Exponential Function Word Problems With Answers

Exponential Function Word Problems with Answers

Exponential functions are mathematical models that describe situations where a quantity grows or decays at a rate proportional to its current value. Understanding these functions is crucial in various fields, from finance and biology to physics and computer science. This article explores a variety of exponential function word problems, providing detailed solutions and explanations to enhance your understanding and problem-solving skills. We'll cover different scenarios and strategies to tackle these types of problems effectively.

Understanding the Basics: The Exponential Function Formula

Before diving into word problems, let's solidify our understanding of the general exponential function formula:

y = ab^x

Where:

• y represents the final amount or value.
• a represents the initial amount or value.
• b represents the growth or decay factor (1 + r for growth, 1 - r for decay, where 'r' is the rate).
• x represents the time or number of periods.

Remember that 'r' should be expressed as a decimal (e.g., 5% = 0.05). If b > 1, we have exponential growth; if 0 < b < 1, we have exponential decay.

Types of Exponential Function Word Problems & Solved Examples

Let's now tackle several types of exponential function word problems, progressing in complexity.

1. Compound Interest Problems

Compound interest is a classic application of exponential functions. Interest earned is added to the principal, and subsequent interest is calculated on the accumulated amount.

Problem 1: You deposit $1,000 into a savings account that earns 5% annual interest compounded annually. How much money will you have in the account after 3 years?

Solution:

Here, a = $1000 (initial amount), r = 0.05 (interest rate), and x = 3 (number of years). The formula for compound interest is a variation of the general exponential function:

A = P(1 + r)^t

Where:

• A = the future value of the investment/loan, including interest
• P = the principal investment amount (the initial deposit or loan amount)
• r = the annual interest rate (decimal)
• n = the number of times that interest is compounded per year
• t = the number of years the money is invested or borrowed for

In this case, n=1 (compounded annually). Therefore:

A = 1000(1 + 0.05)³ = 1000(1.05)³ ≈ $1157.63

Answer: You will have approximately $1157.63 in the account after 3 years.

Problem 2: Suppose you invest $5,000 in an account that pays 4% annual interest compounded quarterly. How much will you have after 5 years?

Solution:

Here, a = $5000, r = 0.04, t = 5, and n = 4 (compounded quarterly). The formula becomes:

A = P(1 + r/n)^(nt)

A = 5000(1 + 0.04/4)^(4*5) = 5000(1.01)²⁰ ≈ $6083.26

Answer: You will have approximately $6083.26 after 5 years.

2. Population Growth Problems

Exponential functions are frequently used to model population growth.

Problem 3: The population of a city is currently 100,000 and is growing at a rate of 2% per year. What will the population be in 10 years?

Solution:

Here, a = 100,000, r = 0.02, and x = 10. Using the formula y = ab^x:

y = 100000(1 + 0.02)¹⁰ = 100000(1.02)¹⁰ ≈ 121899

Answer: The population will be approximately 121,899 in 10 years.

Problem 4: A bacterial colony starts with 500 bacteria and doubles in size every hour. How many bacteria will there be after 6 hours?

Solution:

Here, the growth factor is 2 (doubling), so b = 2. The initial amount is a = 500, and x = 6 hours.

y = 500 * 2⁶ = 500 * 64 = 32000

Answer: There will be 32,000 bacteria after 6 hours.

3. Radioactive Decay Problems

Radioactive decay follows an exponential decay model.

Problem 5: A radioactive substance has a half-life of 10 years. If you start with 100 grams, how much will remain after 30 years?

Solution:

The half-life means that after 10 years, half the substance remains. Therefore, the decay factor b = 0.5. a = 100 grams, and x = 30 years (3 half-lives).

y = 100 * (0.5)^30/10 = 100 * (0.5)³ = 100 * 0.125 = 12.5

Answer: 12.5 grams will remain after 30 years.

Problem 6: A certain element decays exponentially. If 100 grams decay to 25 grams in 10 years, what is the half-life of the element?

Solution:

We know that y = 25, a = 100, and x = 10. We need to find b:

25 = 100 * b¹⁰ 0.25 = b¹⁰ b = (0.25)^1/10 ≈ 0.87055

Since b = 1 - r, we can find r: r = 1 - b ≈ 0.12945. Now we can find the half-life by solving for x when y = 50:

50 = 100 * (0.87055)^x 0.5 = (0.87055)^x Taking the logarithm of both sides:

log(0.5) = x * log(0.87055) x ≈ 5

Answer: The half-life is approximately 5 years.

4. Cooling/Heating Problems (Newton's Law of Cooling)

Newton's Law of Cooling states that the rate of change of an object's temperature is proportional to the difference between its temperature and the ambient temperature. This can be modeled with an exponential function.

Problem 7: A cup of coffee cools from 90°C to 70°C in 10 minutes in a room at 20°C. What will its temperature be after 20 minutes?

Solution: This problem requires a slightly different approach. Newton's Law of Cooling is often expressed as:

T(t) = T_a + (T₀ - T_a)e^-kt

Where:

• T(t) is the temperature at time t
• T_a is the ambient temperature
• T₀ is the initial temperature
• k is a cooling constant

First, we need to find k. Using the given information:

70 = 20 + (90 - 20)e^-10k 50 = 70e^-10k e^-10k = 5/7 -10k = ln(5/7) k ≈ 0.0336

Now, we can find the temperature after 20 minutes:

T(20) = 20 + (90 - 20)e^-20(0.0336) ≈ 54.2°C

Answer: The temperature will be approximately 54.2°C after 20 minutes.

Advanced Exponential Function Word Problems

The following problems require a deeper understanding of logarithmic and exponential properties.

Problem 8: The value of a car depreciates exponentially. After 3 years, its value is $12,000, and after 5 years, its value is $9,000. What was the initial value of the car?

Solution:

Let V(t) be the value of the car after t years. We have:

V(t) = Ae^-kt

We have two equations:

12000 = Ae^-3k 9000 = Ae^-5k

Dividing the first equation by the second:

(12000/9000) = e^2k 4/3 = e^2k ln(4/3) = 2k k ≈ 0.1438

Substitute k back into either equation to find A. Using the first equation:

12000 = Ae^-3(0.1438) A ≈ $18,000

Answer: The initial value of the car was approximately $18,000.

These examples illustrate the diverse applications of exponential functions in solving real-world problems. Remember to carefully identify the initial value, growth/decay factor, and time period to correctly apply the appropriate formula and techniques. Practice solving a wide range of problems to solidify your understanding and build confidence in tackling more complex scenarios. Remember to always check your answers and ensure they are realistic within the context of the problem. Consistent practice is key to mastering exponential function word problems.