Exponential Function Growth And Decay Word Problems

Let's dive into the world of exponential functions, exploring how they model growth and decay in various real-world scenarios. Exponential functions are powerful tools for understanding phenomena that change at a rate proportional to their current value. They are widely used in fields like finance, biology, physics, and even social sciences. By understanding the principles behind exponential growth and decay, you can analyze and predict trends in various contexts.

Understanding Exponential Functions

An exponential function takes the general form:

f(x) = a * b^x

Where:

• f(x) is the value of the function at a given time x.
• a is the initial value of the function (the value when x = 0).
• b is the base, representing the growth or decay factor.
• x is the independent variable, often representing time.

Exponential Growth: Occurs when b > 1. The function's value increases exponentially as x increases. The larger the value of b, the faster the growth.

Exponential Decay: Occurs when 0 < b < 1. The function's value decreases exponentially as x increases, approaching zero. The closer b is to zero, the faster the decay.

Key Components in Growth and Decay Word Problems

To effectively solve exponential growth and decay word problems, it's crucial to identify these key components:

• Initial Value (a): This is the starting amount or quantity. It's the value of the dependent variable at time zero.

• Growth/Decay Rate (r): This is the percentage increase (for growth) or decrease (for decay) per time period. When working with the base b, remember that:

  • For growth: b = 1 + r
  • For decay: b = 1 - r

• Time (t): This is the duration over which the growth or decay occurs. It must be consistent with the growth/decay rate's time period.

• The Dependent Variable (y or f(t)): This is the amount or quantity after time t. It's what you're trying to find or what's given in the problem.

Solving Exponential Growth Word Problems

Let's tackle some examples to illustrate how to solve exponential growth word problems.

Example 1: Bacterial Growth

A culture of bacteria initially contains 500 bacteria. The bacteria population doubles every hour. How many bacteria are present after 6 hours?

• Identify the components:
  • Initial value (a): 500
  • Growth rate: Doubles every hour, so b = 2
  • Time (t): 6 hours
• Set up the equation:
  • f(t) = a * b^t
  • f(6) = 500 * 2^6
• Solve for f(6):
  • f(6) = 500 * 64
  • f(6) = 32,000

Therefore, there are 32,000 bacteria after 6 hours.

Example 2: Investment Growth

You invest $1000 in an account that earns 5% interest compounded annually. How much money will you have after 10 years?

• Identify the components:
  • Initial value (a): $1000
  • Growth rate (r): 5% = 0.05. Therefore, b = 1 + 0.05 = 1.05
  • Time (t): 10 years
• Set up the equation:
  • f(t) = a * b^t
  • f(10) = 1000 * (1.05)^10
• Solve for f(10):
  • f(10) = 1000 * 1.62889 (approximately)
  • f(10) = $1628.89 (approximately)

Therefore, you will have approximately $1628.89 after 10 years.

Example 3: Population Growth

The population of a town is increasing at a rate of 3% per year. If the initial population is 10,000, what will the population be in 15 years?

• Identify the components:
  • Initial value (a): 10,000
  • Growth rate (r): 3% = 0.03. Therefore, b = 1 + 0.03 = 1.03
  • Time (t): 15 years
• Set up the equation:
  • f(t) = a * b^t
  • f(15) = 10000 * (1.03)^15
• Solve for f(15):
  • f(15) = 10000 * 1.55797 (approximately)
  • f(15) = 15,579.7 (approximately)

Therefore, the population will be approximately 15,580 in 15 years (round to the nearest whole number since we're dealing with people).

Solving Exponential Decay Word Problems

Now, let's examine exponential decay word problems.

Example 1: Radioactive Decay

The half-life of a radioactive substance is 50 years. If you start with 100 grams of the substance, how much will remain after 200 years?

Understanding Half-Life: Half-life is the time it takes for half of the substance to decay. This gives us a direct way to find the decay factor, b. After one half-life, the amount remaining is half of the initial amount.

• Identify the components:

  • Initial value (a): 100 grams
  • Half-life: 50 years. This means after 50 years, we have 50 grams remaining. We use this to find b.
  • Time (t): 200 years

• Find the decay factor (b):

  • We know that after 50 years, the amount is halved. So, f(50) = 50.
  • Using the general formula: 50 = 100 * b^50
  • Divide both sides by 100: 0.5 = b^50
  • Take the 50th root of both sides: b = (0.5)^(1/50) ≈ 0.98614

• Set up the equation:

  • f(t) = a * b^t
  • f(200) = 100 * (0.98614)^200

• Solve for f(200):

  • f(200) = 100 * 0.0625 (approximately)
  • f(200) = 6.25 grams

Therefore, approximately 6.25 grams of the substance will remain after 200 years. Notice that 200 years is four half-lives (200/50 = 4). So, the initial amount is halved four times: 100 -> 50 -> 25 -> 12.5 -> 6.25.

Example 2: Depreciation

A car is purchased for $25,000. It depreciates at a rate of 15% per year. What will its value be after 5 years?

• Identify the components:
  • Initial value (a): $25,000
  • Decay rate (r): 15% = 0.15. Therefore, b = 1 - 0.15 = 0.85
  • Time (t): 5 years
• Set up the equation:
  • f(t) = a * b^t
  • f(5) = 25000 * (0.85)^5
• Solve for f(5):
  • f(5) = 25000 * 0.443705 (approximately)
  • f(5) = $11,092.63 (approximately)

Therefore, the car's value will be approximately $11,092.63 after 5 years.

