What is exponential growth in the context of Cambridge IGCSE Mathematics?

Exponential growth refers to a process where the quantity increases at a rate proportional to its current value. In Cambridge IGCSE Mathematics, this is typically modeled using the formula y = a(1 + r)^t, where 'a' is the initial amount, 'r' is the growth rate, and 't' is the time period. This concept is crucial for understanding phenomena such as population growth and compound interest.

How does exponential decay differ from exponential growth in Mathematics?

Exponential decay is the opposite of exponential growth. It describes a process where a quantity decreases at a rate proportional to its current value. The formula used is y = a(1 - r)^t, where 'a' is the initial amount, 'r' is the decay rate, and 't' is the time period. This concept is often applied to scenarios like radioactive decay and depreciation of assets in Cambridge IGCSE Mathematics.

How can I identify exponential growth and decay in a mathematical problem?

To identify exponential growth or decay, look for problems that involve a quantity changing at a rate proportional to its current value. If the problem involves an increase, it's exponential growth, and if it involves a decrease, it's exponential decay. Key indicators include phrases like 'grows by a percentage' or 'decreases by a percentage' over time, which are common in Cambridge IGCSE Mathematics problems.

What are some real-life examples of exponential growth and decay?

Real-life examples of exponential growth include population growth, where the number of individuals increases rapidly over time, and compound interest, where the amount of money grows exponentially. Exponential decay examples include radioactive decay, where unstable atoms lose energy over time, and the depreciation of car value, where the car's worth decreases exponentially. These concepts are frequently explored in Cambridge IGCSE Mathematics.

Why is understanding exponential growth and decay important in Mathematics?

Understanding exponential growth and decay is crucial because these concepts model many real-world phenomena. They help students predict future outcomes and understand the dynamics of change over time. In Cambridge IGCSE Mathematics, mastering these concepts is essential for solving complex problems related to finance, science, and economics, providing a solid foundation for further studies.

Why is "Using $(1 + r)^{n}$ for decay (or vice versa)" a common mistake in Exponential Growth and Decay?

Why it happens: Students forget to subtract for decay. How to avoid it: Decay → (1 - r). Growth → (1 + r). Always check whether the quantity is increasing or decreasing. Source: 0580/42 — every recent series

Why is "Using the rate as a whole number ($r = 5$ instead of $0.05$)" a common mistake in Exponential Growth and Decay?

Why it happens: Students paste percentage values directly into the formula. How to avoid it: Always divide by 100 first.

Why is "Multiplying by the rate instead of using the power" a common mistake in Exponential Growth and Decay?

Why it happens: Treating exponential growth like simple proportion: 5000 × 1.04 × 5 is wrong. How to avoid it: Power, not multiplication: (1.04)^5.

Why is "Rounding mid-calculation" a common mistake in Exponential Growth and Decay?

Why it happens: Students round (1.04)^5 to 1.16 early. How to avoid it: Keep at least 4 d.p. in the multiplier; round only at the end.

Cambridge IGCSE Mathematics (0580)

Exponential Growth and Decay

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Detailed notes on Number for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Exponential Growth and Decay — Cambridge IGCSE 0580 Maths Extended (2026)

When something grows or shrinks by a fixed PERCENTAGE every period — bacterial populations, radioactive samples, depreciating cars — the result is exponential. One formula, two directions, repeated multiplier method.

At a glance

Growth: $A = A_{0}(1 + r/100)^{n}$ . Decay: $A = A_{0}(1 - r/100)^{n}$ .
$A_{0}$ = starting value. $r$ = % rate per period. $n$ = number of periods.
Same engine as compound interest, applied to populations, prices, mass, etc.
When rate is given as a decimal (e.g. $0.04$ for 4%), use $A = A_{0}(1 + 0.04)^{n}$ directly.
Half-life of decay = time for the quantity to halve.
Doubling time of growth = time for the quantity to double.
Solve for $n$ using logs (Extended only): $n = \dfrac{\log(A/A_{0})}{\log(1 + r/100)}$ .

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E1.17 — Use exponential growth and decay (e.g. compound interest, depreciation, population growth, radioactive decay).
Apply the formula $A = A_{0}(1 + r/100)^{n}$ for growth and $A = A_{0}(1 - r/100)^{n}$ for decay.

Exponential growth — the formula

Repeated percentage increase compounds. The total after $n$ periods is the original times the multiplier, raised to the $n$ -th power.

