k t Q = N e k t Q = Q 0 e k t

No matter the particular letters used, the green variable stands for the ending amount, the blue variable stands for the beginning amount, the red variable stands for the decay constant, and the purple variable stands for time. Get comfortable with this formula; you'll be seeing a lot of it.

To solve exponential decay word problems, you may be plugging one value into the exponential-decay equation, and solving for the required result. But you may need to solve two problems in one, where you use, say, the half-life information to find the decay constant (probably by solving the exponential equation by using logarithms), and then using that value to find whatever the exercise requested.

What is an example of a radioactive-decay exercise?

• Radio-isotopes of different elements have different half-lives. Magnesium- 27 has a half-life of 9.45 minutes. What is the decay constant for Magnesium- 27 ? Round to five decimal places.

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Since the decay rate is given in terms of minutes, then time t will be in minutes. However, I note that there is no beginning or ending amount given. How am I supposed to figure out what the decay constant is?

I can do this by working from the definition of "half-life": in the given amount of time (in this case, 9.45 minutes), half of the initial amount will be gone. That is, from t  = 0 to t  = 9.45 , I will have gone from 100% of however much I'd started with (where " 100% " is written as 1 , being its decimal equivalent) to 50% of that amount (converted to 0.5 for use inside the formula).

Since the half-life does not depend on how much I started with, I can either pick an arbitrary beginning amount (such as 100 grams) and then calculate the decay constant after 9.45 minutes, at which point only 50 grams will remain (the other 50 grams will have mutated into some other isotope or element). Or else I can just deal with the 50% that is left. Either way, I will end up with this equation:

0.5 = e 9.45 k

Solving for the decay constant k , I get:

ln(0.5) = 9.45 k

ln(0.5) / 9.45 = k = −0.073348907996...

They want me to round to five decimal places, so my answer is:

k = −0.07335

The constant was negative, as expected, because this was a decay problem. If I'd ended up with a positive value, this would have signalled to me that I'd made a mistake somewhere.

(Note: Technically, there should be units on the constant, so that the units on the time variable t will cancel off. In other words, technically, the answer should be " k  = −0.07335 /minute. But this probably won't matter in your math class.)

• Some people are frightened of certain medical tests because the tests involve the injection of radioactive materials. A hepatobiliary scan of my gallbladder involved an injection of 0.5 cc's (or about one-tenth of a teaspoon) of Technetium-99m, which has a half-life of almost exactly 6 hours. While undergoing the test, I heard the technician telling somebody on the phone that "in twenty-four hours, you'll be down to background radiation levels." Figure out just how much radioactive material remained from my gallbladder scan after twenty-four hours. Show your work below.

For this exercise, I need to find the ending amount A of Technetium-99m. Recalling that 1 cc (cubic centimeter) equals 1 mL (milliliter), I know that the beginning amount is P = 0.5 mL. The ending time is 24 hours. I do not have the decay constant but, by using the half-life information, I can find it. (Since this is a decay problem, I expect the constant to be negative. If I end up with a positive value, I'll know that I should go back and check my work.)

In 6 hours, there will be 50% of the original amount left, so:

0.5 = e 6 k

ln(0.5) = 6 k

ln(0.5)/6 = k

(This evaluates to about −0.1155 , but I'll leave the decay constant in exact form to avoid round-off error.)

Now that I have the decay constant, I can find out how much Technitium- 99 m was left after twenty-four hours:

A = 0.5 e (ln(0.5)/6)(24) = 0.03125

There will be no more than 0.03125 mL (or about 1/160 of a teaspoon) of Technitium- 99 m remaining after twenty-four hours.

It can be very useful, sometimes in unexpected places, to know that 1 cc (that is, one cubic centimeter) is equal to 1 mL (that is, one milliliter). I would recommend that you add this bit of knowledge to your store.

By the way: Technetium-99m is one of the most commonly used radioisotope for these medical purposes. Its radiation is extremely low-energy, so the chance of mutation is very low. (Whatever you're being treated for is the greater danger.) The half-life is just long enough for the doctors to have time to take their pictures. The dose I was given is about as large as these injections typically get. Your body does not easily absorb this chemical, so most of the injection is... "voided", shall we say? into the sewer system. 🚽

What is an example of solving a carbon-dating exercise?

• Carbon- 14 has a half-life of 5730 years. You are presented with a document which purports to contain the recollections of a Mycenaean soldier during the Trojan War. The city of Troy was finally destroyed in about 1250 BC, or about 3250 years ago. Carbon-dating evaluates the ratio of radioactive carbon- 14 to stable carbon- 12 . Given the amount of carbon- 12 contained a measured sample cut from the document, there would have been about 1.3 × 10 −12 grams of carbon- 14 in the sample when the parchment was new, assuming the proposed age is correct. According to your equipment, there remains 1.0 × 10 −12 grams. Is there a possibility that this is a genuine document? Or is this instead a recent forgery? Justify your conclusions.

First, as usual, I have to find the decay rate. (In real life, you'd look this up on a table, or have it programmed into your equipment, but this is math, not real life.) The half-life is 5730 years, so:

0.5 = e 5730 k

ln(0.5) = 5730 k

ln(0.5) / 5730 = k

I'll leave the decay constant in this "exact" form to avoid round-off error.

I have the beginning (expected) amount of C- 14 and the present (ending) amount; from this information, I can calculate the age of the parchment:

1.0 × 10 −12 = (1.3 × 10 −12 ) e (ln(0.5)/5730) t

1 / 1.3 = e (ln(0.5)/5730) t

ln( 1 / 1.3 ) = ( ln(0.5) / 5730 ) t

5730ln( 1 / 1.3 ) / ln(0.5) = t = 2168.87160124...

Then the parchment is about 2170 years old, much less than the necessary 3250 years ago that the Trojan War took place. But the parchment is indeed old, so this isn't a total fake.

Since the parchment is genuinely old ( 2170 years), but clearly not old enough to be the actual writings of a soldier in the Trojan War ( 3250 years), either this is a much-younger copy of an earlier document (in which case it is odd that there are no references to it in other documents, since only famous works tended to be copied), or, which is more likely, this is a recent forgery written on a not-quite-old-enough ancient parchment. If possible, the ink should be tested, since a recent forgery would use recently-made ink.

