Exponential Decay And Growth Word Problems

Exponential growth and decay are mathematical concepts that describe the increase or decrease of a quantity over time. They are widely used in various fields, including finance, biology, physics, and more. Word problems involving exponential growth and decay often require understanding the underlying principles and applying them to real-world scenarios.

Understanding Exponential Growth and Decay

Exponential growth occurs when the rate of increase of a quantity is proportional to the current value of the quantity. In other words, the larger the quantity, the faster it grows. This phenomenon is often observed in populations, investments, and other areas where growth is self-reinforcing.

Exponential decay, on the other hand, occurs when the rate of decrease of a quantity is proportional to the current value of the quantity. In this case, the smaller the quantity, the slower it decays. This is commonly seen in radioactive decay, drug metabolism, and other processes where the quantity diminishes over time.

Formulas for Exponential Growth and Decay

The general formula for exponential growth and decay is:

y = a(1 + r)^t

Where:

• y is the final amount after time t.
• a is the initial amount.
• r is the rate of growth (if positive) or decay (if negative), expressed as a decimal.
• t is the time period.

For continuous exponential growth or decay, the formula is:

y = ae^(kt)

Where:

• y is the final amount after time t.
• a is the initial amount.
• e is the base of the natural logarithm (approximately 2.71828).
• k is the continuous growth rate (if positive) or decay rate (if negative).
• t is the time period.

Solving Exponential Growth Word Problems

Exponential growth word problems typically involve scenarios where a quantity increases over time at a constant percentage rate. Here are some examples:

Example 1: Population Growth

A population of bacteria doubles every hour. If the initial population is 100, how many bacteria will there be after 6 hours?

Solution:

In this problem, the population doubles every hour, so the growth rate is 100% or 1. The initial population is 100, and we want to find the population after 6 hours.

Using the formula for exponential growth:

y = a(1 + r)^t

Where:

• y is the final population.
• a is the initial population (100).
• r is the growth rate (1).
• t is the time period (6 hours).

Plugging in the values:

y = 100(1 + 1)^6
y = 100(2)^6
y = 100(64)
y = 6400

Therefore, after 6 hours, there will be 6400 bacteria.

Example 2: Investment Growth

An investment of $5,000 grows at an annual rate of 8% compounded annually. How much will the investment be worth after 10 years?

Solution:

In this problem, the investment grows at an annual rate of 8%, so the growth rate is 0.08. The initial investment is $5,000, and we want to find the value of the investment after 10 years.

Using the formula for exponential growth:

y = a(1 + r)^t

Where:

• y is the final value of the investment.
• a is the initial investment ($5,000).
• r is the growth rate (0.08).
• t is the time period (10 years).

Plugging in the values:

y = 5000(1 + 0.08)^10
y = 5000(1.08)^10
y ≈ 5000(2.1589)
y ≈ 10794.62

Therefore, after 10 years, the investment will be worth approximately $10,794.62.

Example 3: Continuous Compounding

Suppose you invest $1,000 in an account that pays 5% annual interest compounded continuously. How much will you have after 5 years?

Solution:

Using the formula for continuous compounding:

y = ae^(kt)

Where:

• y is the final amount.
• a is the initial amount ($1,000).
• e is the base of the natural logarithm (approximately 2.71828).
• k is the continuous growth rate (0.05).
• t is the time period (5 years).

Plugging in the values:

y = 1000 * e^(0.05 * 5)
y = 1000 * e^(0.25)
y ≈ 1000 * 1.284
y ≈ 1284.03

Therefore, after 5 years, you will have approximately $1,284.03.

Solving Exponential Decay Word Problems

Exponential decay word problems involve scenarios where a quantity decreases over time at a constant percentage rate.

Example 1: Radioactive Decay

A radioactive substance has a half-life of 50 years. If you start with 100 grams of the substance, how much will remain after 150 years?

Solution:

First, determine the decay rate. Since the substance has a half-life of 50 years, half of the substance remains after 50 years. Using the formula:

y = a(1 + r)^t

After 50 years (t=50), y = 0.5a, so:

0.  5a = a(1 + r)^50
1.  5 = (1 + r)^50

Taking the 50th root of both sides:

(0.5)^(1/50) = 1 + r
r ≈ -0.0137

So, the decay rate is approximately -0.0137. Now, we want to find the amount remaining after 150 years:

y = 100(1 - 0.0137)^150
y = 100(0.9863)^150
y ≈ 100 * 0.125
y ≈ 12.5

Therefore, after 150 years, approximately 12.5 grams of the substance will remain.

Alternatively, since 150 years is three half-lives (150/50 = 3), you can repeatedly halve the initial amount:

• After 50 years: 100 / 2 = 50 grams
• After 100 years: 50 / 2 = 25 grams
• After 150 years: 25 / 2 = 12.5 grams

Example 2: Depreciation

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

Solution:

In this problem, the car depreciates at a rate of 15%, so the decay rate is -0.15. The initial value is $25,000, and we want to find the value after 5 years.