When a quantity grows by a fixed percentage $r\%$ each period, the value after $n$ periods is

$A = A_{0}\left(1 + \dfrac{r}{100}\right)^{n}.$

A₀ is where you start, (1 + r/100) is the per-period multiplier, and n is how many times it is applied.

This is the same compound-interest formula, repurposed.

Worked. "A bacterial colony of $400$ doubles in size every hour. Find the population after $5$ hours."

$r = 100\%$ per hour → multiplier $= 2$ .
$A = 400 \times 2^{5} = 400 \times 32 = 12{,}800$ .

Worked. "A town's population is $50{,}000$ and grows at $2.5\%$ per year. Find the population after $10$ years."

$A = 50{,}000 \times (1.025)^{10} \approx 50{,}000 \times 1.2801 \approx 64{,}004$ .

Why a power, not a multiplier and add? Each period multiplies by the same factor, so $n$ periods means multiplying $n$ times — that's $(\text{multiplier})^{n}$ .

Growth multiplies by more than 1 and accelerates; decay multiplies by less than 1 and flattens towards zero.

$A = A_{0}(1 + r/100)^{n}$ .
Each period the quantity is multiplied by $(1 + r/100)$ .
$n$ periods → raise to the power $n$ .
Always identify what $A_{0}$ , $r$ , $n$ refer to in the question.

Exponential decay

Same formula, MINUS sign on the rate. Used for depreciation, cooling, radioactive decay.

When a quantity decreases by a fixed percentage $r\%$ each period:

$A = A_{0}\left(1 - \dfrac{r}{100}\right)^{n}.$

The multiplier is now LESS than 1 (e.g. $0.85$ for a $15\%$ decline), so each period the quantity shrinks.

Worked. "A car is bought for $24,000 and depreciates by $15\%$ per year. Value after $4$ years?"

$A = 24000 \times (0.85)^{4} = 24000 \times 0.5220 \approx 12529.13$ , so $12,529.13.

Worked. "A radioactive sample contains $1{,}600\,\text{g}$ and decays by $3\%$ per day. How much remains after $20$ days?"

$A = 1600 \times (0.97)^{20} \approx 1600 \times 0.5438 \approx 870.0\,\text{g}$ .

Decay never reaches zero. The multiplier is positive, so the quantity tends to zero but never actually gets there. In real life, "after a million years there is still 0.0001 g" is the model's answer.

$A = A_{0}(1 - r/100)^{n}$ .
Multiplier $< 1$ .
Examples: depreciation, cooling, radioactive decay.
Asymptotically approaches zero — never reaches it.

Half-life and doubling time

The natural way to describe an exponential change. Time for the quantity to halve (decay) or double (growth).

Half-life ( $T_{1/2}$ ) is the time for the quantity to halve. After $n$ half-lives, the original $A_{0}$ is reduced to $A_{0}/2^{n}$ .

Doubling time ( $T_{2}$ ) is the time for the quantity to double. After $n$ doublings, $A_{0}$ has grown to $A_{0} \times 2^{n}$ .

Worked. "An isotope has a half-life of $5$ days. Starting with $80\,\text{g}$ , how much remains after $20$ days?"

$20 / 5 = 4$ half-lives.
$A = 80 / 2^{4} = 80 / 16 = 5\,\text{g}$ .

Every half-life cuts the quantity in half — count how many fit into the total time.

Worked. "A bacteria culture doubles every $30$ minutes. Starting at $200$ , how many after $3$ hours?"

$3$ hours = $180$ minutes = $6$ doublings.
$A = 200 \times 2^{6} = 200 \times 64 = 12{,}800$ .

Each half-hour doubles the count; after 6 doublings 200 becomes 12,800.

Connecting half-life to a percentage rate. A $3\%$ -per-day decay has half-life $T_{1/2} = \dfrac{\log(0.5)}{\log(0.97)} \approx \dfrac{-0.301}{-0.01323} \approx 22.76\,\text{days}.$ So a $3\%$ daily decline halves the quantity every $\approx 23$ days. (Logs are needed for this, see Going deeper.)

Half-life: time to halve (decay).
Doubling time: time to double (growth).
After $n$ half-lives: $A = A_{0} / 2^{n}$ .
After $n$ doublings: $A = A_{0} \times 2^{n}$ .