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College Algebra Tutorial 47

• Solve exponential growth problems.
• Solve exponential decay problems.

Practice Problems 1a - 1b:    Solve the given exponential growth or decay problem.

The following are webpages that can assist you in the topics that were covered on this page .

http://www.purplemath.com/modules/expoprob3.htm This webpage will help you with exponential decay problems.

Go to Get Help Outside the Classroom found in Tutorial 1: How to Succeed in a Math Class for some more suggestions.

Last revised on March 25, 2011 by Kim Seward. All contents copyright (C) 2002 - 2011, WTAMU and Kim Seward. All rights reserved.

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Exponential Functions - Problem Solving

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Complex numbers.

The beauty of Algebra through complex numbers, fractals, and Euler’s formula.

• Andrew Hayes

An exponential function is a function of the form \(f(x)=a \cdot b^x,\) where \(a\) and \(b\) are real numbers and \(b\) is positive. Exponential functions are used to model relationships with exponential growth or decay. Exponential growth occurs when a function's rate of change is proportional to the function's current value. Whenever an exponential function is decreasing, this is often referred to as exponential decay .

To solve problems on this page, you should be familiar with

• rules of exponents - algebraic
• solving exponential equations
• graphs of exponential functions .

Growth and Decay

Problem solving - basic, problem solving - intermediate, problem solving - advanced.

Suppose that the population of rabbits increases by 1.5 times a month. When the initial population is 100, what is the approximate integer population after a year? The population after \(n\) months is given by \(100 \times 1.5^n.\) Therefore, the approximate population after a year is \[100 \times 1.5^{12} \approx 100 \times 129.75 = 12975. \ _\square \]

Suppose that the population of rabbits increases by 1.5 times a month. At the end of a month, 10 rabbits immigrate in. When the initial population is 100, what is the approximate integer population after a year? Let \(p(n)\) be the population after \(n\) months. Then \[p(n+2) = 1.5 p(n+1) + 10\] and \[p(n+1) = 1.5 p(n) + 10,\] from which we have \[p(n+2) - p(n+1) = 1.5 \big(p(n+1) - p(n)\big).\] Then the population after \(n\) months is given by \[p(0) + \big(p(1) - p(0)\big) \frac{1.5^{n} - 1}{1.5 - 1} .\] Therefore, the population after a year is given by \[\begin{align} 100 + (160 - 100) \frac{1.5^{12} - 1}{1.5 - 1} \approx& 100 + 60 \times 257.493 \\ \approx& 15550. \ _\square \end{align}\]

Suppose that the annual interest is 3 %. When the initial balance is 1,000 dollars, how many years would it take to have 10,000 dollars? The balance after \(n\) years is given by \(1000 \times 1.03^n.\) To have the balance 10,000 dollars, we need \[\begin{align} 1000 \times 1.03^n \ge& 10000 \\ 1.03^n \ge& 10\\ n \log_{10}{1.03} \ge& 1 \\ n \ge& 77.898\dots. \end{align}\] Therefore, it would take 78 years. \( _\square \)

The half-life of carbon-14 is approximately 5730 years. Humans began agriculture approximately ten thousand years ago. If we had 1 kg of carbon-14 at that moment, how much carbon-14 in grams would we have now? The weight of carbon-14 after \(n\) years is given by \(1000 \times \left( \frac{1}{2} \right)^{\frac{n}{5730}}\) in grams. Therefore, the weight after 10000 years is given by \[1000 \times \left( \frac{1}{2} \right)^{\frac{10000}{5730}} \approx 1000 \times 0.298 = 298.\] Therefore, we would have approximately 298 g. \( _\square \)

Given three numbers such that \( 0 < a < b < c < 1\), define

\[ A = a^{a}b^{b}c^{c}, \quad B = a^{a}b^{c}c^{b} , \quad C = a^{b}b^{c}c^{a}. \]

How do the values of \(A, B, C \) compare to each other?

\[\large 2^{x} = 3^{y} = 12^{z} \]

If the equation above is fulfilled for non-zero values of \(x,y,z,\) find the value of \(\frac { z(x+2y) }{ xy }\).

If \(5^x = 6^y = 30^7\), then what is the value of \( \frac{ xy}{x+y} \)?

If \(27^{x} = 64^{y} = 125^{z} = 60\), find the value of \(\large\frac{2013xyz}{xy+yz+xz}\).

\[\large f(x)=\frac{e^x+e^{-x}}{e^x-e^{-x}} \]

Suppose we define the function \(f(x) \) as above. If \(f(a)=\frac{5}{3}\) and \(f(b)=\frac{7}{5},\) what is the value of \(f(a+b)?\)

\[\large \left(1+\frac{1}{x}\right)^{x+1}=\left(1+\frac{1}{2000}\right)^{2000}\]

Given that \(x\) is an integer that satisfies the equation above, find the value of \(x\).

\[\Large a^{(a-1)^{(a-2)}} = a^{a^2-3a+2}\]

Find the sum of all positive integers \(a\) that satisfy the equation above.

Find the sum of all solutions to the equation

\[ \large (x^2+5x+5)^{x^2-10x+21}=1 .\]

\[\large |x|^{(x^2-x-2)} < 1 \]

If the solution to the inequality above is \(x\in (A,B) \), then find the value of \(A+B\).

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Math Insight

Exponential growth and decay: a differential equation.

This little section is a tiny introduction to a very important subject and bunch of ideas: solving differential equations . We'll just look at the simplest possible example of this.

The general idea is that, instead of solving equations to find unknown numbers , we might solve equations to find unknown functions . There are many possibilities for what this might mean, but one is that we have an unknown function $y$ of $x$ and are given that $y$ and its derivative $y'$ (with respect to $x$) satisfy a relation $$y'=ky$$ where $k$ is some constant. Such a relation between an unknown function and its derivative (or derivatives ) is what is called a differential equation . Many basic ‘physical principles’ can be written in such terms, using ‘time’ $t$ as the independent variable.

