Quick revision notes · Mathematics 0580 · Chapter 34 of 37
Exponential Growth And Decay
Repeated percentage change: growth, decay, working backwards, finding the rate and the time.
3 sectionsWorked examples
Common mistakesIGCSE: whole chapter is Extended tier
At a glance
Repeated percentage change: growth, decay, working backwards, finding the rate and the time.
1Growth and decay multipliers
- Repeated percentage change multiplies by the same number each period.
- Growth of r% → multiplier (1 + r/100). Decay of r% → multiplier (1 − r/100).
- After n periods: amount = starting amount × multipliern.
- Typical contexts: population, bacteria, compound interest (growth); depreciation, radioactive decay, cooling (decay).
| Change | Multiplier |
|---|---|
| +4% | 1.04 |
| +12.5% | 1.125 |
| −15% | 0.85 |
| −2.5% | 0.975 |
| doubles | 2 |
| halves | 0.5 |
Formulas
A = P(1 + r/100)ngrowth
A = P(1 − r/100)ndecay
Worked example
A car worth $12 000 loses 15% of its value each year. Find its value after 3 years.
- Multiplier 0.85.
- 12 000 × 0.853 = 7369.5.
$7369.50
2Working backwards and finding the rate
- Starting amount = final amount ÷ multipliern.
- To find the rate: (final ÷ start) = multipliern, so take the n-th root to get the multiplier.
- Multiplier 1.1 means 10% growth; 0.92 means 8% decay.
Worked example
A value grows from 400 to 484 in 2 years at a constant percentage rate. Find the rate.
- 484 ÷ 400 = 1.21.
- Multiplier = √1.21 = 1.1.
10% per year
3Finding the number of periods
- No logarithms needed: try n = 1, 2, 3, … on the calculator until the amount passes the target.
- Give the first whole period that passes it, and show the values either side.
- Half-life questions: each half-life multiplies the amount by 0.5.
How to do it
- Write amount = P × multipliern.
- Try values of n.
- Write the value just below and just above the target.
- Answer with the first n that passes it.
Worked example
$800 is invested at 3% compound interest. After how many whole years is it first worth more than $1000?
- 800 × 1.037 = 983.9…
- 800 × 1.038 = 1013.4…
8 years
✗Common mistakes
| Don’t write | Write instead |
|---|---|
| 15% depreciation for 2 years = 30% off | Multiply by 0.85 twice: 0.7225 (27.75% off) |
| Using 1.15 for a 15% decrease | Decrease: 0.85 |
| Giving the year just before the target is reached | Give the first whole period that passes it |
| Dividing by 2 for the rate from a 2-year change | Take the square root of the overall multiplier |
★In the exam
- Write the multiplier first; most method marks come from it.
- Show the trial values when finding the number of years.
- Read whether the question wants the amount, the increase/decrease, or the percentage.