Spotting an exponential question

Cambridge phrases exponential growth/decay in many disguises. Train yourself to recognise the cue.

If you see ANY of these phrasings, reach for the exponential formula:

"Increases / decreases by $r\%$ each year/month/day."
"Doubles every... / halves every..."
"Compound interest" / "depreciation".
"Population growth at a constant rate of $r\%$ ."
"Radioactive decay" with a half-life.

Same formula both ways — the cue words tell you whether to add or subtract the rate.

If the rate is given in DIFFERENT periods (e.g. annual rate, but you want to know the value after $30$ months), keep the units consistent: convert $30$ months to $30/12 = 2.5$ years before substituting.

Compound interest is exponential growth. "Compound interest at $5\%$ per year for $10$ years" = "exponential growth at $5\%$ per year for $10$ years". Same formula.

Depreciation is exponential decay. "Depreciates by $20\%$ per year" = "decays by $20\%$ per year". Same formula.

Cue words: "each", "per", "every", "compound", "depreciate", "half-life".
Match the units of $n$ to the period in which $r$ is quoted.
Compound interest = exponential growth.
Depreciation = exponential decay.

Quick recap

Growth: $A = A_{0}(1 + r/100)^{n}$ . Decay: $A = A_{0}(1 - r/100)^{n}$ .
Same engine as compound interest, applied to populations / prices / mass.
Each period multiplies by $(1 \pm r/100)$ ; $n$ periods → raise to the $n$ .
Half-life = time to halve. Doubling time = time to double.
After $n$ half-lives, value is $A_{0} / 2^{n}$ .
Use logs to solve for $n$ when the value is given.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Exponential growth — repeated multiplication by a factor greater than 1: $A = A_{0}(1 + r/100)^{n}$ .
Exponential decay — repeated multiplication by a factor less than 1: $A = A_{0}(1 - r/100)^{n}$ .
Half-life — the time for an exponentially decaying quantity to halve.
Doubling time — the time for an exponentially growing quantity to double.
Multiplier — the factor $1 \pm r/100$ that applies one period of change.

How it’s examined

Exponential growth and decay typically appear once per Paper 4, often as a 4-6 mark multi-step problem (compute the value after $n$ years; find when the value reaches a target). Paper 2 includes them as 2-3 mark single-application questions. Examiner reports flag two recurring slips: using simple-interest reasoning ( $A_{0} + n \cdot \text{interest}$ ) for compound questions, and unit mismatches between $r$ and $n$ .

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E1.17); 0580/22 May/Jun 2024 — Q14 (population growth); 0580/42 Oct/Nov 2024 — Q14 (depreciation, multi-step); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Exponential Growth and Decay

Step-by-step solutions to past-paper-style questions on exponential growth and decay, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Exponential Growth and Decay

Almost every exponential growth and decay exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — find the amount/mass/population after $n$ periods, find the rate — with one exponential model to apply.

How to approach it

Write the multiplier ( $1+r$ for growth, $1-r$ for decay, or $2$ for doubling), apply $A=P\times(\text{multiplier})^{n}$ , and take an $n$ th root when the rate is the unknown.

Common trap

Examiner reports flag using $(1+r)$ for decay, multiplying by the rate instead of raising it to a power, and rounding the multiplier too early.

Multi-step problem

Recognise it by

Two linked tasks — predict a value over a long horizon and then find when a threshold is first crossed.

How to approach it

Handle each part in turn: apply the model for the stated number of periods, then trial successive values of $n$ , checking both $n$ and $n-1$ against the threshold.

Common trap

Examiner reports flag confusing 'full years' with rounded years — the quantity must strictly exceed the threshold, so verify the year before as well.

Show that / prove

Recognise it by

The words show that a quantity first crosses a threshold during a stated year.

How to approach it

Compute the value at the end of both bounding years ( $A_{n-1}$ and $A_{n}$ ) and state that one is above and the other below the threshold.

Common trap

Examiner reports flag candidates computing only the year after the crossing and omitting the year before — both sides of the boundary are required.