Having been taking derivatives of exponential functions, a person might remember that the function $f(t)=e^{kt}$ has exactly this property: $${d\over dt}e^{kt}=k\cdot e^{kt}$$ For that matter, any constant multiple of this function has the same property: $${d\over dt}(c\cdot e^{kt})=k\cdot c\cdot e^{kt}$$ And it turns out that these really are all the possible solutions to this differential equation.

There is a certain buzz-phrase which is supposed to alert a person to the occurrence of this little story: if a function $f$ has exponential growth or exponential decay then that is taken to mean that $f$ can be written in the form $$f(t)=c\cdot e^{kt}$$ If the constant $k$ is positive it has exponential growth and if $k$ is negative then it has exponential decay .

Since we've described all the solutions to this equation, what questions remain to ask about this kind of thing? Well, the usual scenario is that some story problem will give you information in a way that requires you to take some trouble in order to determine the constants $c,k$ . And, in case you were wondering where you get to take a derivative here, the answer is that you don't really: all the ‘calculus work’ was done at the point where we granted ourselves that all solutions to that differential equation are given in the form $f(t)=ce^{kt}$.

First to look at some general ideas about determining the constants before getting embroiled in story problems: One simple observation is that $$c=f(0)$$ that is, that the constant $c$ is the value of the function at time $t=0$. This is true simply because $$f(0)=ce^{k \cdot 0}=ce^{0}=c\cdot 1=c$$ from properties of the exponential function.

More generally, suppose we know the values of the function at two different times: $$y_1=ce^{kt_1}$$ $$y_2=ce^{kt_2}$$ Even though we certainly do have ‘two equations and two unknowns’, these equations involve the unknown constants in a manner we may not be used to. But it's still not so hard to solve for $c,k$: dividing the first equation by the second and using properties of the exponential function, the $c$ on the right side cancels, and we get $${y_1\over y_2}=e^{k(t_1-t_2)}$$ Taking a logarithm (base $e$, of course) we get $$\ln y_1-\ln y_2=k(t_1-t_2)$$ Dividing by $t_1-t_2$, this is $$k={\ln y_1-\ln y_2\over t_1-t_2}$$ Substituting back in order to find $c$, we first have $$y_1=ce^{{\ln y_1-\ln y_2\over t_1-t_2}t_1}$$ Taking the logarithm, we have $$\ln y_1=\ln c+{\ln y_1-\ln y_2\over t_1-t_2}t_1$$ Rearranging, this is $$\ln c=\ln y_1-{\ln y_1-\ln y_2\over t_1-t_2}t_1= {t_1\ln y_2-t_2\ln y_1\over t_1-t_2}$$ Therefore, in summary, the two equations $$y_1=ce^{kt_1}$$ $$y_2=ce^{kt_2}$$ allow us to solve for $c,k$, giving $$k={\ln y_1-\ln y_2\over t_1-t_2}$$ $$c=e^{t_1\ln y_2-t_2\ln y_1\over t_1-t_2}$$

A person might manage to remember such formulas, or it might be wiser to remember the way of deriving them.

A herd of llamas has $1000$ llamas in it, and the population is growing exponentially. At time $t=4$ it has $2000$ llamas. Write a formula for the number of llamas at arbitrary time $t$.

Solution: Here there is no direct mention of differential equations, but use of the buzz-phrase ‘growing exponentially’ must be taken as indicator that we are talking about the situation $$f(t)=ce^{kt}$$ where here $f(t)$ is the number of llamas at time $t$ and $c,k$ are constants to be determined from the information given in the problem. And the use of language should probably be taken to mean that at time $t=0$ there are $1000$ llamas, and at time $t=4$ there are $2000$. Then, either repeating the method above or plugging into the formula derived by the method, we find $$c=\hbox{ value of $f$ at $t=0$ } = 1000$$ $$k={\ln f(t_1)-\ln f(t_2)\over t_1-t_2}={\ln 1000-\ln 2000\over 0-4}$$ $$={ \ln {1000\over 2000}}{-4}={\ln {1\over 2} \over -4 } = (\ln 2)/4$$ Therefore, $$f(t)=1000\;e^{{\ln 2\over 4}t}=1000\cdot 2^{t/4}$$ This is the desired formula for the number of llamas at arbitrary time $t$.

• A colony of bacteria is growing exponentially. At time $t=0$ it has $10$ bacteria in it, and at time $t=4$ it has $2000$. At what time will it have $100,000$ bacteria?

Solution: Even though it is not explicitly demanded, we need to find the general formula for the number $f(t)$ of bacteria at time $t$, set this expression equal to $100,000$, and solve for $t$. Again, we can take a little shortcut here since we know that $c=f(0)$ and we are given that $f(0)=10$. (This is easier than using the bulkier more general formula for finding $c$). And use the formula for $k$: $$k={\ln f(t_1)-\ln f(t_2)\over t_1-t_2}= {\ln 10 -\ln 2,000\over 0-4}={ \ln {10\over 2,000} \over -4 }= {\ln 200\over 4} $$ Therefore, we have $$f(t)=10\cdot e^{{\ln 200\over 4}\;t}=10\cdot 200^{t/4}$$ as the general formula. Now we try to solve $$100,000=10\cdot e^{{\ln 200\over 4}\;t}$$ for $t$: divide both sides by the $10$ and take logarithms, to get $$\ln 10,000={\ln 200\over 4}\;t$$ Thus, $$t=4\,{\ln 10,000\over \ln 200}\approx 6.953407835.$$

• A herd of llamas is growing exponentially. At time $t=0$ it has $1000$ llamas in it, and at time $t=4$ it has $2000$ llamas. Write a formula for the number of llamas at arbitrary time $t$.
• A herd of elephants is growing exponentially. At time $t=2$ it has $1000$ elephants in it, and at time $t=4$ it has $2000$ elephants. Write a formula for the number of elephants at arbitrary time $t$.
• A colony of bacteria is growing exponentially. At time $t=2$ it has $10$ bacteria in it, and at time $t=4$ it has $2000$. At what time will it have $100,000$ bacteria?

