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Exponential Decay

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

Exponential decay uses a repeated multiplier between 0 and 1. This lesson is part of Precalculus: Advanced Functions, so the emphasis is on interpreting behavior, choosing the right representation, and explaining the result clearly rather than memorizing isolated algebra moves.

What you'll learn

Why it matters: Exponential and logarithmic models describe growth, decay, sound, pH, finance, and scientific scales.

Worked example

Problem. A quantity decreases by 50% each period. What multiplier represents this decay?

  1. Worked Example: First identify exactly what the question is asking: A quantity decreases by 50% each period. What multiplier represents this decay?
  2. For exponential situations, identify the starting value and the repeated multiplier before calculating.
  3. A 50% decrease leaves 50% of the amount.
  4. 50% as a decimal is 0.5.
  5. So the repeated multiplier is 0.5.
  6. Check the result by substituting or estimating: the response should match 0.5 and make sense in the original problem.

Answer: 0.5

Practice problems

1. A quantity decreases by 50% each period. What multiplier represents this decay?

Show solution
  1. Warm-up: First identify exactly what the question is asking: A quantity decreases by 50% each period. What multiplier represents this decay?
  2. For exponential situations, identify the starting value and the repeated multiplier before calculating.
  3. A 50% decrease leaves 50% of the amount.
  4. 50% as a decimal is 0.5.
  5. So the repeated multiplier is 0.5.
  6. Check the result by substituting or estimating: the response should match 0.5 and make sense in the original problem.

Answer: 0.5

2. For A(x) = 48*(1/2)^x, find A(4).

Show solution
  1. Warm-up: First identify exactly what the question is asking: For A(x) = 48*(1/2)^x, find A(4).
  2. For fractions, use equivalent forms, common denominators, or reciprocals depending on the operation being used.
  3. Substitute x = 4: A(4) = 48*(1/2)^4.
  4. (1/2)^4 = 1/16, so divide 48 by 16.
  5. 48/16 = 3.
  6. Check the result by substituting or estimating: the response should match 3 and make sense in the original problem.

Answer: 3

3. When a quantity has a multiplier of 1/2 each period, the time for it to drop to half its amount is called its:

Choices: half-life · doubling time · growth rate · asymptote

Show solution
  1. A multiplier of 1/2 halves the amount every period.
  2. The period over which it halves has a special name.
  3. That period is the half-life.

Answer: half-life

4. In A(x) = a*b^x, a base of b = 0.8 produces:

Choices: exponential decay · exponential growth · a constant function · linear growth

Show solution
  1. Decay happens when the base b satisfies 0 < b < 1.
  2. Here b = 0.8, which is between 0 and 1.
  3. So the model represents exponential decay.

Answer: exponential decay

5. For A(x) = 200*(0.5)^x, find the initial value A(0).

Show solution
  1. Core Practice: First identify exactly what the question is asking: For A(x) = 200*(0.5)^x, find the initial value A(0).
  2. Use inverse operations to isolate the unknown, and keep both sides balanced at every step.
  3. The initial value is the output when x = 0.
  4. Any nonzero base to the 0 power is 1, so (0.5)^0 = 1.
  5. A(0) = 200*1 = 200.
  6. Check the result by substituting or estimating: the response should match 200 and make sense in the original problem.

Answer: 200

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