1Calculate exponential growth (population)
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q8• growth, population
Question
A town has a population of $24{,}000$ growing at $3\%$ per year. Find the population after $5$ years (to the nearest integer).
Step-by-step solution
1. Step 1
  Use $A = P(1 + r)^{n}$ with $r = 0.03$ .
  $A = 24000 \times (1.03)^{5}$
2. Step 2
  $(1.03)^{5} = 1.1592740\dots$
  $A = 24000 \times 1.1592740 = 27822.58\dots$
Answer
$27{,}823$ people
2Calculate exponential decay (depreciation)
ExtendedDirect calculation• decay, depreciation
Question
A car bought for $28,000 depreciates by $12\%$ each year. Find its value after $4$ years.
Step-by-step solution
1. Step 1
  Decay multiplier: $1 - 0.12 = 0.88$ .
  $A = 28000 \times (0.88)^{4}$
2. Step 2
  $(0.88)^{4} = 0.5996\dots$
  $A = 28000 \times 0.5996 = 16789.45$
Answer
$16,789.45
Examiner tip
Decay uses $(1 - r)$ , not $(1 + r)$ . Subtracting $12\%$ four times directly is also wrong.
3Find the time required (trial-and-improvement)
ExtendedDirect calculation• growth, time
Question
An investment of $5,000 grows at $4\%$ per year. After how many full years will it exceed $7,500?
Step-by-step solution
1. Step 1
  Try $n = 8$ : $5000 \times 1.04^{8} = 5000 \times 1.3686 = 6843.16$ . Not yet.
2. Step 2
  Try $n = 9$ : $5000 \times 1.04^{9} = 5000 \times 1.4233 = 7116.86$ . Not yet.
3. Step 3
  Try $n = 11$ : $5000 \times 1.04^{11} = 5000 \times 1.5395 = 7697.61$ . Yes — exceeds $7,500.
4. Step 4
  Check $n = 10$ : $5000 \times 1.04^{10} = 5000 \times 1.4802 = 7401.22$ . Below.
Answer
After $11$ full years
Examiner tip
Trial-and-improvement is fully accepted on 0580. Show each trial clearly — examiners credit the working.
4Apply exponential decay to half-life
ExtendedDirect calculation• decay, half-life
Question
A radioactive sample of mass $80\,\text{g}$ decays at $5\%$ per year. Find the mass after $10$ years.
Step-by-step solution
1. Step 1
  $A = 80 \times (0.95)^{10}$ .
2. Step 2
  $(0.95)^{10} = 0.5987\dots$
  $A = 80 \times 0.5987 = 47.90$
Answer
$47.9\,\text{g}$ (3 s.f.)
5Bacterial doubling — short calculation
CoreDirect calculation• growth, doubling
Question
A culture starts with $200$ bacteria. The number doubles every hour. How many bacteria are there after $4$ hours?
Step-by-step solution
1. Step 1
  Doubling each hour means a multiplier of $2$ per hour. After $n$ hours: $A = 200 \times 2^{n}$ .
  $A = 200 \times 2^{4}$
2. Step 2
  Evaluate $2^{4} = 16$ .
  $A = 200 \times 16 = 3200$
Answer
$3{,}200$ bacteria
Examiner tip
On a 'doubles every hour' question, the multiplier is $2$ (not $1 + 0.02$ ). Examiners credit candidates who write the formula explicitly before evaluating.
6Decay over a short period
CoreDirect calculation• decay, context
Question
The amount of a drug in a patient's blood decays by $20\%$ each hour. The initial dose is $400\,\text{mg}$ . Find the amount remaining after $3$ hours.
Step-by-step solution
1. Step 1
  Decay multiplier: $1 - 0.20 = 0.80$ .
2. Step 2
  Apply for $3$ hours.
  $A = 400 \times (0.80)^{3} = 400 \times 0.512$
3. Step 3
  Evaluate.
  $A = 204.8\,\text{mg}$
Answer
$204.8\,\text{mg}$
7Compound interest framed as exponential growth
ExtendedDirect calculation• Adapted from 0580/22 Oct/Nov 2023 Q14• compound interest, growth, context
Question
An investment of $3,200 grows at $2.5\%$ per year, compounded annually. (a) Write down the multiplier for one year. (b) Find the value of the investment after $8$ years.
Step-by-step solution
1. Step 1
  (a) Multiplier $= 1 + 0.025 = 1.025$ .
2. Step 2
  (b) Apply for $n = 8$ years.
  $A = 3200 \times (1.025)^{8}$
3. Step 3