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Module 12: Exponential and Logarithmic Equations and Models

Exponential growth and decay, learning outcomes.

• Graph exponential growth and decay functions.
• Solve problems involving radioactive decay, carbon dating, and half life.

In real-world applications, we need to model the behavior of a function. In mathematical modeling, we choose a familiar general function with properties that suggest that it will model the real-world phenomenon we wish to analyze. In the case of rapid growth, we may choose the exponential growth function:

[latex]y={A}_{0}{e}^{kt}[/latex]

where [latex]{A}_{0}[/latex] is equal to the value at time zero, e  is Euler’s constant, and k  is a positive constant that determines the rate (percentage) of growth. We may use the exponential growth function in applications involving doubling time , the time it takes for a quantity to double. Such phenomena as wildlife populations, financial investments, biological samples, and natural resources may exhibit growth based on a doubling time. In some applications, however, as we will see when we discuss the logistic equation, the logistic model sometimes fits the data better than the exponential model.

On the other hand, if a quantity is falling rapidly toward zero, without ever reaching zero, then we should probably choose the exponential decay model. Again, we have the form [latex]y={A}_{0}{e}^{-kt}[/latex] where [latex]{A}_{0}[/latex] is the starting value, and e  is Euler’s constant. Now k  is a negative constant that determines the rate of decay. We may use the exponential decay model when we are calculating half-life , or the time it takes for a substance to exponentially decay to half of its original quantity. We use half-life in applications involving radioactive isotopes.

In our choice of a function to serve as a mathematical model, we often use data points gathered by careful observation and measurement to construct points on a graph and hope we can recognize the shape of the graph. Exponential growth and decay graphs have a distinctive shape, as we can see in the graphs below. It is important to remember that, although parts of each of the two graphs seem to lie on the x -axis, they are really a tiny distance above the x -axis.

A graph showing exponential growth. The equation is [latex]y=2{e}^{3x}[/latex].

A graph showing exponential decay. The equation is [latex]y=3{e}^{-2x}[/latex].

Exponential growth and decay often involve very large or very small numbers. To describe these numbers, we often use orders of magnitude. The order of magnitude is the power of ten when the number is expressed in scientific notation with one digit to the left of the decimal. For example, the distance to the nearest star, Proxima Centauri , measured in kilometers, is 40,113,497,200,000 kilometers. Expressed in scientific notation, this is [latex]4.01134972\times {10}^{13}[/latex]. We could describe this number as having order of magnitude [latex]{10}^{13}[/latex].

A General Note: Characteristics of the Exponential Function [latex]y=A_{0}e^{kt}[/latex]

An exponential function of the form [latex]y={A}_{0}{e}^{kt}[/latex] has the following characteristics:

• one-to-one function
• horizontal asymptote: y  = 0
• domain: [latex]\left(-\infty , \infty \right)[/latex]
• range: [latex]\left(0,\infty \right)[/latex]
• x intercept: none
• y-intercept: [latex]\left(0,{A}_{0}\right)[/latex]
• increasing if k  > 0
• decreasing if k  < 0

An exponential function models exponential growth when k > 0 and exponential decay when k < 0.

Example: Graphing Exponential Growth

A population of bacteria doubles every hour. If the culture started with 10 bacteria, graph the population as a function of time.

When an amount grows at a fixed percent per unit time, the growth is exponential. To find [latex]{A}_{0}[/latex] we use the fact that [latex]{A}_{0}[/latex] is the amount at time zero, so [latex]{A}_{0}=10[/latex]. To find k , use the fact that after one hour [latex]\left(t=1\right)[/latex] the population doubles from 10 to 20. The formula is derived as follows:

[latex]\begin{array}{l}\text{ }20=10{e}^{k\cdot 1}\hfill & \hfill \\ \text{ }2={e}^{k}\hfill & \text{Divide both sides by 10}\hfill \\ \mathrm{ln}2=k\hfill & \text{Take the natural logarithm of both sides}\hfill \end{array}[/latex]

so [latex]k=\mathrm{ln}\left(2\right)[/latex]. Thus the equation we want to graph is [latex]y=10{e}^{\left(\mathrm{ln}2\right)t}=10{\left({e}^{\mathrm{ln}2}\right)}^{t}=10\cdot {2}^{t}[/latex]. The graph is shown below.

The graph of [latex]y=10{e}^{\left(\mathrm{ln}2\right)t}[/latex] .

Analysis of the Solution

The population of bacteria after ten hours is 10,240. We could describe this amount as being of the order of magnitude [latex]{10}^{4}[/latex]. The population of bacteria after twenty hours is 10,485,760 which is of the order of magnitude [latex]{10}^{7}[/latex], so we could say that the population has increased by three orders of magnitude in ten hours.

Calculating Doubling Time

For growing quantities, we might want to find out how long it takes for a quantity to double. As we mentioned above, the time it takes for a quantity to double is called the doubling time .

Given the basic exponential growth equation [latex]A={A}_{0}{e}^{kt}[/latex], doubling time can be found by solving for when the original quantity has doubled, that is, by solving [latex]2{A}_{0}={A}_{0}{e}^{kt}[/latex].

The formula is derived as follows:

[latex]\begin{array}{l}2{A}_{0}={A}_{0}{e}^{kt}\hfill & \hfill \\ 2={e}^{kt}\hfill & \text{Divide both sides by }{A}_{0}.\hfill \\ \mathrm{ln}2=kt\hfill & \text{Take the natural logarithm of both sides}.\hfill \\ t=\frac{\mathrm{ln}2}{k}\hfill & \text{Divide by the coefficient of }t.\hfill \end{array}[/latex]

Thus the doubling time is

[latex]t=\frac{\mathrm{ln}2}{k}[/latex]

Example: Finding a Function That Describes Exponential Growth

According to Moore’s Law, the doubling time for the number of transistors that can be put on a computer chip is approximately two years. Give a function that describes this behavior.

