Growth and Decay (Higher)

This lesson is Higher tier. It generalises the compound-interest idea into exponential growth and decay — any situation where a quantity is repeatedly multiplied by a fixed factor each time period. Examples in OCR GCSE Mathematics (J560) include population growth, radioactive and medicinal decay, bacterial growth and investment returns. This lesson covers the general growth and decay formula, the iterative (year-by-year) method, the efficient multiplier method, and $n$n-year problems including finding the number of periods.

This lesson is rich in AO1 technique with the formula $A = A_0 \times k^n$A=A0​×kn, AO2 reasoning when you interpret the meaning of the multiplier, and AO3 problem-solving in multi-stage and "how many periods" questions.

Key Vocabulary

Term	Meaning
Exponential growth	Repeated multiplication by a factor greater than $1$1
Exponential decay	Repeated multiplication by a factor between $0$0 and $1$1
Growth/decay factor ($k$k)	The fixed multiplier per time period
Initial amount ($A_0$A0​)	The starting value, before any change
Iterative method	Working out each step in turn, one period at a time
$n$n	The number of time periods

The Growth and Decay Formula

The previous lesson on compound interest was really a special case of a much bigger idea: exponential change. Whenever a quantity is multiplied by the same factor every period — whether that factor is more than $1$1 (growth) or less than $1$1 (decay) — you get exponential behaviour, and a single formula describes it all. The defining feature of exponential change is that the quantity changes by a fixed percentage (and hence a fixed multiplier) each period, not by a fixed amount. That is what makes the graph curve ever more steeply for growth, or flatten ever more gently for decay.

All exponential growth and decay (including compound interest and depreciation) follow:

$A = A_0 \times k^n,$A=A0​×kn,

where $A_0$A0​ is the initial amount, $k$k is the growth or decay factor per period, and $n$n is the number of periods. The whole method reduces to three short steps: identify $A_0$A0​, decide the factor $k$k, and raise it to the power $n$n.

• For growth of $r\%$r% per period, $k = 1 + \dfrac{r}{100}$k=1+100r​ (so $k > 1$k>1).
• For decay of $r\%$r% per period, $k = 1 - \dfrac{r}{100}$k=1−100r​ (so $0 < k < 1$0<k<1).

Change per period	Factor $k$k	Type
$+8\%$+8%	$1.08$1.08	growth
$+150\%$+150% (multiply by $2.5$2.5)	$2.5$2.5	growth
$-12\%$−12%	$0.88$0.88	decay
Halving each period	$0.5$0.5	decay

Growth Problems

Exponential growth uses a factor $k$k greater than $1$1. The same quantity is multiplied by $k$k each period, so the increase gets bigger and bigger over time — a hallmark of growth that is "compound" rather than steady. Real examples include savings, bacterial cultures, and the early stage of anything that spreads. The method is identical to compound interest: write the factor, raise it to the power of the number of periods, and multiply by the starting amount.

Worked Example 1

A colony of $500$500 bacteria grows by $20\%$20% every hour. Work out the population after $4$4 hours.

The growth factor is $k = 1.2$k=1.2:

$A = 500 \times 1.2^4 = 500 \times 2.0736 = 1{,}036.8.$A=500×1.24=500×2.0736=1,036.8.

Since the population must be a whole number, this is about $1{,}037$1,037 bacteria.

Answer: approximately $1{,}037$1,037 bacteria.

Worked Example 1b

A lake's algae population doubles every $3$3 days. It starts at $5{,}000$5,000 cells. Work out the population after $9$9 days.

Doubling every $3$3 days means a factor of $2$2 applied once for each $3$3-day period. In $9$9 days there are $9 \div 3 = 3$9÷3=3 such periods, so:

$A = 5{,}000 \times 2^3 = 5{,}000 \times 8 = 40{,}000.$A=5,000×23=5,000×8=40,000.

Answer: $40{,}000$40,000 cells.

Common error: using $n = 9$n=9 instead of $n = 3$n=3. The power is the number of doubling periods, not the number of days — read the period carefully.

Worked Example 2

An investment of £6,000 grows at $4.5\%$4.5% per year. Work out its value after $6$6 years.

The growth factor is $k = 1.045$k=1.045:

$A = 6{,}000 \times 1.045^6 = 6{,}000 \times 1.302260 = 7{,}813.56.$A=6,000×1.0456=6,000×1.302260=7,813.56.

So the value is about £7,813.56.

Answer: £7,813.56. This is exactly a compound-interest calculation; the growth-and-decay formula simply gives it a more general name.

Worked Example 2b

A rare stamp collection is valued at £8,000 and increases in value by $6\%$6% each year. Work out its value after $4$4 years, to the nearest pound.

