Exponential Growth and Decay Practice Worksheet - Part Two with six real-world math problems.

Let's solve each problem step by step using exponential growth and decay formulas.

The general formula for exponential change is:

$$
A = A_0 \cdot (1 + r)^t
$$

Where:
- $ A $ = final amount
- $ A_0 $ = initial amount
- $ r $ = rate of change (positive for growth, negative for decay)
- $ t $ = time

For decay, the formula becomes:
$$
A = A_0 \cdot (1 - r)^t
$$

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1. An adult takes 850 mg of ibuprofen. Each hour, the amount of ibuprofen in the system decreases by about 7%. How much ibuprofen is left after 5 hours?

This is exponential decay with:
- $ A_0 = 850 $ mg
- $ r = 0.07 $ (7% per hour)
- $ t = 5 $

$$
A = 850 \cdot (1 - 0.07)^5 = 850 \cdot (0.93)^5
$$

Calculate $ 0.93^5 $:
$$
0.93^5 \approx 0.6983
$$

$$
A \approx 850 \cdot 0.6983 \approx 593.56 \text{ mg}
$$

✔ Answer: About 593.6 mg of ibuprofen remains after 5 hours.

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2. A truck bought new for $31,000 depreciates at a rate of 5% per year. How much is the truck worth in 5 years?

Again, exponential decay:
- $ A_0 = 31,000 $
- $ r = 0.05 $
- $ t = 5 $

$$
A = 31,000 \cdot (1 - 0.05)^5 = 31,000 \cdot (0.95)^5
$$

$ 0.95^5 \approx 0.7738 $

$$
A \approx 31,000 \cdot 0.7738 \approx 23,987.80
$$

✔ Answer: The truck is worth about $23,988 after 5 years.

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3. On a scratch-off ticket for the state lottery, you win $10,000. Because you don't need the money right now, you decide to put it in a CD at the bank that pays 4% per year. How much will it be worth after 4 years?

Exponential growth:
- $ A_0 = 10,000 $
- $ r = 0.04 $
- $ t = 4 $

$$
A = 10,000 \cdot (1 + 0.04)^4 = 10,000 \cdot (1.04)^4
$$

$ 1.04^4 \approx 1.1699 $

$$
A \approx 10,000 \cdot 1.1699 = 11,699
$$

✔ Answer: The CD will be worth about $11,699 after 4 years.

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4. After college, you decide that you really want to buy a boat, so you start putting $100 into a savings account every month. The account earns 10% interest compounded monthly. How much will you have in your account after 5 years?

⚠️ This is not simple exponential growth — it's a savings annuity because you're making regular deposits.

Use the future value of an ordinary annuity formula:

$$
A = P \cdot \frac{(1 + r)^n - 1}{r}
$$

Where:
- $ P = 100 $ (monthly deposit)
- $ r = \frac{10\%}{12} = \frac{0.10}{12} \approx 0.008333 $ (monthly interest rate)
- $ n = 5 \times 12 = 60 $ months

$$
A = 100 \cdot \frac{(1 + 0.008333)^{60} - 1}{0.008333}
$$

First, compute $ (1.008333)^{60} \approx 1.6453 $

Then:
$$
A = 100 \cdot \frac{1.6453 - 1}{0.008333} = 100 \cdot \frac{0.6453}{0.008333} \approx 100 \cdot 77.43 \approx 7,743
$$

✔ Answer: You’ll have about $7,743 after 5 years.

> Note: If the question had said "you deposit $100 once" and it grows at 10% compounded monthly, then it would be different. But since it says “every month,” this is an annuity.

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5. Cancer cells in a patient’s body can multiply by about 12% per month. If a person has an estimated 10,000 cancer cells in their lungs, how many do they have after one year?

Exponential growth:
- $ A_0 = 10,000 $
- $ r = 0.12 $ (12% per month)
- $ t = 12 $ months

$$
A = 10,000 \cdot (1 + 0.12)^{12} = 10,000 \cdot (1.12)^{12}
$$

$ 1.12^{12} \approx 3.1058 $

$$
A \approx 10,000 \cdot 3.1058 = 31,058
$$

✔ Answer: About 31,058 cancer cells after one year.

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6. Certain kinds of bacteria can double every day. If we place only one of these types of bacteria in a petri dish for a month, how many bacteria will we have?

Assuming:
- Bacteria double every day → growth factor = 2
- Start with 1 bacterium
- Time = 30 days (a month)

$$
A = 1 \cdot 2^{30}
$$

$ 2^{10} = 1,024 $
$ 2^{20} = (2^{10})^2 = 1,048,576 $
$ 2^{30} = (2^{10})^3 = 1,073,741,824 $

✔ Answer: There will be 1,073,741,824 bacteria after 30 days.

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✔ Final Answers Summary:

1. Ibuprofen: ~593.6 mg
2. Truck value: ~$23,988
3. CD investment: ~$11,699
4. Savings account: ~$7,743
5. Cancer cells: ~31,058
6. Bacteria: 1,073,741,824

Let me know if you'd like these rounded differently or explained with graphs!