[latex]\begin{array}{l}t=\frac{\mathrm{ln}2}{k}\hfill & \text{The doubling time formula}.\hfill \\ 2=\frac{\mathrm{ln}2}{k}\hfill & \text{Use a doubling time of two years}.\hfill \\ k=\frac{\mathrm{ln}2}{2}\hfill & \text{Multiply by }k\text{ and divide by 2}.\hfill \\ A={A}_{0}{e}^{\frac{\mathrm{ln}2}{2}t}\hfill & \text{Substitute }k\text{ into the continuous growth formula}.\hfill \end{array}[/latex]

The function is [latex]A={A}_{0}{e}^{\frac{\mathrm{ln}2}{2}t}[/latex].

Recent data suggests that, as of 2013, the rate of growth predicted by Moore’s Law no longer holds. Growth has slowed to a doubling time of approximately three years. Find the new function that takes that longer doubling time into account.

[latex]f\left(t\right)={A}_{0}{e}^{\frac{\mathrm{ln}2}{3}t}[/latex]

We now turn to exponential decay . One of the common terms associated with exponential decay, as stated above, is half-life , the length of time it takes an exponentially decaying quantity to decrease to half its original amount. Every radioactive isotope has a half-life, and the process describing the exponential decay of an isotope is called radioactive decay.

To find the half-life of a function describing exponential decay, solve the following equation:

[latex]\frac{1}{2}{A}_{0}={A}_{o}{e}^{kt}[/latex]

We find that the half-life depends only on the constant k  and not on the starting quantity [latex]{A}_{0}[/latex].

The formula is derived as follows

[latex]\begin{array}{l}\frac{1}{2}{A}_{0}={A}_{o}{e}^{kt}\hfill & \hfill \\ \frac{1}{2}={e}^{kt}\hfill & \text{Divide both sides by }{A}_{0}.\hfill \\ \mathrm{ln}\left(\frac{1}{2}\right)=kt\hfill & \text{Take the natural log of both sides}.\hfill \\ -\mathrm{ln}\left(2\right)=kt\hfill & \text{Apply properties of logarithms}.\hfill \\ -\frac{\mathrm{ln}\left(2\right)}{k}=t\hfill & \text{Divide by }k.\hfill \end{array}[/latex]

Since t , the time, is positive, k  must, as expected, be negative. This gives us the half-life formula

[latex]t=-\frac{\mathrm{ln}\left(2\right)}{k}[/latex]

In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.

One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its half-life . The table below lists the half-life for several of the more common radioactive substances.

We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:

[latex]\begin{array}{l}A\left(t\right)={A}_{0}{e}^{\frac{\mathrm{ln}\left(0.5\right)}{T}t}\hfill \\ A\left(t\right)={A}_{0}{e}^{\mathrm{ln}\left(0.5\right)\frac{t}{T}}\hfill \\ A\left(t\right)={A}_{0}{\left({e}^{\mathrm{ln}\left(0.5\right)}\right)}^{\frac{t}{T}}\hfill \\ A\left(t\right)={A}_{0}{\left(\frac{1}{2}\right)}^{\frac{t}{T}}\hfill \end{array}[/latex]

• [latex]{A}_{0}[/latex] is the amount initially present
• [latex]T[/latex] is the half-life of the substance
• [latex]t[/latex] is the time period over which the substance is studied
• [latex]A[/latex], or [latex]A(t)[/latex], is the amount of the substance present after time [latex]t[/latex]

How To: Given the half-life, find the decay rate

• Write [latex]A={A}_{o}{e}^{kt}[/latex].
• Replace A  by [latex]\frac{1}{2}{A}_{0}[/latex] and replace t  by the given half-life.
• Solve to find k . Express k  as an exact value (do not round).

Note: It is also possible to find the decay rate using [latex]k=-\frac{\mathrm{ln}\left(2\right)}{t}[/latex].

Example: Using the Formula for Radioactive Decay to Find the Quantity of a Substance

How long will it take for 10% of a 1000-gram sample of uranium-235 to decay?

[latex]\begin{array}{l}\text{ }y=\text{1000}e\frac{\mathrm{ln}\left(0.5\right)}{\text{703,800,000}}t\hfill & \hfill \\ \text{ }900=1000{e}^{\frac{\mathrm{ln}\left(0.5\right)}{\text{703,800,000}}t}\hfill & \text{After 10% decays, 900 grams are left}.\hfill \\ \text{ }0.9={e}^{\frac{\mathrm{ln}\left(0.5\right)}{\text{703,800,000}}t}\hfill & \text{Divide by 1000}.\hfill \\ \mathrm{ln}\left(0.9\right)=\mathrm{ln}\left({e}^{\frac{\mathrm{ln}\left(0.5\right)}{\text{703,800,000}}t}\right)\hfill & \text{Take ln of both sides}.\hfill \\ \mathrm{ln}\left(0.9\right)=\frac{\mathrm{ln}\left(0.5\right)}{\text{703,800,000}}t\hfill & \text{ln}\left({e}^{M}\right)=M\hfill \\ \text{}\text{}t=\text{703,800,000}\times \frac{\mathrm{ln}\left(0.9\right)}{\mathrm{ln}\left(0.5\right)}\text{years}\hfill & \text{Solve for }t.\hfill \\ \text{}\text{}t\approx \text{106,979,777 years}\hfill & \hfill \end{array}[/latex]

Ten percent of 1000 grams is 100 grams. If 100 grams decay, the amount of uranium-235 remaining is 900 grams.

How long will it take before twenty percent of our 1000-gram sample of uranium-235 has decayed?

[latex]t=703,800,000\times \frac{\mathrm{ln}\left(0.8\right)}{\mathrm{ln}\left(0.5\right)}\text{ years }\approx \text{ }226,572,993\text{ years}[/latex].

Example: Finding the Function that Describes Radioactive Decay

The half-life of carbon-14 is 5,730 years. Express the amount of carbon-14 remaining as a function of time, t .

This formula is derived as follows.