The growth factor is $k = 1.06$k=1.06:

$A = 8{,}000 \times 1.06^4 = 8{,}000 \times 1.262477 = 10{,}099.82.$A=8,000×1.064=8,000×1.262477=10,099.82.

To the nearest pound, this is £10,100.

Answer: £10,100.

Decay Problems

Exponential decay works in exactly the same way as growth, but the factor $k$k is between $0$0 and $1$1, so the quantity shrinks each period without ever quite reaching zero. The single most common slip is to use the loss percentage as the factor instead of the remaining percentage: a $12\%$12% decay leaves $88\%$88%, so $k = 0.88$k=0.88, not $0.12$0.12. Keep asking "what fraction is left after each period?" and the factor follows.

Worked Example 3

A radioactive sample has mass $800$800 g and decays by $12\%$12% each year. Work out the mass remaining after $5$5 years.

The decay factor is $k = 0.88$k=0.88:

$A = 800 \times 0.88^5 = 800 \times 0.527732 = 422.19 \text{ g}.$A=800×0.885=800×0.527732=422.19 g.

Answer: $422.19$422.19 g (to $2$2 d.p.). The sample has lost roughly half its mass in five years, but it never disappears entirely — exponential decay approaches zero without ever reaching it.

Worked Example 4

A cup of coffee cools so that its temperature above room temperature falls by $30\%$30% each minute. It starts $80^{\circ}\text{C}$80∘C above room temperature. Work out the temperature above room temperature after $3$3 minutes.

The decay factor is $k = 0.7$k=0.7:

$A = 80 \times 0.7^3 = 80 \times 0.343 = 27.44.$A=80×0.73=80×0.343=27.44.

So it is about $27.44^{\circ}\text{C}$27.44∘C above room temperature.

Answer: $27.44^{\circ}\text{C}$27.44∘C above room temperature.

Common error: subtracting $30\%$30% once and stopping. Decay is repeated each period, so apply the factor $n$n times.

Worked Example 4b

A car's value depreciates by $14\%$14% each year. It is worth £20,000 now. Work out its value after $3$3 years, to the nearest pound.

A $14\%$14% loss leaves $86\%$86%, so the decay factor is $k = 0.86$k=0.86:

$A = 20{,}000 \times 0.86^3 = 20{,}000 \times 0.636056 = 12{,}721.12.$A=20,000×0.863=20,000×0.636056=12,721.12.

To the nearest pound, this is £12,721.

Answer: £12,721.

Common error: using $0.14$0.14 as the factor. The factor is the proportion that remains ($0.86$0.86), not the proportion lost.

The Iterative Method

The iterative method works out each period in turn, applying the multiplier step by step rather than as a single power. When the rate is the same every period, the iterative method and the formula give identical answers — the formula is just a shortcut for repeated multiplication. But the iterative method is essential in two situations: when the rate changes from one period to the next (so no single power applies), and when you want to confirm a formula answer or build a table to find when a quantity crosses a threshold.

Worked Example 5

A fish population of $2{,}000$2,000 falls by $10\%$10% in year 1, then by $25\%$25% in year 2 (due to a drought). Work out the population after $2$2 years.

Year 1: $2{,}000 \times 0.9 = 1{,}800$2,000×0.9=1,800.

Year 2: $1{,}800 \times 0.75 = 1{,}350$1,800×0.75=1,350.

Answer: $1{,}350$1,350 fish. Here the single-formula method does not apply because the two factors differ; iteration handles it cleanly. You could also combine the two factors as $2{,}000 \times 0.9 \times 0.75 = 2{,}000 \times 0.675 = 1{,}350$2,000×0.9×0.75=2,000×0.675=1,350, which shows the iterative method is just step-by-step multiplication.

Worked Example 5b

A savings account contains £4,000. It grows by $3\%$3% in the first year, then the rate rises to $5\%$5% in the second year. Work out the value after $2$2 years.

The rate differs each year, so apply the factors in turn:

Year 1: $4{,}000 \times 1.03 = 4{,}120$4,000×1.03=4,120.

Year 2: $4{,}120 \times 1.05 = 4{,}326$4,120×1.05=4,326.

Answer: £4,326. As a single line, $4{,}000 \times 1.03 \times 1.05 = 4{,}326$4,000×1.03×1.05=4,326.

Extended Worked Examples

A particularly common Higher question asks "after how many periods" a quantity passes a target. The neat trick is that the starting amount can be divided out, leaving a simple condition on the factor alone. For growth from $A_0$A0​ to a target $T$T, you need $k^n > \dfrac{T}{A_0}$kn>A0​T​; for decay, $k^n < \dfrac{T}{A_0}$kn<A0​T​. Then test whole-number powers either side of the boundary — no logarithms are required at GCSE.

Worked Example 6: Finding the number of periods (growth)