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MathChapter 12: Composition, Recursion, and Exponential Functions

Notes

Exponential Growth and Decay Notes

Translate percentages, doubling times, and half-lives into source-aligned exponential models and evaluate or compare them accurately.

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Repeated percentage change is multiplicative. Growth uses a factor above one, decay uses a factor between zero and one, doubling counts factors of two, and half-life counts factors of one half.

Learning objectives

  • Convert a percent increase or decrease to the correct multiplier.
  • Build and evaluate \(P(1\pm r)^t\) models.
  • Interpret \(t/d\) as the number of doubling or half-life intervals.
  • Compare two exponential models without premature rounding.

Percentage growth and decay

Annual or per-period percentage models
\[\text{growth: }A=P(1+r)^t,\qquad \text{decay: }A=P(1-r)^t\]

Write \(r\) as a decimal and match \(t\) to the model's period unit.

Translate situations into exponential multipliers and models
SituationRateMultiplierModel
Increase each period\(r\)\(1+r\)\(P(1+r)^t\)
Decrease each period\(r\)\(1-r\)\(P(1-r)^t\)
Double every \(d\) unitsNot a percent input\(2\) per interval\(P\cdot2^{t/d}\)
Half-life \(d\)Not a percent input\(\frac12\) per interval\(P(\frac12)^{t/d}\)
Worked example

Build and evaluate a decay model

A device worth \(24000\) loses \(9\%\) of its value each year. Find its modeled value after \(4\) years.

  1. Initial value

    \(P=24000\).

  2. Decay factor

    \(1-r=1-0.09=0.91\).

  3. Substitute

    \(A=24000(0.91)^4\).

  4. Calculate once

    \(A\approx16457.99\).

The modeled value is approximately \(16457.99\).

Doubling-time growth

Doubling model
\[A=P\cdot2^{t/d}\]

The exponent \(t/d\) counts how many doubling intervals have elapsed.

Doubling timelineEvery interval of length d multiplies the current amount by two.
  1. \(t=0:\ P\)

    Initial amount

  2. \(t=d:\ 2P\)

    Multiply the current amount by two.

  3. \(t=2d:\ 4P\)

    Multiply by two again.

  4. \(t=3d:\ 8P\)

    Three doublings produce eight times the start.

Half-life decay

Half-life model
\[A=P\left(\frac12\right)^{t/d}\]

Every interval of length \(d\) halves the amount currently remaining.

Half-life timelineEvery interval of length d multiplies the amount then remaining by one half.
  1. \(t=0:\ P\)

    Initial amount

  2. \(t=d:\ \frac P2\)

    One half remains.

  3. \(t=2d:\ \frac P4\)

    One half of the previous amount remains.

  4. \(t=3d:\ \frac P8\)

    Three half-lives leave one eighth.

  5. \(t=4d:\ \frac P{16}\)

    Four half-lives leave one sixteenth.

Find an unknown rate without logarithms

Worked example

Use a solvable square relationship

An amount grows from \(1000\) to \(1440\) in two annual periods at rate \(r\). Find \(r\).

  1. Set up

    \(1440=1000(1+r)^2\).

  2. Divide

    \((1+r)^2=1.44\).

  3. Use the positive factor

    \(1+r=1.2\), because a growth multiplier is positive.

  4. Solve

    \(r=0.2=20\%\).

The annual growth rate is \(20\%\).

Build a model from context

  1. Identify the start

    This becomes \(P\), so the model must return \(P\) at \(t=0\).

  2. Identify the change

    Convert a percent to a decimal, or identify doubling/half-life language.

  3. Build the factor

    Use \(1+r\), \(1-r\), \(2\), or \(\frac12\) as appropriate.

  4. Count periods

    Use \(t\) for per-period models and \(t/d\) for doubling or half-life intervals.

  5. Validate

    Check the model at \(t=0\) and after one stated interval.

Common mistakes and traps

  • Writing \(6\%\) as \(6\) instead of \(0.06\).
  • Using \(r\) rather than \(1+r\) or \(1-r\).
  • Replacing repeated percentage change with a linear model.
  • Treating doubling as adding the original amount every interval.
  • Subtracting the same fixed amount for each half-life.
  • Using \(td\) instead of \(t/d\) in a doubling or half-life exponent.
  • Rounding intermediate exponential values too early.
Mini check

Choose the correct multiplier

A quantity decreases by \(14\%\) each month. Which multiplier belongs in its model?

  1. \(0.14\)
  2. \(0.86\)
  3. \(1.14\)
  4. \(14\)
Show answer and explanation

Answer: \(0.86\)

A decrease keeps \(100\%-14\%=86\%\) each month, so the factor is \(0.86\).

Key takeaways

Key takeaways

What to remember

  • Percent growth uses \(1+r\); percent decay uses \(1-r\).
  • Doubling uses \(2^{t/d}\); half-life uses \((\frac12)^{t/d}\).
  • At \(t=0\), every correctly built model returns the initial value \(P\).
  • Keep full precision until the requested final rounding step.
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Put these notes into practice

Apply the ideas with SAT-style questions, then reinforce key details with flashcards.