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Cambridge IGCSE Mathematics 0580 · Number

Exponential growth and decay

Quantities that change by the same percentage again and again: depreciation, population change, bacteria and half-life. You will use a multiplier and a power to jump straight to the answer, work backwards to an earlier value, find how long a change takes, and find the rate from two values.

E1.17Extended onlyFree lesson

What the syllabus says

E1.17 (Extended)

Use exponential growth and decay.

Notes and examples

e.g. depreciation, population change.
Knowledge of $e$ is not required.

Quoted from the Cambridge IGCSE Mathematics 0580 syllabus for 2025 to 2027.

Growth and decay by a constant percentage Extended

Some quantities grow or shrink by the same percentage in each time period. A population might grow by 2% every year; a car might lose 15% of its value every year. Each year the change is a percentage of the current value, not of the original, so the amount added or lost changes every year.

The tool for this is the multiplier. Increasing by 10% leaves you with $100 % + 10 % = 110 %$ of what you had, so you multiply by $1.1$ . Doing that every year means multiplying by $1.1$ again and again, and repeated multiplication is a power:

$value after n periods = starting value \times (multiplier)^{n}$

This is exponential growth when the multiplier is more than 1, and exponential decay when it is between 0 and 1. Compare it with linear change, where the same amount is added each time. The table below starts both at £1000. At first they look alike, but the exponential row pulls away, because each 10% is taken of a bigger number.

Adding £100 a year compared with growing by 10% a year, starting from £1000
Year	0	1	2	3	4	5
Add £100 each year (linear)	1000	1100	1200	1300	1400	1500
Multiply by 1.1 each year (exponential)	1000	1100	1210	1331	1464.10	1610.51

Linear change adds the same amount each year. Exponential change multiplies by the same number each year, so the yearly increase itself grows: 100, then 110, then 121, then 133.10.

Finding the multiplier Extended

Write the percentage you end up with as a decimal.

An increase of $r %$ : multiplier $1 + \frac{r}{100}$ . So 6% up is $1.06$ , and $3.5 %$ up is $1.035$ .
A decrease of $r %$ : multiplier $1 - \frac{r}{100}$ . So 12% down is $0.88$ , and $3.5 %$ down is $0.965$ .
Doubling is a multiplier of 2. Halving is a multiplier of $\frac{1}{2}$ : this is how a half-life works, where a radioactive substance loses half of its mass in each equal period of time.

Reading a multiplier backwards is just as useful: a multiplier of $0.92$ means 92% is left, which is a decrease of 8%.

Count the periods carefully. If something doubles every 20 minutes, then in 2 hours it doubles $120 \div 20 = 6$ times, so the multiplier for the whole 2 hours is $2^{6} = 64$ .

Check yourself Extended

Write down the multiplier for a decrease of $7.5 %$ .

Answer it in the portal, where your answer is checked as you type →

Worked example 1 Extended

A car costs £22 000 when new. Its value depreciates by 15% each year. Calculate its value after 4 years.

Depreciation means losing value. Each year 85% of the value is left, so the multiplier is $0.85$ .
After 4 years: $22 000 \times {0.85}^{4}$ .
${0.85}^{4} = 0.522 006 25$ , so the value is $22 000 \times 0.522 006 25 = 11 484.1375$ .
Money is given to the nearest penny: £11 484.14.
A common wrong answer is to take $4 \times 15 % = 60 %$ off, giving £8800. That treats every year's 15% as 15% of the price when new.

Answer: £11 484.14

Worked example 2 Extended

A colony of bacteria doubles in size every 30 minutes. At first there are 500 bacteria. How many are there after 3 hours? (No calculator.)

In 3 hours there are $180 \div 30 = 6$ periods of 30 minutes.
The colony doubles 6 times, so multiply by $2^{6} = 64$ .
$500 \times 64 = 32 000$ bacteria.

Answer: $32 000$

Population change and other contexts Extended

The same method works in every context; only the words change.

Depreciation: cars, machines and computers lose a percentage of their value each year.
Population change: towns, animals, or bacteria growing or falling by a percentage each year, week or hour.
Half-life and doubling: the multiplier is $\frac{1}{2}$ or 2 for each period.
Compound interest on savings is the same idea, with money growing by a percentage each year.

The time period in the rate must match the number of periods you use. A rate "per week" with a time in days means you must first work out how many weeks there are.

Round sensibly at the end: money to the nearest penny, a population or a number of animals to a whole number or to 3 significant figures. Never round the multiplier, and never round a value from one year and then carry on with it. Use the power on your calculator in one step.

Worked example 3 Extended

The population of a town is $48 500$ . It is expected to grow by $2.4 %$ each year. Estimate the population after 10 years, correct to 3 significant figures.

Multiplier: $1 + 0.024 = 1.024$ .
After 10 years: $48 500 \times {1.024}^{10} = 61 481.05 \dots$
Correct to 3 significant figures: $61 500$ .