  Evaluate $(1.025)^{8} = 1.2184\dots$
  $A = 3200 \times 1.21840 = 3898.89$
Answer
(a) Multiplier $= 1.025$ (b) $3,898.89
Examiner tip
The 2023 mark scheme awards a method mark for the correct multiplier. Stating $1.025$ explicitly is worth a mark even if the final arithmetic slips.
8Find the decay rate (reverse)
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q4• decay, reverse, rate
Question
A car worth $24,000 when new is worth $15,737.79 after $3$ years. Assuming constant percentage depreciation, find the annual rate of depreciation.
Step-by-step solution
1. Step 1
  Use $A = P(1 - r)^{n}$ and substitute the known values.
  $15737.79 = 24000 \times (1 - r)^{3}$
2. Step 2
  Divide both sides by $24000$ .
  $(1 - r)^{3} = 0.65574\dots$
3. Step 3
  Take the cube root.
  $1 - r = \sqrt[3]{0.65574} = 0.8700$
4. Step 4
  So $r = 1 - 0.87 = 0.13$ , i.e. $13\%$ per year.
Answer
$13\%$ per year
Examiner tip
The 2024 mark scheme awards method marks for explicitly showing the cube root. Candidates who simply quote the answer without working lose 2 of the 3 marks.
9Projected population — predict over a long horizon
ChallengeMulti-step problem• Adapted from 0580/42 Oct/Nov 2024 Q5• growth, population, projection
Question
A village has a population of $1{,}850$ in $2025$ , growing at $2.4\%$ per year. (a) Predict its population in $2040$ . (b) After how many full years will the population first exceed $3{,}000$ ?
Step-by-step solution
1. Step 1
  (a) Years between $2025$ and $2040$ : $n = 15$ .
  $P_{2040} = 1850 \times (1.024)^{15}$
2. Step 2
  Evaluate $(1.024)^{15} = 1.4279\dots$
  $P_{2040} = 1850 \times 1.4279 = 2641.65$
3. Step 3
  Round to the nearest integer: $2{,}642$ people.
4. Step 4
  (b) Solve $1850 \times 1.024^{n} > 3000$ , i.e. $1.024^{n} > 1.6216$ .
5. Step 5
  Try $n = 20$ : $1.024^{20} = 1.6084 < 1.6216$ . Try $n = 21$ : $1.024^{21} = 1.6470 > 1.6216$ . ✓
Answer
(a) Approximately $2{,}642$ people in $2040$ . (b) After $21$ full years (i.e. in $2046$ ).
Examiner tip
The 2024 examiner report notes that candidates often confuse 'full years' with rounded years. The population must exceed the threshold, so always check both the value at $n$ and $n - 1$ before stating the answer.
10Show that a decaying quantity first drops below a threshold at year $n$
ChallengeShow that / prove• Adapted from 0580/42 May/Jun 2023 Q22• show that, decay, threshold, reasoning
Question
A radioactive sample has initial mass $500\,\text{g}$ and decays at $8\%$ per year. Show that the mass first drops below $200\,\text{g}$ during the $11$ th year (i.e. between the end of year $10$ and end of year $11$ ).
Step-by-step solution
1. Step 1
  The mass after $n$ complete years is $A_{n} = 500 \times (0.92)^{n}$ . We must check $A_{10}$ and $A_{11}$ .
2. Step 2
  Compute $A_{10}$ .
  $A_{10} = 500 \times (0.92)^{10} = 500 \times 0.43439 = 217.20$
3. Step 3
  At the end of year $10$ , the mass is $217.20\,\text{g}$ , which is still above $200\,\text{g}$ .
4. Step 4
  Compute $A_{11}$ .
  $A_{11} = 500 \times (0.92)^{11} = 500 \times 0.39964 = 199.82$
5. Step 5
  At the end of year $11$ the mass is $199.82\,\text{g}$ , which is below $200\,\text{g}$ . Since $A_{10} > 200$ and $A_{11} < 200$ , the mass first falls below $200\,\text{g}$ during the $11$ th year. QED.
Answer
$A_{10} = 217.20\,\text{g} > 200$ and $A_{11} = 199.82\,\text{g} < 200$ , so the mass first drops below $200\,\text{g}$ during year $11$ .
Examiner tip
'Show that' questions require both sides of the boundary to be tested. Examiner reports flag that candidates often compute only $A_{11}$ and omit $A_{10}$ , losing the comparison mark.

