NCEA Level 2Exponential modelsAS 91261

Exponential growth and decay: interest, depreciation, half-life

What this lesson covers

  • Growth or decay by a fixed percentage each period is multiplication by the same factor each period.
  • Finding how long something takes means solving for an index, which needs logarithms.
Key idea

Same percentage, every period

  • Interest on interest. Value lost on value.
  • Each period multiplies by the same factor.
  • That is exponential: start × factorⁿ.
Key idea

Turning a percentage into a factor

  • Growing 4.5%: multiply by 1.045.
  • Losing 15%: multiply by 0.85.
  • Doubling every 3 hours: 2^(t/3).
SituationEach period
Grows 4.5% a year× 1.045
Loses 15% a year× 0.851 − 0.15
Doubles every 3 hours× 2 every 3 hours2^(t/3)
Worked example

Compound interest

  • $5000 at 4.5% a year.
  • Compounded yearly.
  1. factor = 1 + 0.045 = 1.045grows 4.5%
  2. A = 5000 × 1.045ⁿstart × factorⁿ
  3. A = 5000 × 1.045⁶n = 6
  4. = $6511.30
Worked example

Depreciation

  • A $28 000 car.
  • Loses 15% a year.
  1. factor = 1 − 0.15 = 0.85loses 15%
  2. V = 28 000 × 0.85ⁿstart × factorⁿ
  3. V = 28 000 × 0.85⁴n = 4
  4. = 14 616.175
  5. = $14 616.18nearest cent
Key idea

Not 15% × 4

  • Not 60% off.
  • 15% of what is left.
  • Keeps 52.2%, not 40%.
  1. 4 × 15% = 60% off: keeps 40%not of the original
  2. 0.85 × 0.85 × 0.85 × 0.85each year × 0.85
  3. = 0.85⁴ = 0.5220
  4. keeps 52.2%, not 40%
Worked example

How long to double at 6%?

  • Solve 1.06ⁿ = 2 with logs: 11.9 years.
  • Doubling again takes the same time.
  • That constant doubling time is the signature.
How long to double at 6%?
Worked example

The working for doubling

  • Logs bring n down.
  • Round up, and say why.
  1. 1.06ⁿ = 2value doubles
  2. log 1.06ⁿ = log 2log both sides
  3. n log 1.06 = log 2power law
  4. n = log 2 ÷ log 1.06÷ log 1.06
  5. = 11.896
  6. 1.06¹¹ = 1.8983 < 2not yet doubled
  7. 1.06¹² = 2.0122 > 2doubled
  8. 12 years
Worked example

Growth with a doubling time

  • Divide by 200 first.
  • Logs bring t/3 down.
  1. N = 200 × 2^(t/3)doubles every 3 h
  2. 200 × 2^(t/3) = 10 000N = 10 000
  3. 2^(t/3) = 50÷ 200
  4. (t/3) log 2 = log 50logs, power law
  5. t/3 = log 50 ÷ log 2÷ log 2
  6. = 5.6439
  7. t = 3 × 5.6439 ≈ 16.9 hours
Worked example

Decay: when is it below 10 mg?

  • Starts at 80 mg.
  • Measured on the hour.
  1. 80 × 0.8ᵗ = 10loses 20% an hour
  2. 0.8ᵗ = 0.125÷ 80
  3. t log 0.8 = log 0.125logs, power law
  4. t = log 0.125 ÷ log 0.8
  5. = 9.32 hours
  6. 9 h: 80 × 0.8⁹ = 10.74 mgabove 10
  7. 10 h: 80 × 0.8¹⁰ = 8.59 mgbelow 10
  8. first below 10 mg: 10 hours
Key idea

Where the marks are

  • Achieved: evaluate a given model.
  • Merit: form the model and solve for time.
  • Excellence: interpret, including the rounding.
Summary

What to take away

  • A percentage becomes a factor; time comes from logs.
Grow r%multiply by 1 + r/100
Lose r%multiply by 1 − r/100
Modelstart × factorⁿ
How longsolve for the index with logs
Whole periodsround up, and say why
Your turn

Your turn: four questions

  • Pause here and answer each one.
  • Questions 3 and 4 need logs.
  • Think about rounding in context.
Question
1$3000 at 5% compound interest a year. Value after 8 years?
2A $1200 phone loses 25% a year. Value after 3 years?
3At 7% a year, whole years to triple?
4A town of 5000 grows 3% a year. When does it pass 8000?
Show answer

1. 3000 × 1.05⁸ = $4432.37. 2. 1200 × 0.75³ = $506.25. 3. 1.07ⁿ = 3, n = log 3 ÷ log 1.07 = 16.24, so 17 years. 4. 1.03ⁿ = 1.6, n = 15.90, so in the 16th year (after 16 years).

Worked example

The answers: questions 1 and 2

  1. 1. A = 3000 × 1.05ⁿgrows 5%: × 1.05
  2. A = 3000 × 1.05⁸n = 8
  3. = $4432.37
  4. 2. V = 1200 × 0.75ⁿloses 25%: × 0.75
  5. V = 1200 × 0.75³n = 3
  6. = 1200 × 0.421875
  7. = $506.25
Worked example

The answers: question 3

  1. 3. P × 1.07ⁿ = 3Ptripled
  2. 1.07ⁿ = 3÷ P
  3. n log 1.07 = log 3logs, power law
  4. n = log 3 ÷ log 1.07
  5. = 16.24
  6. 1.07¹⁶ = 2.9522 < 3not yet tripled
  7. 17 years
Worked example

The answers: question 4

  1. 4. 5000 × 1.03ⁿ = 8000grows 3%: × 1.03
  2. 1.03ⁿ = 8000 ÷ 5000 = 1.6÷ 5000
  3. n log 1.03 = log 1.6logs, power law
  4. n = log 1.6 ÷ log 1.03
  5. = 15.90
  6. 15 years: 5000 × 1.03¹⁵ ≈ 7790below 8000
  7. 16 years: 5000 × 1.03¹⁶ ≈ 8024above 8000
  8. passes 8000 after 16 years

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