Exponential Function Growth And Decay Word Problems

The world is full of phenomena that change over time. Some things grow, like populations of bacteria or investments in a bank account. Other things decay, like radioactive substances or the temperature of a hot cup of coffee. Often, these changes follow a predictable pattern called exponential growth or decay. Understanding how to model these phenomena with exponential functions is a powerful tool in many fields, from finance to physics.

Exponential functions are characterized by their rapid rate of change. Unlike linear functions that change at a constant rate, exponential functions change at a rate proportional to their current value. This means that as the value increases, the rate of increase also increases, leading to explosive growth. Conversely, as the value decreases, the rate of decrease also decreases, leading to a gradual decay towards zero. This article delves into the fascinating world of exponential function growth and decay word problems, providing you with the knowledge and skills to solve them.

Understanding Exponential Functions

Before we jump into the problems, let's review the basics of exponential functions. An exponential function has the general form:

y = a * b^(x)

Where:

• y is the final amount
• a is the initial amount
• b is the growth or decay factor
• x is the time or number of periods

When b > 1, the function represents exponential growth. This means that the value of y increases as x increases. The larger the value of b, the faster the growth rate.

When 0 < b < 1, the function represents exponential decay. This means that the value of y decreases as x increases. The closer the value of b is to 0, the faster the decay rate.

It's important to note that in many real-world scenarios, the growth or decay is expressed as a percentage. In such cases, the growth or decay factor b can be calculated as follows:

• For growth: b = 1 + r, where r is the growth rate expressed as a decimal
• For decay: b = 1 - r, where r is the decay rate expressed as a decimal

For example, if something grows at a rate of 5% per year, then r = 0.05 and b = 1.05. If something decays at a rate of 10% per year, then r = 0.10 and b = 0.90.

Solving Exponential Growth Word Problems

Let's start with some examples of exponential growth word problems.

Example 1:

A population of bacteria doubles every hour. If the initial population is 100, how many bacteria will there be after 5 hours?

Solution:

1. Identify the variables:

   • a (initial amount) = 100
   • b (growth factor) = 2 (since the population doubles)
   • x (time) = 5 hours

2. Plug the variables into the exponential growth formula:

   • y = a * b^(x)
   • y = 100 * 2^(5)
   • y = 100 * 32
   • y = 3200

3. Answer: After 5 hours, there will be 3200 bacteria.

Example 2:

An investment of $5000 earns 8% interest compounded annually. How much will the investment be worth after 10 years?

Solution:

1. Identify the variables:

   • a (initial amount) = $5000
   • r (growth rate) = 8% = 0.08
   • b (growth factor) = 1 + r = 1 + 0.08 = 1.08
   • x (time) = 10 years

2. Plug the variables into the exponential growth formula:

   • y = a * b^(x)
   • y = 5000 * (1.08)^(10)
   • y = 5000 * 2.158924997 (approximately)
   • y = 10794.62 (approximately)

3. Answer: After 10 years, the investment will be worth approximately $10,794.62.

Example 3:

The population of a town is increasing at a rate of 3% per year. If the population in 2020 was 20,000, what will the population be in 2030?

Solution:

1. Identify the variables:

   • a (initial amount) = 20,000
   • r (growth rate) = 3% = 0.03
   • b (growth factor) = 1 + r = 1 + 0.03 = 1.03
   • x (time) = 2030 - 2020 = 10 years

2. Plug the variables into the exponential growth formula:

   • y = a * b^(x)
   • y = 20000 * (1.03)^(10)
   • y = 20000 * 1.343916379 (approximately)
   • y = 26878.33 (approximately)

3. Answer: The population in 2030 will be approximately 26,878.

Solving Exponential Decay Word Problems

Now, let's move on to exponential decay word problems.

Example 1:

A radioactive substance has a half-life of 20 years. If you start with 100 grams of the substance, how much will remain after 60 years?

Solution:

Half-life is the time it takes for half of a substance to decay. To solve this problem, we first need to determine the decay factor. Since the substance halves every 20 years, after 20 years, you'll have 50% left. Thus, the decay factor applies over a 20 year period.

1. Identify the variables:

   • a (initial amount) = 100 grams
   • x (time) = 60 years
   • Since we are dealing with half-life, it's easiest to think of x as the number of half-lives that have occurred. Since the half life is 20 years and the total time is 60 years, the number of half-lives is 60/20 = 3. Therefore, x = 3 (number of half-lives)
   • b (decay factor) = 0.5 (since the substance halves)

2. Plug the variables into the exponential decay formula:

   • y = a * b^(x)
   • y = 100 * (0.5)^(3)
   • y = 100 * 0.125
   • y = 12.5

3. Answer: After 60 years, 12.5 grams of the substance will remain.

Example 2:

The value of a car depreciates at a rate of 15% per year. If the car initially costs $25,000, what will its value be after 5 years?

Solution:

1. Identify the variables:

   • a (initial amount) = $25,000
   • r (decay rate) = 15% = 0.15
   • b (decay factor) = 1 - r = 1 - 0.15 = 0.85
   • x (time) = 5 years

2. Plug the variables into the exponential decay formula:

   • y = a * b^(x)
   • y = 25000 * (0.85)^(5)
   • y = 25000 * 0.443705281 (approximately)
   • y = 11092.63 (approximately)

3. Answer: After 5 years, the car's value will be approximately $11,092.63.

Example 3:

A hot cup of coffee cools at a rate of 8% per minute. If the initial temperature of the coffee is 90°C, what will its temperature be after 10 minutes?