Key Formulae — Exponential Growth and Decay

The formulae you need to memorise for exponential growth and decay on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Exponential growth
$A = P(1 + r)^{n}$
$P$
starting amount
$r$
growth rate per period (decimal: $5\% = 0.05$ )
$n$
number of periods
$A$
final amount after $n$ periods
When to use
Populations, savings, any quantity that grows by a fixed percentage each period.
Exponential decay
$A = P(1 - r)^{n}$
$r$
decay rate per period (decimal)
When to use
Depreciation of cars, radioactive decay, drug clearance — any percentage decrease per period.

Key Definitions and Keywords — Exponential Growth and Decay

Definitions to memorise and the exact keywords mark schemes credit for exponential growth and decay answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Exponential growth
Examiner keyword
An increase by a fixed percentage of the current value at each step.
Exponential decay
Examiner keyword
A decrease by a fixed percentage of the current value at each step.
Depreciation
The loss in value of an asset over time, usually modelled as exponential decay.
Multiplier
Examiner keyword
$1 + r$ for growth, $1 - r$ for decay. Multiplying by this each period gives the new value.

Common Mistakes and Misconceptions — Exponential Growth and Decay

The traps other students keep falling into on exponential growth and decay questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Using $(1 + r)^{n}$ for decay (or vice versa)
0580/42 — every recent series
Why it happens
Students forget to subtract for decay.
How to avoid it
Decay → $(1 - r)$ . Growth → $(1 + r)$ . Always check whether the quantity is increasing or decreasing.
✕Using the rate as a whole number ( $r = 5$ instead of $0.05$ )
Why it happens
Students paste percentage values directly into the formula.
How to avoid it
Always divide by $100$ first.
✕Multiplying by the rate instead of using the power
Why it happens
Treating exponential growth like simple proportion: $5000 \times 1.04 \times 5$ is wrong.
How to avoid it
Power, not multiplication: $(1.04)^{5}$ .
✕Rounding mid-calculation
Why it happens
Students round $(1.04)^{5}$ to $1.16$ early.
How to avoid it
Keep at least 4 d.p. in the multiplier; round only at the end.

Exponential Growth and Decay — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Cambridge IGCSE Mathematics (0580)

Exponential Growth and Decay

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Short Notes - Exponential Growth and Decay

Detailed Study Notes

Detailed notes on Number for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Exponential Growth and Decay — Cambridge IGCSE 0580 Maths Extended (2026)

At a glance

Growth: $A = A_{0}(1 + r/100)^{n}$ . Decay: $A = A_{0}(1 - r/100)^{n}$ .
$A_{0}$ = starting value. $r$ = % rate per period. $n$ = number of periods.
Same engine as compound interest, applied to populations, prices, mass, etc.
When rate is given as a decimal (e.g. $0.04$ for 4%), use $A = A_{0}(1 + 0.04)^{n}$ directly.
Half-life of decay = time for the quantity to halve.
Doubling time of growth = time for the quantity to double.
Solve for $n$ using logs (Extended only): $n = \dfrac{\log(A/A_{0})}{\log(1 + r/100)}$ .

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E1.17 — Use exponential growth and decay (e.g. compound interest, depreciation, population growth, radioactive decay).
Apply the formula $A = A_{0}(1 + r/100)^{n}$ for growth and $A = A_{0}(1 - r/100)^{n}$ for decay.

Exponential growth — the formula

Repeated percentage increase compounds. The total after $n$ periods is the original times the multiplier, raised to the $n$ -th power.

When a quantity grows by a fixed percentage $r\%$ each period, the value after $n$ periods is

$A = A_{0}\left(1 + \dfrac{r}{100}\right)^{n}.$

A₀ is where you start, (1 + r/100) is the per-period multiplier, and n is how many times it is applied.

This is the same compound-interest formula, repurposed.

Worked. "A bacterial colony of $400$ doubles in size every hour. Find the population after $5$ hours."

$r = 100\%$ per hour → multiplier $= 2$ .
$A = 400 \times 2^{5} = 400 \times 32 = 12{,}800$ .

Worked. "A town's population is $50{,}000$ and grows at $2.5\%$ per year. Find the population after $10$ years."

$A = 50{,}000 \times (1.025)^{10} \approx 50{,}000 \times 1.2801 \approx 64{,}004$ .

Why a power, not a multiplier and add? Each period multiplies by the same factor, so $n$ periods means multiplying $n$ times — that's $(\text{multiplier})^{n}$ .

Growth multiplies by more than 1 and accelerates; decay multiplies by less than 1 and flattens towards zero.

