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Cambridge IGCSE Mathematics — 0580 Extended

Topic 1.14: Number — Exponential Growth & Decay

The exponential model

When a quantity changes by the same percentage each period, you multiply repeatedly — this is exponential change.

\[\text{new amount} = \text{original} \times (\text{multiplier})^n\]

where \(n\) = number of time periods (years, months, etc.).

  • Growth: multiplier > 1 (e.g. 5% increase → × 1.05).
  • Decay: multiplier < 1 (e.g. 12% decrease → × 0.88).

You do not need \(e^x\) at IGCSE — use the multiplier form above.

Method

  1. Identify original amount and number of periods \(n\).
  2. Write the multiplier (1 + rate for growth, 1 − rate for decay).
  3. Calculate original \(\times\) (multiplier)\(^{n}\).
  4. Round to the required accuracy (often 2 d.p. for money).

Exponential vs simple (linear) change

Which model?

Same fixed amount added each year

Linear (simple interest): \(A = P + kn\).

Same percentage applied each year

Exponential (compound): \(A = P \times (\text{multiplier})^n\).

Compare £1000 at 10% simple interest and 10% compound interest for 3 years.

Worked solution: 1000 at 10 percent for 3 years is 1300 simple and 1331 compound
Simple adds £100 each year; compound multiplies by 1.10 each year

Depreciation and growth

Depreciation is exponential decay — a car, machine or value loses a fixed percentage of its current value each year, not a fixed amount.

A car worth £18 000 depreciates by 15% per year. Find its value after 4 years.

Worked solution: 18000 times 0.85 to the power 4 equals 9396.11 pounds
Decay multiplier 0.85: \(18\,000 \times (0.85)^4 = £9396.11\)

A population of 50 000 increases by 3% per year. Estimate the population after 6 years.

Worked solution: 50000 times 1.03 to the power 6 is about 59703
Growth multiplier 1.03: \(50\,000 \times (1.03)^6 \approx 59\,703\)

Exam context

Exponential decay also models radioactive decay, cooling, and any situation where the change depends on the current amount. Always check whether the question says "each year on its current value" (exponential) or "of the original value" (simple).

Paper 2 (non-calculator)

Powers of friendly multipliers can be built in steps: \(0.80^2 = 0.64\), then \(0.64 \times 0.80 = 0.512\). For 5% growth, \(1.05^2 = 1.1025\).

Try this

A laptop costs £800 and loses 20% of its value each year. Find its value after 3 years.

Show answer
Answer

  1. 20% of current value each year → multiplier \(0.80\)

    \[ 800 \times (0.80)^{3} \]

  2. Build the power in steps

    \[ 0.80^{2} = 0.64,\quad 0.64 \times 0.80 = 0.512 \]

  3. Final value

    \[ 800 \times 0.512 = £409.60 \]

Exam Traps

  • "Of its current value each year" is exponential. "Of the original value each year" is simple (linear) — they are not the same.
  • A 15% decrease is \(\times 0.85\), not \(\times 0.15\) and not \(\times 1.15\).

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