Answer: $61 500$

Working backwards, and finding the time or the rate Extended

An earlier value. If the value now is the result of multiplying by $m$ each year for $n$ years, then the starting value is the value now divided by $m^{n}$ . For example, if a house is worth £315 000 after rising 5% a year for 2 years, it was worth $315 000 \div {1.05}^{2} = £ 285 714.29$ two years ago. Multiplying by ${0.95}^{2}$ instead is wrong, because a 5% fall does not undo a 5% rise.

How long it takes. To find when a value first passes a target, keep multiplying by the multiplier on your calculator and count. Type the starting value, press =, then type $\times 0.88$ = and keep pressing = (the calculator repeats the last operation using ANS). Write down the values either side of the target so the examiner can see your method. You do not need logarithms at IGCSE.

The rate. If a value is multiplied by $m$ each year for $n$ years, the overall multiplier is $m^{n}$ . So $m$ is the $n$ th root of the overall multiplier:

$m = (\frac{final value}{starting value})^{\frac{1}{n}}$

Then the percentage rate is $(m - 1) \times 100$ for growth, or $(1 - m) \times 100$ for decay.

Worked example 4 Extended

A machine is bought for £9000. It loses 12% of its value each year. After how many complete years is its value first less than half of what it cost?

Half of £9000 is £4500. The multiplier is $0.88$ .
After 5 years: $9000 \times {0.88}^{5} = 4749.59$ , which is still more than £4500.
After 6 years: $9000 \times {0.88}^{6} = 4179.64$ , which is less than £4500.
So the value is first less than half after 6 years. Showing both values, either side of £4500, earns the method marks.

Answer: 6 years

Worked example 5 Extended

£5000 is invested and grows by the same percentage each year. After 2 years it is worth £6050. Find the percentage rate. (No calculator.)

Overall multiplier: $\frac{6050}{5000} = 1.21$ .
Over 2 years, $m^{2} = 1.21$ , so $m = \sqrt{1.21} = 1.1$ .
A multiplier of $1.1$ is an increase of 10% each year.

Answer: 10%

The shape of exponential graphs Extended

Plot the value against the number of periods and you get a curve, not a straight line. A growth curve gets steeper and steeper, because each step adds a percentage of a bigger amount. A decay curve falls steeply at first and then levels off: it gets closer and closer to zero but never reaches it, because a percentage of something positive is always left.

Check yourself Extended

A motorbike is bought for £15 000. It depreciates by 8% each year. Calculate its value after 3 years.

Answer it in the portal, where your answer is checked as you type →

Check yourself Extended

There are 300 bacteria at first, and the number doubles every 20 minutes. How many bacteria are there after 2 hours?

Answer it in the portal, where your answer is checked as you type →

Key points

Growth or decay by $r %$ each period: value $=$ start $\times$ multiplier $^{n}$ .
Increase by $r %$ : multiplier $1 + \frac{r}{100}$ . Decrease by $r %$ : multiplier $1 - \frac{r}{100}$ . Halving is $\frac{1}{2}$ , doubling is 2.
Exponential change multiplies by the same number each time; linear change adds the same amount each time.
To go back in time, divide by multiplier $^{n}$ .
To find how long, try successive whole numbers of periods and show the values either side of the target.
To find the rate, $m = (\frac{final}{start})^{\frac{1}{n}}$ .
Round only the final answer, to the accuracy the context needs.

Where marks are lost

Treating repeated percentage change as simple: 15% a year for 4 years is not a 60% decrease.
Using $0.15$ instead of $0.85$ for a 15% decrease, or $1.5$ instead of $1.05$ for a 5% increase.
Working backwards by multiplying by the opposite percentage: to undo a 7% fall, divide by $0.93$ ; do not multiply by $1.07$ .
Rounding the value each year, which drifts away from the accurate answer.
Miscounting the periods: doubling every 20 minutes for 2 hours is 6 doublings, not 3 or 2.
Giving the year before the target is passed, or not showing the values either side of it.
Giving a money answer to 3 significant figures (£11 500) when the nearest penny is expected.

Related IGCSE Mathematics 0580 lessons

E1.18

All IGCSE Mathematics 0580 lessons →

An original lesson written to the Cambridge IGCSE Mathematics 0580 syllabus (2025 to 2027). The worked examples and questions are our own, not Cambridge questions. Back to every IGCSE Mathematics 0580 lesson.

Exponential growth and decay

Growth and decay by a constant percentage Extended

Finding the multiplier Extended

Check yourself Extended

Worked example 1 Extended

Worked example 2 Extended

Population change and other contexts Extended

Worked example 3 Extended

Working backwards, and finding the time or the rate Extended

Worked example 4 Extended

Worked example 5 Extended

The shape of exponential graphs Extended

Check yourself Extended

Check yourself Extended

Key points

Where marks are lost

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Surds

Money

Time

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