$A = A_{0}(1 + r/100)^{n}$ .
Each period the quantity is multiplied by $(1 + r/100)$ .
$n$ periods → raise to the power $n$ .
Always identify what $A_{0}$ , $r$ , $n$ refer to in the question.

Exponential decay

Same formula, MINUS sign on the rate. Used for depreciation, cooling, radioactive decay.

When a quantity decreases by a fixed percentage $r\%$ each period:

$A = A_{0}\left(1 - \dfrac{r}{100}\right)^{n}.$

The multiplier is now LESS than 1 (e.g. $0.85$ for a $15\%$ decline), so each period the quantity shrinks.

Worked. "A car is bought for $24,000 and depreciates by $15\%$ per year. Value after $4$ years?"

$A = 24000 \times (0.85)^{4} = 24000 \times 0.5220 \approx 12529.13$ , so $12,529.13.

Worked. "A radioactive sample contains $1{,}600\,\text{g}$ and decays by $3\%$ per day. How much remains after $20$ days?"

$A = 1600 \times (0.97)^{20} \approx 1600 \times 0.5438 \approx 870.0\,\text{g}$ .

$A = A_{0}(1 - r/100)^{n}$ .
Multiplier $< 1$ .
Examples: depreciation, cooling, radioactive decay.
Asymptotically approaches zero — never reaches it.

Half-life and doubling time

The natural way to describe an exponential change. Time for the quantity to halve (decay) or double (growth).

Half-life ( $T_{1/2}$ ) is the time for the quantity to halve. After $n$ half-lives, the original $A_{0}$ is reduced to $A_{0}/2^{n}$ .

Doubling time ( $T_{2}$ ) is the time for the quantity to double. After $n$ doublings, $A_{0}$ has grown to $A_{0} \times 2^{n}$ .

Worked. "An isotope has a half-life of $5$ days. Starting with $80\,\text{g}$ , how much remains after $20$ days?"

$20 / 5 = 4$ half-lives.
$A = 80 / 2^{4} = 80 / 16 = 5\,\text{g}$ .

Every half-life cuts the quantity in half — count how many fit into the total time.

Worked. "A bacteria culture doubles every $30$ minutes. Starting at $200$ , how many after $3$ hours?"

$3$ hours = $180$ minutes = $6$ doublings.
$A = 200 \times 2^{6} = 200 \times 64 = 12{,}800$ .

Each half-hour doubles the count; after 6 doublings 200 becomes 12,800.

Half-life: time to halve (decay).
Doubling time: time to double (growth).
After $n$ half-lives: $A = A_{0} / 2^{n}$ .
After $n$ doublings: $A = A_{0} \times 2^{n}$ .

Spotting an exponential question

Cambridge phrases exponential growth/decay in many disguises. Train yourself to recognise the cue.

If you see ANY of these phrasings, reach for the exponential formula:

"Increases / decreases by $r\%$ each year/month/day."
"Doubles every... / halves every..."
"Compound interest" / "depreciation".
"Population growth at a constant rate of $r\%$ ."
"Radioactive decay" with a half-life.

Same formula both ways — the cue words tell you whether to add or subtract the rate.

Compound interest is exponential growth. "Compound interest at $5\%$ per year for $10$ years" = "exponential growth at $5\%$ per year for $10$ years". Same formula.

Depreciation is exponential decay. "Depreciates by $20\%$ per year" = "decays by $20\%$ per year". Same formula.

Cue words: "each", "per", "every", "compound", "depreciate", "half-life".
Match the units of $n$ to the period in which $r$ is quoted.
Compound interest = exponential growth.
Depreciation = exponential decay.

Quick recap

Growth: $A = A_{0}(1 + r/100)^{n}$ . Decay: $A = A_{0}(1 - r/100)^{n}$ .
Same engine as compound interest, applied to populations / prices / mass.
Each period multiplies by $(1 \pm r/100)$ ; $n$ periods → raise to the $n$ .
Half-life = time to halve. Doubling time = time to double.
After $n$ half-lives, value is $A_{0} / 2^{n}$ .
Use logs to solve for $n$ when the value is given.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Exponential growth — repeated multiplication by a factor greater than 1: $A = A_{0}(1 + r/100)^{n}$ .
Exponential decay — repeated multiplication by a factor less than 1: $A = A_{0}(1 - r/100)^{n}$ .
Half-life — the time for an exponentially decaying quantity to halve.
Doubling time — the time for an exponentially growing quantity to double.
Multiplier — the factor $1 \pm r/100$ that applies one period of change.