Solution:

1. Identify the variables:

   • a (initial amount) = 90°C
   • r (decay rate) = 8% = 0.08
   • b (decay factor) = 1 - r = 1 - 0.08 = 0.92
   • x (time) = 10 minutes

2. Plug the variables into the exponential decay formula:

   • y = a * b^(x)
   • y = 90 * (0.92)^(10)
   • y = 90 * 0.434388555 (approximately)
   • y = 39.09 (approximately)

3. Answer: After 10 minutes, the coffee's temperature will be approximately 39.09°C.

More Complex Problems and Variations

Now that you have a grasp of the basic principles, let's tackle some more complex scenarios and variations of exponential growth and decay problems.

1. Finding the Time (x):

Sometimes, instead of finding the final amount (y), you might be asked to find the time (x) it takes for something to reach a certain value. This requires using logarithms to solve for x.

Example:

A population of bacteria starts at 500 and grows at a rate of 10% per hour. How long will it take for the population to reach 2000?

Solution:

1. Identify the variables:

   • a (initial amount) = 500
   • y (final amount) = 2000
   • r (growth rate) = 10% = 0.10
   • b (growth factor) = 1 + r = 1 + 0.10 = 1.10
   • x (time) = unknown

2. Plug the variables into the exponential growth formula:

   • y = a * b^(x)
   • 2000 = 500 * (1.10)^(x)

3. Solve for x:

   • Divide both sides by 500: 4 = (1.10)^(x)
   • Take the logarithm of both sides (you can use any base, but the natural logarithm or base-10 logarithm are common): ln(4) = ln((1.10)^(x))
   • Use the logarithm property ln(a^b) = b * ln(a): ln(4) = x * ln(1.10)
   • Divide both sides by ln(1.10): x = ln(4) / ln(1.10)
   • Calculate: x ≈ 14.89

4. Answer: It will take approximately 14.89 hours for the population to reach 2000.

2. Finding the Growth/Decay Rate (r):

Sometimes, you might be given the initial and final amounts, as well as the time, and asked to find the growth or decay rate.

Example:

An investment of $1000 grows to $1600 in 8 years. What is the annual growth rate?

Solution:

1. Identify the variables:

   • a (initial amount) = $1000
   • y (final amount) = $1600
   • x (time) = 8 years
   • b (growth factor) = unknown
   • r (growth rate) = unknown

2. Plug the variables into the exponential growth formula:

   • y = a * b^(x)
   • 1600 = 1000 * b^(8)

3. Solve for b:

   • Divide both sides by 1000: 1.6 = b^(8)
   • Take the 8th root of both sides: b = (1.6)^(1/8)
   • Calculate: b ≈ 1.0616

4. Solve for r:

   • Since b = 1 + r, then r = b - 1
   • r = 1.0616 - 1 = 0.0616

5. Convert to percentage:

   • r = 0.0616 * 100% = 6.16%

6. Answer: The annual growth rate is approximately 6.16%.

3. Continuous Growth/Decay:

In some cases, growth or decay happens continuously, rather than at discrete intervals (like annually or hourly). In these cases, we use the following formula:

y = a * e^(kt)

Where:

• y is the final amount
• a is the initial amount
• e is the mathematical constant approximately equal to 2.71828
• k is the continuous growth or decay rate
• t is the time

If k > 0, it represents continuous growth. If k < 0, it represents continuous decay.

Example:

A population of bacteria grows continuously at a rate of 15% per hour. If the initial population is 200, how many bacteria will there be after 6 hours?

Solution:

1. Identify the variables:

   • a (initial amount) = 200
   • k (continuous growth rate) = 15% = 0.15
   • t (time) = 6 hours

2. Plug the variables into the continuous growth formula:

   • y = a * e^(kt)
   • y = 200 * e^(0.15 * 6)
   • y = 200 * e^(0.9)
   • y = 200 * 2.459603111 (approximately)
   • y = 491.92 (approximately)

3. Answer: After 6 hours, there will be approximately 491.92 bacteria.

Common Mistakes to Avoid

• Confusing growth and decay: Always carefully read the problem to determine whether the quantity is increasing or decreasing.
• Incorrectly calculating the growth/decay factor: Remember that for growth, b = 1 + r, and for decay, b = 1 - r.
• Forgetting to convert percentages to decimals: Always divide the percentage by 100 before using it in the formula.
• Using the wrong formula: Make sure you use the correct formula for exponential growth/decay based on the problem's context. When dealing with continuous growth/decay, remember to use the formula with e.
• Rounding too early: Avoid rounding intermediate calculations, as this can lead to inaccuracies in the final answer. Round only at the very end.
• Not understanding half-life: Remember that half-life problems require you to think about how many "half-life periods" have elapsed.

Conclusion

Exponential growth and decay are powerful mathematical concepts that describe a wide range of phenomena in the real world. By understanding the underlying formulas and practicing with various word problems, you can develop the skills to model and analyze these phenomena. Remember to carefully identify the variables, choose the correct formula, and avoid common mistakes. With practice, you'll become proficient in solving exponential growth and decay word problems. Now that you have mastered the art of solving exponential growth and decay problems, you can confidently apply these skills in various fields, from finance and biology to physics and environmental science. Keep practicing, and you'll be amazed at how often these concepts appear in the world around you.

How do you think these concepts could be applied to current global issues like climate change or population growth? Are you interested in trying to model some real-world scenarios using exponential functions?