Using the formula for exponential decay:

y = a(1 + r)^t

Where:

• y is the final value of the car.
• a is the initial value ($25,000).
• r is the decay rate (-0.15).
• t is the time period (5 years).

Plugging in the values:

y = 25000(1 - 0.15)^5
y = 25000(0.85)^5
y ≈ 25000 * 0.4437
y ≈ 11092.56

Therefore, after 5 years, the car will be worth approximately $11,092.56.

Example 3: Drug Metabolism

A drug is eliminated from the body at a rate of 20% per hour. If the initial dose is 100 mg, how much of the drug will remain in the body after 3 hours?

Solution:

In this problem, the drug is eliminated at a rate of 20%, so the decay rate is -0.20. The initial dose is 100 mg, and we want to find the amount remaining after 3 hours.

Using the formula for exponential decay:

y = a(1 + r)^t

Where:

• y is the final amount of the drug.
• a is the initial dose (100 mg).
• r is the decay rate (-0.20).
• t is the time period (3 hours).

Plugging in the values:

y = 100(1 - 0.20)^3
y = 100(0.8)^3
y = 100 * 0.512
y = 51.2

Therefore, after 3 hours, 51.2 mg of the drug will remain in the body.

Key Considerations When Solving Word Problems

When tackling exponential growth and decay word problems, keep the following points in mind:

1. Identify the initial amount (a): This is the starting quantity before any growth or decay occurs.

2. Determine the growth or decay rate (r or k): Pay close attention to whether the rate is given as a percentage or a decimal. If the problem involves continuous growth or decay, use the continuous rate (k). If the problem gives a percentage, convert it to a decimal by dividing by 100.

3. Determine the time period (t): Make sure the time period is consistent with the rate. For example, if the rate is annual, the time period should be in years.

4. Choose the correct formula: Use the formula for discrete exponential growth or decay: y = a(1 + r)^t. Use the formula for continuous exponential growth or decay: y = ae^(kt).

5. Pay attention to wording: Words like "doubles," "triples," or "halves" indicate specific growth or decay rates. For example, "doubles" means the growth rate is 100% (r = 1), and "halves" means after a specific time the remaining amount is 50% of the original.

6. Solve for the unknown variable: Once you have identified the values for a, r (or k), and t, plug them into the appropriate formula and solve for the unknown variable (usually y).

7. Check your answer: Does the answer make sense in the context of the problem? For example, if you are solving for the population of bacteria after a certain amount of time, the answer should be a positive number.

Advanced Applications and Problem-Solving Strategies

Here are some more complex scenarios and strategies:

Finding the Time Required

Sometimes, you may need to find the time required for a quantity to reach a certain level. This involves solving for 't' in the exponential growth or decay formulas, often requiring logarithms.

Example: How long will it take for an investment of $1,000 to double if it earns 6% annual interest compounded continuously?

Solution:

We want to find 't' when y = 2a (double the initial amount).

2a = ae^(kt)
2 = e^(0.06t)

Take the natural logarithm of both sides:

ln(2) = 0.06t
t = ln(2) / 0.06
t ≈ 11.55 years

So it will take approximately 11.55 years for the investment to double.

Using Half-Life

Half-life problems often appear in the context of radioactive decay but can apply to any scenario where a quantity decreases by half over a period.

Example: Carbon-14 has a half-life of 5,730 years. A fossil contains 30% of its original carbon-14. How old is the fossil?

Solution:

First, find the decay rate 'k' using the half-life:

0.  5 = e^(k * 5730)
ln(0.5) = k * 5730
k = ln(0.5) / 5730
k ≈ -0.000121

Now, we know that the fossil contains 30% of its original carbon-14, so:

0.  3 = e^(-0.000121 * t)
ln(0.3) = -0.000121 * t
t = ln(0.3) / -0.000121
t ≈ 9952 years

So the fossil is approximately 9,952 years old.

Applications in Finance

Exponential growth and decay are crucial in financial calculations such as compound interest, present value, and future value.

Example: You want to have $10,000 in 8 years. How much do you need to invest now if the account earns 7% annual interest compounded continuously?

Solution:

We need to find the initial amount 'a' given the final amount y = $10,000, the interest rate k = 0.07, and the time period t = 8 years.

10000 = ae^(0.07 * 8)
10000 = ae^(0.56)
a = 10000 / e^(0.56)
a ≈ 10000 / 1.7507
a ≈ 5712.21

Therefore, you need to invest approximately $5,712.21 now to have $10,000 in 8 years.

Conclusion

Exponential growth and decay word problems are prevalent in various fields and require a solid understanding of the underlying formulas and principles. By carefully identifying the initial amount, growth or decay rate, time period, and choosing the appropriate formula, you can successfully solve these problems. Remember to pay attention to wording, check your answers, and practice with various examples to improve your problem-solving skills. Understanding these concepts not only helps in academic settings but also provides valuable insights into real-world phenomena.