Example 3: Cooling

A cup of coffee is initially 90°C. The room temperature is 20°C. According to Newton's Law of Cooling, the coffee cools at a rate proportional to the difference between its temperature and the room temperature. If the coffee cools to 70°C in 10 minutes, what will its temperature be after 20 minutes?

Newton's Law of Cooling: This law states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature (i.e. the temperature of its surroundings). The formula is:

`T(t) = T_s + (T_0 - T_s) * e^(-kt)`

Where:
*   `T(t)` is the temperature at time *t*
*   `T_s` is the surrounding temperature
*   `T_0` is the initial temperature
*   `k` is a constant that depends on the object and its surroundings

• Identify the components:

  • Surrounding temperature (Ts): 20°C
  • Initial temperature (T0): 90°C
  • Temperature at t=10 minutes (T(10)): 70°C
  • Time (t): 20 minutes (we want to find T(20))

• First, find the constant k:

  • 70 = 20 + (90 - 20) * e^(-10k)
  • 50 = 70 * e^(-10k)
  • 50/70 = e^(-10k)
  • 5/7 = e^(-10k)
  • Take the natural logarithm of both sides: ln(5/7) = -10k
  • k = -ln(5/7) / 10 ≈ 0.03365

• Now, find the temperature at t=20 minutes:

  • T(20) = 20 + (90 - 20) * e^(-0.03365 * 20)
  • T(20) = 20 + 70 * e^(-0.673)
  • T(20) = 20 + 70 * 0.5099 (approximately)
  • T(20) = 20 + 35.693 (approximately)
  • T(20) = 55.693°C (approximately)

Therefore, the coffee's temperature will be approximately 55.69°C after 20 minutes.

General Strategies for Solving Exponential Word Problems

Here's a summary of strategies for tackling these problems:

1. Read Carefully: Understand the context of the problem. What is growing or decaying? What are the units involved?
2. Identify Key Information: Extract the initial value, growth/decay rate (or half-life), and time period from the problem statement.
3. Determine the Formula: Choose the appropriate formula for exponential growth (f(t) = a * b^t where b = 1+r) or decay (f(t) = a * b^t where b = 1-r, or using half-life to find b). For Newton's Law of Cooling, use T(t) = T_s + (T_0 - T_s) * e^(-kt).
4. Set Up the Equation: Substitute the identified values into the chosen formula.
5. Solve for the Unknown: Use algebraic techniques and a calculator to solve for the desired variable.
6. Check Your Answer: Does the answer make sense in the context of the problem? For example, in a decay problem, the final amount should be less than the initial amount.
7. Units: Ensure your answer has the correct units (e.g., grams, dollars, degrees Celsius, people).

Common Mistakes to Avoid

• Incorrectly Identifying Growth vs. Decay: Make sure you correctly determine whether the quantity is increasing (growth) or decreasing (decay). This affects how you calculate the base, b.
• Using the Wrong Formula: Be careful to use the correct formula for the specific situation. For example, don't use a simple exponential decay formula when Newton's Law of Cooling is required.
• Incorrectly Calculating the Base (b): Remember that for growth, b = 1 + r, and for decay, b = 1 - r. If given a half-life, you need to calculate b using the half-life information.
• Inconsistent Time Units: Ensure that the time unit used in the growth/decay rate matches the time unit for which you're calculating the final amount. If the rate is annual, the time must be in years. If the rate is monthly, the time must be in months.
• Rounding Errors: Avoid rounding intermediate calculations excessively, as this can lead to inaccuracies in the final answer. Keep several decimal places during calculations and round the final answer to an appropriate level of precision.
• Forgetting the Initial Value: Always include the initial value (a) in your equation. It's a crucial factor in determining the final amount.
• Misinterpreting the Question: Make sure you understand what the question is asking for. Are you asked to find the final amount, the time it takes to reach a certain amount, or the growth/decay rate?

Advanced Exponential Models

Beyond the basic exponential growth and decay models, there are more complex scenarios. Here are a few:

• Logistic Growth: This model describes growth that is initially exponential but slows down as it approaches a carrying capacity. The carrying capacity represents the maximum sustainable population or value. The formula is more complex than simple exponential growth.
• Compounded Interest (More Frequently Than Annually): When interest is compounded n times per year, the formula becomes: A = P(1 + r/n)^(nt), where A is the final amount, P is the principal, r is the annual interest rate, n is the number of times interest is compounded per year, and t is the number of years.
• Continuous Compounding: This is the theoretical limit of compounding as the number of compounding periods approaches infinity. The formula is: A = Pe^(rt), where e is Euler's number (approximately 2.71828).

These advanced models introduce more variables and require a deeper understanding of calculus and financial mathematics.

The Importance of Exponential Functions

Exponential functions are not just mathematical abstractions; they are fundamental tools for understanding and modeling the world around us. From predicting population growth and managing investments to understanding radioactive decay and modeling the spread of diseases, exponential functions provide valuable insights into a wide range of phenomena. By mastering the concepts and techniques presented in this article, you will be well-equipped to analyze and solve a variety of real-world problems involving exponential growth and decay. Understanding these concepts empowers you to make informed decisions in various aspects of life, from personal finance to understanding scientific advancements.