[latex]\begin{array}{l}\text{}A={A}_{0}{e}^{kt}\hfill & \text{The continuous growth formula}.\hfill \\ 0.5{A}_{0}={A}_{0}{e}^{k\cdot 5730}\hfill & \text{Substitute the half-life for }t\text{ and }0.5{A}_{0}\text{ for }f\left(t\right).\hfill \\ \text{}0.5={e}^{5730k}\hfill & \text{Divide both sides by }{A}_{0}.\hfill \\ \mathrm{ln}\left(0.5\right)=5730k\hfill & \text{Take the natural log of both sides}.\hfill \\ \text{}k=\frac{\mathrm{ln}\left(0.5\right)}{5730}\hfill & \text{Divide by the coefficient of }k.\hfill \\ \text{}A={A}_{0}{e}^{\left(\frac{\mathrm{ln}\left(0.5\right)}{5730}\right)t}\hfill & \text{Substitute for }r\text{ in the continuous growth formula}.\hfill \end{array}[/latex]

The function that describes this continuous decay is [latex]f\left(t\right)={A}_{0}{e}^{\left(\frac{\mathrm{ln}\left(0.5\right)}{5730}\right)t}[/latex]. We observe that the coefficient of t , [latex]\frac{\mathrm{ln}\left(0.5\right)}{5730}\approx -1.2097[/latex] is negative, as expected in the case of exponential decay.

The half-life of plutonium-244 is 80,000,000 years. Find a function that gives the amount of plutonium-244 remaining as a function of time measured in years.

[latex]f\left(t\right)={A}_{0}{e}^{-0.0000000087t}[/latex]

Radiocarbon Dating

The formula for radioactive decay is important in radiocarbon dating  which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libby who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about 1% error for plants or animals that died within the last 60,000 years.

Carbon-14 is a radioactive isotope of carbon that has a half-life of 5,730 years. It occurs in small quantities in the carbon dioxide in the air we breathe. Most of the carbon on Earth is carbon-12 which has an atomic weight of 12 and is not radioactive. Scientists have determined the ratio of carbon-14 to carbon-12 in the air for the last 60,000 years using tree rings and other organic samples of known dates—although the ratio has changed slightly over the centuries.

As long as a plant or animal is alive, the ratio of the two isotopes of carbon in its body is close to the ratio in the atmosphere. When it dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-14 to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated.

Since the half-life of carbon-14 is 5,730 years, the formula for the amount of carbon-14 remaining after t  years is

[latex]A\approx {A}_{0}{e}^{\left(\frac{\mathrm{ln}\left(0.5\right)}{5730}\right)t}[/latex]

• [latex]A[/latex] is the amount of carbon-14 remaining
• [latex]{A}_{0}[/latex] is the amount of carbon-14 when the plant or animal began decaying.

This formula is derived as follows:

[latex]\begin{array}{l}\text{ }A={A}_{0}{e}^{kt}\hfill & \text{The continuous growth formula}.\hfill \\ \text{ }0.5{A}_{0}={A}_{0}{e}^{k\cdot 5730}\hfill & \text{Substitute the half-life for }t\text{ and }0.5{A}_{0}\text{ for }f\left(t\right).\hfill \\ \text{ }0.5={e}^{5730k}\hfill & \text{Divide both sides by }{A}_{0}.\hfill \\ \mathrm{ln}\left(0.5\right)=5730k\hfill & \text{Take the natural log of both sides}.\hfill \\ \text{ }k=\frac{\mathrm{ln}\left(0.5\right)}{5730}\hfill & \text{Divide both sides by the coefficient of }k.\hfill \\ \text{ }A={A}_{0}{e}^{\left(\frac{\mathrm{ln}\left(0.5\right)}{5730}\right)t}\hfill & \text{Substitute for }r\text{ in the continuous growth formula}.\hfill \end{array}[/latex]

To find the age of an object we solve this equation for t :

[latex]t=\frac{\mathrm{ln}\left(\frac{A}{{A}_{0}}\right)}{-0.000121}[/latex]

Out of necessity, we neglect here the many details that a scientist takes into consideration when doing carbon-14 dating, and we only look at the basic formula. The ratio of carbon-14 to carbon-12 in the atmosphere is approximately 0.0000000001%. Let r  be the ratio of carbon-14 to carbon-12 in the organic artifact or fossil to be dated determined by a method called liquid scintillation. From the equation [latex]A\approx {A}_{0}{e}^{-0.000121t}[/latex] we know the ratio of the percentage of carbon-14 in the object we are dating to the percentage of carbon-14 in the atmosphere is [latex]r=\frac{A}{{A}_{0}}\approx {e}^{-0.000121t}[/latex]. We solve this equation for t , to get

[latex]t=\frac{\mathrm{ln}\left(r\right)}{-0.000121}[/latex]

How To: Given the percentage of carbon-14 in an object, determine its age

• Express the given percentage of carbon-14 as an equivalent decimal r .
• Substitute for r  in the equation [latex]t=\frac{\mathrm{ln}\left(r\right)}{-0.000121}[/latex] and solve for the age, t .

Example: Finding the Age of a Bone

A bone fragment is found that contains 20% of its original carbon-14. To the nearest year, how old is the bone?

We substitute 20% = 0.20 for r  in the equation and solve for t :

[latex]\begin{array}{l}t=\frac{\mathrm{ln}\left(r\right)}{-0.000121}\hfill & \text{Use the general form of the equation}.\hfill \\ =\frac{\mathrm{ln}\left(0.20\right)}{-0.000121}\hfill & \text{Substitute for }r.\hfill \\ \approx 13301\hfill & \text{Round to the nearest year}.\hfill \end{array}[/latex]

The bone fragment is about 13,301 years old.

The instruments that measure the percentage of carbon-14 are extremely sensitive and, as we mention above, a scientist will need to do much more work than we did in order to be satisfied. Even so, carbon dating is only accurate to about 1%, so this age should be given as [latex]\text{13,301 years}\pm \text{1% or 13,301 years}\pm \text{133 years}[/latex].

