Exponential growth and decay
Section: Number 2 | Syllabus: Cambridge IGCSE Mathematics (0580)
Exponential Growth and Decay
Extended Only
Exponential growth and decay happen when a quantity is multiplied by the same growth or decay factor repeatedly, once per time period - exactly the same idea as compound interest, applied to contexts like depreciation and population change.
- After n time periods, a quantity becomes: value = original × (multiplier)^n
- For growth (e.g. population increase), the multiplier is 1 + (R/100)
- For decay (e.g. depreciation of a car or machine), the multiplier is 1 - (R/100)
- The same multiplier is applied every period - each period's change is based on the previous period's value, not the original value
Worked Example: Depreciation
- Question: A machine is bought for 15 000. It depreciates by 20% each year. Find its value after 3 years.
- Step 1: The decay multiplier is 1 - (20/100) = 0.8
- Step 2: Apply it 3 times: 15\,000 × 0.8^3 = 15\,000 × 0.512
- Answer: 7680
Worked Example: Population Growth
- Question: A town has a population of 24 000. The population grows by 6% each year. Find the population after 4 years, correct to the nearest hundred.
- Step 1: The growth multiplier is 1 + (6/100) = 1.06
- Step 2: Apply it 4 times: 24\,000 × 1.06^4 = 24\,000 × 1.26247... = 30\,299.4...
- Answer: 30 300 (nearest hundred)
Common Mistakes
MistakeTreating repeated percentage change as a simple total, e.g. for 20% depreciation over 3 years, taking 3 × 20% = 60% off the original value directly (15 000 × 0.4 = 6000)
Fixeach year's decay applies to the previous year's (already-reduced) value, not the original - the correct method is 15 000 × 0.8³ = 7680, not a simple 60% reduction
MistakeLeaving a rounded answer with decimal places in a context that must be a whole number, e.g. giving a population as 30 299.4
Fixround appropriately for the context - a population of people must be a whole number, so round to the nearest whole number or, as requested, the nearest hundred
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