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Exponential Growth and Decay

A free Algebra I lesson from the “Exponential Functions and Sequences” unit, with a worked example and practice problems including step-by-step solutions.

For exponential growth at rate r per period: y = a * (1 + r)^t. For decay (loss) at rate r per period: y = a * (1 - r)^t. Here a is the starting amount and t is the number of periods. Compound interest, population growth, and depreciation all fit one of these forms.

What you'll learn

Why it matters: Bank accounts (compound interest), car values (depreciation), medicine doses (half-life), and bacteria cultures all use this model.

Worked example

Problem. $1000 invested at 5% annual compound interest. Value after 3 years?

  1. y = 1000 * (1.05)^3.
  2. (1.05)^3 = 1.157625.
  3. y = 1000 * 1.157625 = 1157.625.

Answer: 1157.625

Practice problems

1. $500 at 4% annual compound interest. Value after 2 years? (Round to 2 decimal places.)

Show solution
  1. Warm-up: First identify exactly what the question is asking: $500 at 4% annual compound interest. Value after 2 years? (Round to 2 decimal places.)
  2. For decimals, keep place value aligned and use estimation to make sure the decimal point is reasonable.
  3. y = 500 * (1.04)^2 = 500 * 1.0816 = 540.80.
  4. Check the result by substituting or estimating: the response should match 540.80 and make sense in the original problem.

Answer: 540.80

2. A car worth $20,000 depreciates 10% per year. Value after 1 year?

Show solution
  1. Warm-up: First identify exactly what the question is asking: A car worth $20,000 depreciates 10% per year. Value after 1 year?
  2. Choose the operation or relationship that matches the wording, then carry it out one clear step at a time.
  3. y = 20000 * (1 - 0.10) = 20000 * 0.90 = 18000.
  4. Check the result by substituting or estimating: the response should match 18000 and make sense in the original problem.

Answer: 18000

3. 200 bacteria double every hour. Count after 3 hours?

Show solution
  1. Warm-up: First identify exactly what the question is asking: 200 bacteria double every hour. Count after 3 hours?
  2. Choose the operation or relationship that matches the wording, then carry it out one clear step at a time.
  3. y = 200 * 2^3 = 200 * 8 = 1600.
  4. Check the result by substituting or estimating: the response should match 1600 and make sense in the original problem.

Answer: 1600

4. An 80-gram radioactive sample halves every year. Amount after 3 years?

Show solution
  1. Core Practice: First identify exactly what the question is asking: An 80-gram radioactive sample halves every year. Amount after 3 years?
  2. Choose the operation or relationship that matches the wording, then carry it out one clear step at a time.
  3. y = 80 * (1/2)^3 = 80 * 1/8 = 10.
  4. Check the result by substituting or estimating: the response should match 10 and make sense in the original problem.

Answer: 10

5. $1000 at 8% annual compound. Value after 2 years? (Round to 2 decimal places.)

Show solution
  1. Core Practice: First identify exactly what the question is asking: $1000 at 8% annual compound. Value after 2 years? (Round to 2 decimal places.)
  2. For decimals, keep place value aligned and use estimation to make sure the decimal point is reasonable.
  3. y = 1000 * (1.08)^2 = 1000 * 1.1664 = 1166.40.
  4. Check the result by substituting or estimating: the response should match 1166.40 and make sense in the original problem.

Answer: 1166.40

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