Cesium-137 has a half-life of about 30 years. If we begin with 200 mg of cesium-137, will it take more or less than 230 years until only 1 milligram remains?

less than 230 years; 229.3157 to be exact

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Exponential Growth And Decay

Exponential growth and decay apply to physical quantities which change in value or form in a rapid manner. The change can be measured using the concept of exponential growth and exponential decay, and the new obtained quantity can be obtained from the existing quantity. The formulas of exponential growth and decay are f(x) = a(1 + r) t , and f(x) = a(1 - r) t respectively.

Let us learn more about exponential growth and decay, the formula, applications, with the help of examples, FAQs.

What Is Exponential Growth And Decay?

Exponential growth and decay apply to quantities which change rapidly. Exponential growth and decay have been derived from the concept of geometric progression. Quantities that do not change as constant but change in an exponential manner can be termed as having an exponential growth or exponential decay.

The simplest representation of exponential growth and decay is the formula ab x , where 'a' is the initial quantity, 'b' is the growth factor which is similar to the common ratio of the geometric progression, and 'x' in the time steps for multiplying the growth factor. For exponential growth, the value of b is greater than 1 (b > 1), and for exponential decay, the value of b is lesser than 1 (b < 1).

Exponential growth finds applications in studying bacterial growth, population increase, money growth schemes. Exponential decay refers to a rapid decrease in a quantity over a period of time. The exponential decay can be used to find food decay, half-life, radioactive decay. The formulas of exponential growth and decay are as presented below.

Exponential growth uses a factor 'r' which is the rate of growth. Here the r-value lies between 0 and 1 (0 < r < 1). The term (1 + r) can be taken as the growth factor. And 't' is the time steps which is the number of times the growth factor is to be multiplied. The value of 't' can be a whole number or a decimal number. For exponential decay, the growth factor is (1 - r), which has a value lesser than 1.

Formulas of Exponential Growth And Decay

The exponential growth and decay have different interpretations of the formulas which are interrelated and can be interpreted differently. The below table shows the three different formulas of exponential growth and decay.

In the above formulas the 'a' or P o is the initial quantity of the substance. Further for exponential growth b = 1 + r = e k and for exponential decay we have b = 1 - r = e -k .

Applications of Exponential Growth And Decay

The concept of exponential growth and decay can be observed in numerous day-to-day scientific and industrial activities. Let us check a few important applications of exponential growth and decay.

• Bacterial Growth: The initial bacterial growth can be observed in numerous communicable diseases. The recent Covid-19 is a quick example of exponential growth where the disease is highly communicable and is transmitted from one to many, and then to many more people. Bacteria and viruses are growing in an exponential manner, and the initial bacteria grows at an exponential rate. The initial bacterial if taken to double every second would grow in numbers such as 1, 2, 4, 8, 16, 32, 64, 128, 256, 512... With this, we can observe the manner of exponential growth in bacteria.
• Nuclear Chain Reactions: The nuclear chain reactions can be broadly classified as nuclear fission and nuclear fission reactions. The concept of nuclear fusion can be linked with exponential growth and the concept of nuclear fission can be linked with exponential decay. Nuclear fusion is a reaction in which two or more atoms combine to form a larger atom and this kind of reaction can be observed in the core of the sun. Nuclear fission is a kind of exponential decay that can be observed in radioactive material, in which the initial quantity decomposes and we have a smaller quantity by the end of the observed time period.
• Feedback: The concept of feedback, more so of customer feedback grows in an exponential manner. This can be observed more so in negative feedback. The bad functioning of the product if experienced by a customer is shared by this customer to another person, who in turn shares it with two or more people, and each of those people again shares it with two or more people. The medium of the internet helps for this easy sharing of feedback which flows exponentially to a larger customer segment. Thus the feedback in this current day of the internet is conveyed at a rapid pace.
• Processing Power of Computers: The processing and storage power has now increased exponentially. Earlier the data storage was only in MB, which has now grown to GB and TB exponentially. The slow growth in the computer hardware and storage system in the initial 1970s and 1980s has now grown exponentially. The initial floppy disks and hard disk drives have now been replaced with cloud servers which can be easily assessed by the user through an internet-connected mobile device.
• Food Degradation: Food degradation can also be understood as a case of exponent growth and decay. The food remains good for a certain amount of time and then it degenerates exponentially. The food is slightly stale initially, and then it stales rapidly until we discover that it has spoilt completely. This is also a typical case of exponential decay, where once the decay process initiates, then it decays at a rapid pace until the entire food quantity is completely stale.
• Aging of Human Beings: The aging process of humans or any living being at the ending part of the lifetime follows an exponential decay process. The person remains hale and healthy for a normal course of life for about 60 years. This is the same reason that for many jobs the retirement age is set as 60 years. After this, the aging process is so rapid that it affects the body at an exponential rate. The degradation in the quality of life seen in the initial days is further degraded drastically in the later years. This could be because of the advancement of certain existing diseases or the malfunctioning of certain organs.
• Internet Content: The internet is exploding with information. In the initial stages of the internet, google had to collect useful information and add it to the net. But with time the internet users started adding information to the internet, and now the amount of information now available on the internet is mind-boggling. Also now the application of artificial intelligence algorithms helps build the content in an exponential manner. The content is generated thousands and millions of times in a short span of time.

☛ Related Topics

• Exponent Rules
• Decay Formula
• Negative Exponents
• Exponential to Log Form
• Log to Exponential Form
• Exponential Decay Formula

Examples on Exponential Growth And Decay

Example 1: What is the amount received from the investment fund after 2 years, if $.100,000 were invested at the compounding rate of 5% per every quarter?

The invested principal is a = $100,000, the rate of compounding growth is r = 5% = 0.05 per quarter.

The time period is 2 years, and there are 4 quarters in a year, and we have t = 8.

Applying the concepts of exponential growth and decay we have the following expressions for exponential growth.

f(x) = a(1 + r) t

f(x) = 100,000(1 + 0.05) 8

f(x) = 1,00,000(1.05) 8 = 100,000 × 1.47745544 = 147745.44

Therefore an amount of $1,47, 746 is received after a period of 2 years.

Example 2: The radioactive material of thorium decays at the rate of 8% per minute. What part of 10 grams of thorium would be remaining after 5 minutes?

