6-3 Additional Practice Exponential Growth And Decay Answer Key

6-3 Additional Practice: Exponential Growth and Decay — Answers & Explanations

This comprehensive guide provides detailed answers and explanations for a hypothetical "6-3 Additional Practice" worksheet on exponential growth and decay. Since I don't have access to a specific worksheet with that title, I'll cover a wide range of problems commonly found in this topic at a high school or introductory college level. This will serve as a valuable resource for students looking to check their work and solidify their understanding of exponential functions.

Understanding Exponential Growth and Decay

Before diving into the practice problems, let's review the core concepts. Exponential growth and decay are modeled by equations of the form:

A = A₀ * b^(kt)

Where:

• A is the final amount or value.
• A₀ is the initial amount or value.
• b is the base (often e, the natural logarithm base, or 2).
• k is the growth/decay constant (positive for growth, negative for decay).
• t is time.

Growth: If 'k' is positive, the function represents exponential growth. The amount increases over time at an accelerating rate.

Decay: If 'k' is negative, the function represents exponential decay. The amount decreases over time at a decelerating rate.

Types of Problems and Solved Examples

We'll explore various problem types encountered in exponential growth and decay, providing detailed solutions and explanations.

Problem Type 1: Finding the Final Amount (Given Initial Amount, Rate, and Time)

Example: A population of bacteria starts with 1000 individuals and grows at a rate of 20% per hour. What is the population after 3 hours?

Solution:

Here, A₀ = 1000, k = 0.20 (20% expressed as a decimal), t = 3. We'll use the formula:

A = A₀ * (1 + k)^t

A = 1000 * (1 + 0.20)^3 = 1000 * (1.2)^3 = 1000 * 1.728 = 1728

Answer: The population after 3 hours is 1728 bacteria.

Explanation: We use (1 + k) because the population is increasing. If it were decreasing, we'd use (1 - k).

Problem Type 2: Finding the Initial Amount (Given Final Amount, Rate, and Time)

Example: After 5 years, an investment of $X grew to $15,000 with a 10% annual interest rate (compounded annually). Find the initial investment (X).

Solution:

Here, A = 15000, k = 0.10, t = 5. We need to solve for A₀:

15000 = A₀ * (1 + 0.10)^5

15000 = A₀ * (1.1)^5

15000 = A₀ * 1.61051

A₀ = 15000 / 1.61051 ≈ 9311.62

Answer: The initial investment was approximately $9311.62.

Explanation: We rearrange the formula to solve for A₀.

Problem Type 3: Finding the Growth/Decay Rate (Given Initial and Final Amounts, and Time)

Example: A radioactive substance decays from 100 grams to 50 grams in 10 years. What is the decay rate (k)?

Solution:

Here, A₀ = 100, A = 50, t = 10. We'll use the decay formula:

50 = 100 * (1 - k)^10

0.5 = (1 - k)^10

(0.5)^(1/10) = 1 - k

1 - k ≈ 0.933

k ≈ 0.067

Answer: The decay rate is approximately 6.7% per year.

Explanation: We use logarithms to solve for k. Taking the 10th root is equivalent to raising to the power of 1/10.

Problem Type 4: Finding the Time (Given Initial and Final Amounts, and Rate)

Example: A population grows exponentially at a rate of 5% per year. How long will it take for the population to double?

Solution:

Let's assume A₀ = 1 (it could be any value, as we're looking at the doubling time). Then A = 2. k = 0.05. We need to find t:

2 = 1 * (1 + 0.05)^t

2 = (1.05)^t

log(2) = t * log(1.05)

t = log(2) / log(1.05) ≈ 14.21

Answer: It will take approximately 14.21 years for the population to double.

Explanation: We use logarithms to solve for t. Any base logarithm can be used, as long as it's consistent.

Problem Type 5: Problems Involving Half-Life

Example: A radioactive isotope has a half-life of 200 years. If we start with 1 kg, how much remains after 600 years?

Solution:

Half-life means that half the substance decays in the given time. We can use the decay formula, but it's simpler to think in terms of half-lives.

After 200 years (1 half-life): 1 kg / 2 = 0.5 kg After 400 years (2 half-lives): 0.5 kg / 2 = 0.25 kg After 600 years (3 half-lives): 0.25 kg / 2 = 0.125 kg

Answer: 0.125 kg (or 125 grams) remains after 600 years.

Problem Type 6: Exponential Growth with Continuous Compounding

Example: An investment of $1000 earns 8% interest compounded continuously. What is the value after 5 years?

Solution: For continuous compounding, we use the formula:

A = A₀ * e^(rt)

Where:

• A = final amount
• A₀ = initial amount ($1000)
• r = annual interest rate (0.08)
• t = time in years (5)
• e = Euler's number (approximately 2.71828)

A = 1000 * e^(0.08 * 5) = 1000 * e^0.4 ≈ 1000 * 1.4918 ≈ 1491.82

Answer: The investment will be worth approximately $1491.82 after 5 years.

Problem Type 7: Modeling Real-World Scenarios

Example: The number of users on a new social media platform grows exponentially. On day 1, there are 100 users. On day 7, there are 5000 users. Find the growth rate and predict the number of users on day 10.

Solution:

1. Find k: We have A₀ = 100, A = 5000, t = 6 (days 1 to 7).

5000 = 100 * (1 + k)^6

50 = (1 + k)^6

(50)^(1/6) = 1 + k

k ≈ 1.29 - 1 = 0.29 or 29%

2. Predict users on day 10: t = 9 (days 1 to 10)

A = 100 * (1 + 0.29)^9 ≈ 100 * 10.04 ≈ 10040

Answer: The approximate growth rate is 29% per day, and we predict approximately 10040 users on day 10.

Further Practice and Resources

This guide covered various problem types in exponential growth and decay. To further enhance your understanding, try creating your own problems using different contexts (population growth, radioactive decay, compound interest, etc.). You can also search online for additional practice exercises or consult textbooks for further explanations and examples. Remember to practice regularly to master this crucial topic in mathematics. The key is to understand the underlying concepts and the application of the formulas. By working through diverse problem sets and seeking clarification when needed, you can build a solid foundation in exponential functions.