Exponential Growth and Decay

Grade 10 · mathematics · 100 practice problems · read aloud

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Exponential Growth and Decay

What is it and why is it useful? 🔍

Exponential growth and decay describe quantities that change by a fixed percentage over equal time periods. Unlike linear growth (which adds a constant amount), exponential growth multiplies by a constant factor. This is crucial for modeling real-world phenomena like population growth, radioactive decay, and compound interest.

How to Solve Problems: Step-by-Step

  1. Identify the type: Is it growth (increasing) or decay (decreasing)?
  2. Find the key values: Initial amount (A₀), growth/decay rate (r), and time (t).
  3. Use the formula: A = A₀(1 ± r)t
  4. Solve: Plug in the values and calculate carefully.
  5. Interpret: What does your answer mean in the context of the problem?

Worked Examples

Example 1: Population Growth

A town of 5,000 people grows at 3% per year. What's the population in 7 years?

Step 1: Growth problem (A₀ = 5000, r = 0.03, t = 7)

Step 2: Use A = A₀(1 + r)t

Step 3: A = 5000(1 + 0.03)7 = 5000(1.03)7

Step 4: A ≈ 5000 × 1.2299 ≈ 6,150 people

Example 2: Car Depreciation

A $25,000 car depreciates 15% annually. What's its value in 4 years?

Step 1: Decay problem (A₀ = 25000, r = 0.15, t = 4)

Step 2: Use A = A₀(1 - r)t

Step 3: A = 25000(1 - 0.15)4 = 25000(0.85)4

Step 4: A ≈ 25000 × 0.522 ≈ $13,050

Common Mistakes to Avoid ⚠️

  • Using r instead of (1±r): Don't write 5000(0.03)7 for growth! You must use 1.03.
  • Confusing growth/decay: Use (1 + r) for growth, (1 - r) for decay.
  • Rate conversion: Remember 3% = 0.03, not 3!
  • Order of operations: Calculate the exponent before multiplying.

Tips & Tricks

  • Memory aid: "GROWth means GO up" → use (1 + r)
  • Quick check: For decay, your multiplier (1-r) should be less than 1.
  • Estimate first: In Example 1, 3% for 7 years is roughly 21% total growth, so about 6,000 people. This helps catch calculator errors!

How to Practice

Start with identifying growth vs. decay in real-world scenarios. Create flashcards for the formula and its components. Solve mixed practice problems where you sometimes find the final amount, sometimes the initial amount or rate. Use online graphing tools to visualize how changing the rate affects the curve. Practice with both calculator and non-calculator problems to build number sense.

Practice problems

6 of the 100, worked through step by step — try them before opening the answer.

1 log₂(8) = ?

Hint: Think about what power you need to raise the base to in order to get the argument. For example, log₃(9) asks '3 to what power equals 9?'

Show the answer

Answer: 3

  1. Understand the meaning of log₂(8). The expression log₂(8) means: "To what power must we raise 2 to get 8?"
  2. Write this as an equation. Let x = log₂(8). By definition, this means: 2^x = 8.
  3. Express 8 as a power of 2. We know that 8 = 2 × 2 × 2 = 2^3.
  4. Substitute 8 with 2^3 in the equation. So: 2^x = 2^3.
  5. Since the bases are the same (base 2), we can equate the exponents. Therefore, x = 3.
  6. Conclusion. So, log₂(8) = 3. Final answer: 3

We are asked: log₂(8) = ?

2 log₃(81) = ?

Hint: Think about what power you need to raise the base to in order to get the given number. For example, log₂(16) asks: 2 to what power equals 16?

Show the answer

Answer: 4

  1. We need to find the exponent x such that 3^x = 81
  2. Recognize that 81 = 3 × 3 × 3 × 3 = 3^4
  3. Therefore, 3^x = 3^4
  4. Since the bases are equal, the exponents must be equal: x = 4
  5. The answer is 4

3 A population of 3,125 bacteria decreases by 20% each hour. How many bacteria remain after 3 hours?

Hint: Identify the initial amount and the decay factor. A decrease of 20% means the population retains 80% each hour. Write an exponential decay model and evaluate it at the given time.

Show the answer

Answer: 1,600

  1. Initial population a = 3125. Decay rate = 20% = 0.20, so decay factor b = 1 - 0.20 = 0.80. Model: P(t) = 3125 * (0.80)^t.
  2. For t = 3 hours: P(3) = 3125 * (0.80)^3.
  3. Compute (0.80)^3 = 0.80 * 0.80 * 0.80 = 0.512.
  4. Multiply: 3125 * 0.512 = 1600.

The answer is 1,600 bacteria.

4 A population of 8,200 bacteria grows at a rate of 15% per hour. Write the exponential growth model and find the population after 4 hours.

Hint: Identify the initial amount and the growth factor. The growth factor is 1 plus the decimal form of the growth rate. Then substitute the time into the exponential model.

Show the answer

Answer: 14,342

  1. The initial population is a = 8200. The growth rate is 15% = 0.15, so the growth factor is b = 1 + 0.15 = 1.15. The exponential growth model is P(t) = 8200(1.15)^t.
  2. To find the population after 4 hours, substitute t = 4: P(4) = 8200(1.15)^4.
  3. Calculate (1.15)^4: 1.15^2 = 1.3225, then 1.3225^2 = 1.74900625.
  4. Multiply by 8200: 8200 × 1.74900625 = 14341.85125.
  5. Round to the nearest whole number: 14342.

The answer is 14342.

5 A sample of a radioactive isotope decays such that its mass after t years is given by M(t) = 250(0.5)^(t/9). What is the mass of the sample after 27 years?

Hint: Identify the initial mass and the half-life from the formula. Then substitute the given time into the exponent, remembering that the exponent represents the number of half-lives that have passed.

Show the answer

Answer: 31.25

  1. The formula is M(t) = 250(0.5)^(t/9). Here, 250 is the initial mass in grams, and the half-life is 9 years.
  2. We need the mass after 27 years, so substitute t = 27.
  3. M(27) = 250(0.5)^(27/9)
  4. Simplify the exponent: 27/9 = 3.
  5. So M(27) = 250(0.5)^3
  6. Calculate (0.5)^3 = 0.5 * 0.5 * 0.5 = 0.125
  7. Multiply: 250 * 0.125 = 31.25
  8. The mass after 27 years is 31.25 grams. Final answer: 31.25

6 The population of a town is 2500 and it grows at a rate of 5% per year. Write an exponential model for the population after t years and find the population after 10 years.

Hint: Identify the initial amount and the growth factor. For a percent increase, the growth factor is 1 plus the decimal form of the percent. Then substitute the given time into the exponential model.

Show the answer

Answer: 4072

  1. Identify the initial population a = 2500. The growth rate is 5% per year, so as a decimal it is 0.05. The growth factor b = 1 + 0.05 = 1.05.
  2. The exponential growth model is P(t) = a * b^t = 2500 * (1.05)^t.
  3. To find the population after 10 years, substitute t = 10: P(10) = 2500 * (1.05)^10.
  4. Calculate (1.05)^10. 1.05^2 = 1.1025, 1.05^4 = (1.1025)^2 = 1.21550625, 1.05^8 = (1.21550625)^2 = 1.4774554437890625, then 1.05^10 = 1.05^8 * 1.05^2 = 1.4774554437890625 * 1.1025 = 1.6288946267774414.
  5. Multiply by 2500: 2500 * 1.6288946267774414 = 4072.2365669436035.
  6. Round to the nearest whole number since population is a count: 4072.

The answer is 4072.

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