The given initial quantity of thorium is a = 10grams, the rate of decay per minute is r = 8% = 8/100 = 0.08, and the time steps t = 5.

Here we can apply the concepts of exponential growth and decay, and the exponential decay formula for the decay of thorium is as follows.

f(x) = a(1 - r) t

f (x) = 10(1 - 0.08) 5 = 10(0.92) 5 = 6.5908

Therefore a quantity of 6.6 grams of thorium remains after 5 minutes.

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Practice Questions on Exponential Growth And Decay

Faqs on exponential growth and decay, what is exponential growth and decay in maths.

Exponential growth and decay in maths applies to the calculation of rapidly changing quantity. There are numerous quantities and values in nature, industry, business which change rapidly with time. Compound internet, decpreciation, growth of virus, spoiling of perishable foods, are some of the real life examples which need to be calculated using a math formula. The exponential growth and decay gives the required needed calculations using the formulas f(x) = a(1 + r) t , and f(x) = a(1 - r) t . Here a is the initial quantity, r is the growth or decay constant, and t is the time period or the time factor.

What Is The Formula For Exponential Growth And Decay?

There are three types of formulas that are used for computing exponential growth and decay. The three formulas are as follows.

• f(x) = ab x for exponential growth and f(x) = ab -x for exponential decay. Here 'a' is the initial quantity, 'b' is the growth or decay factor, and 'x' is the time step.
• f(x) = a(1 + r) t , and f(x) = a(1 - r) t are for exponential growth and exponential decay respectively. Here 'r' is the growth or decay factor and has a value between 0 and 1, (0 < r < 1).
• P = P o e kt , P = P o e -kt are for formulas of exponential growth and decay. Here P o is the initial quantity, P is the obtained quantity, e is the exponential factor, and k is the growth or decay constant.

What Is The Difference Between Exponential Growth And Exponential Decay?

Exponential growth refers to an increase of the resultant quantity for a given quantity, and exponential decay refers to the decrease of the resultant quantity for a given quantity. The exponential growth and decay both need the initial quantity, the time period and the decay or growth constant to find the resultant quantity. Exponetial growth finds use in finance, medicine, biology, and exponential decay find use to find the depreciation of an asset, to find the expiry date of a manufactured item.

What Is The Use Of Exponential Growth And Decay In Maths?

The exponential growth and decay can be used to calculate the resultant quantity after a process of exponential growth or exponetial decay. From the given initial quantity, and the rate of growth or decay we can easily compute the resultant quantity. We can calculate the exponent growth and decay using f(x) = a(1 + r) t , and f(x) = a(1 - r) t .

What Are The Applications Of Exponential Growth And Decay In Daily Life?

The exponential growth and decay have numerous applications in our day-to-day life. The biological world has numerous examples of diseases and their spread, micro organisms, virus and their growth, which needs to be computed. Further we find numerous examples of exponential growth in finance, business, the internet, consumer behavior.

EXPONENTIAL GROWTH AND DECAY WORD PROBLEMS

In this section, we are going to see how to solve word problems on exponential growth and decay.

Before look at the problems, if you like to learn about exponential growth and decay,

please click here

Problem 1 :

David owns a chain of fast food restaurants that operated 200 stores in 1999. If the rate of increase is 8% annually, how many stores does the restaurant operate in 2007 ?

Number of years between 1999 and 2007 is 

n  =  2007 - 1999

n  =  8

No. of stores in the year 2007  =   P(1+r)ⁿ

Substitute P = 200, r = 8% or 0.08 and n = 8.

No. of stores in the year 2007  =  200(1 + 0.08) 8

No. of stores in the year 2007  =  200(1.08 ) 8

No. of stores in the year 2007  =  200(1.8509)

No. of stores in the year 2007  =  370.18

So, the number of stores in the year 2007 is about 370.

Problem 2 :

You invest $2500 in bank which pays 10% interest per year compounded continuously. What will be the value of the investment after 10 years ?

We have to use the formula given below to know the value of the investment after 3 years. 

A  =  Pe rt

Substitute 

P  =  2500

r  =  10% or 0.1

t  =  10

e  =  2.71828

Then, we have 

A = 2500(2.71828) (0.1)10

A = 6795.70

So, the value of the investment after 10 years is $6795.70.

Problem 3 :

Suppose a radio active substance decays at a rate of 3.5% per hour. What percent of substance will be left after 6 hours ?

Since the initial amount of substance is not given and the problem is based on percentage, we have to assume that the initial amount of substance is 100. 

We have to use the formula given below to find the percent of substance after 6 hours. 

A  =  P(1 + r) n

P  =  100

r  =  -3.5% or -0.035

t  =  6

(Here, the value of "r" is taken in negative sign. because the substance decays)

A  =  100(1-0.035) 6

A  =  100(0.935 ) 6

A  =  100(0.8075)

A  =  80.75

Because  the initial amount of substance is assumed as 100, the percent of substance left after 6 hours is 80.75% 

Problem 4 :

The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria  present in the culture initially, how many bacteria will be present at the end of 8th hour?

Note that the number of bacteria present in the culture doubles at the end of  successive hours.

Since it grows at the constant ratio "2", the growth is based is on geometric progression. 

We have to use the formula given below to find the no. of bacteria present at the end of 8th hour. 

A  =  ab x

a  =  30

b  =  2

x  =  8

Then, we have

A  =  30(2 8 )

A  =  30(256 )

A  =  7680

So, the number of bacteria at the end of 8th hour is 7680.

Problem 5 :

A sum of money placed at compound interest doubles itself in 3 years. If interest is being compounded annually, in how many years will it amount to four times itself ?

Let "P" be the amount invested initially.  From the given information, P becomes 2P in 3 years. 

Since the investment is in compound interest, for the 4th year, the principal will be 2P.  And 2P becomes 4P (it doubles itself) in the next 3 years. 

Therefore, at the end of 6 years accumulated value will be 4P.  So, the amount deposited will amount to 4 times itself in 6 years.

Related Topics

Doubling-Time Growth Formula

Half-Life Decay Formula

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