How it’s examined

Take this whole topic with you

Step-by-step worked examples — Exponential Growth and Decay

Step-by-step solutions to past-paper-style questions on exponential growth and decay, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Exponential Growth and Decay

Almost every exponential growth and decay exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — find the amount/mass/population after $n$ periods, find the rate — with one exponential model to apply.

How to approach it

Write the multiplier ( $1+r$ for growth, $1-r$ for decay, or $2$ for doubling), apply $A=P\times(\text{multiplier})^{n}$ , and take an $n$ th root when the rate is the unknown.

Common trap

Examiner reports flag using $(1+r)$ for decay, multiplying by the rate instead of raising it to a power, and rounding the multiplier too early.

Multi-step problem

Recognise it by

Two linked tasks — predict a value over a long horizon and then find when a threshold is first crossed.

How to approach it

Handle each part in turn: apply the model for the stated number of periods, then trial successive values of $n$ , checking both $n$ and $n-1$ against the threshold.

Common trap

Examiner reports flag confusing 'full years' with rounded years — the quantity must strictly exceed the threshold, so verify the year before as well.

Show that / prove

Recognise it by

The words show that a quantity first crosses a threshold during a stated year.

How to approach it

Compute the value at the end of both bounding years ( $A_{n-1}$ and $A_{n}$ ) and state that one is above and the other below the threshold.

Common trap

Examiner reports flag candidates computing only the year after the crossing and omitting the year before — both sides of the boundary are required.

1Calculate exponential growth (population)
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q8• growth, population
Question
A town has a population of $24{,}000$ growing at $3\%$ per year. Find the population after $5$ years (to the nearest integer).
Step-by-step solution
1. Step 1
  Use $A = P(1 + r)^{n}$ with $r = 0.03$ .
  $A = 24000 \times (1.03)^{5}$
2. Step 2
  $(1.03)^{5} = 1.1592740\dots$
  $A = 24000 \times 1.1592740 = 27822.58\dots$
Answer
$27{,}823$ people
2Calculate exponential decay (depreciation)
ExtendedDirect calculation• decay, depreciation
Question
A car bought for $28,000 depreciates by $12\%$ each year. Find its value after $4$ years.
Step-by-step solution
1. Step 1
  Decay multiplier: $1 - 0.12 = 0.88$ .
  $A = 28000 \times (0.88)^{4}$
2. Step 2
  $(0.88)^{4} = 0.5996\dots$
  $A = 28000 \times 0.5996 = 16789.45$
Answer
$16,789.45
Examiner tip
Decay uses $(1 - r)$ , not $(1 + r)$ . Subtracting $12\%$ four times directly is also wrong.
3Find the time required (trial-and-improvement)
ExtendedDirect calculation• growth, time
Question
An investment of $5,000 grows at $4\%$ per year. After how many full years will it exceed $7,500?
Step-by-step solution
1. Step 1
  Try $n = 8$ : $5000 \times 1.04^{8} = 5000 \times 1.3686 = 6843.16$ . Not yet.
2. Step 2
  Try $n = 9$ : $5000 \times 1.04^{9} = 5000 \times 1.4233 = 7116.86$ . Not yet.
3. Step 3
  Try $n = 11$ : $5000 \times 1.04^{11} = 5000 \times 1.5395 = 7697.61$ . Yes — exceeds $7,500.
4. Step 4
  Check $n = 10$ : $5000 \times 1.04^{10} = 5000 \times 1.4802 = 7401.22$ . Below.
Answer
After $11$ full years
Examiner tip
Trial-and-improvement is fully accepted on 0580. Show each trial clearly — examiners credit the working.
4Apply exponential decay to half-life
ExtendedDirect calculation• decay, half-life
Question
A radioactive sample of mass $80\,\text{g}$ decays at $5\%$ per year. Find the mass after $10$ years.
Step-by-step solution
1. Step 1
  $A = 80 \times (0.95)^{10}$ .
2. Step 2
  $(0.95)^{10} = 0.5987\dots$
  $A = 80 \times 0.5987 = 47.90$
Answer
$47.9\,\text{g}$ (3 s.f.)
5Bacterial doubling — short calculation
CoreDirect calculation• growth, doubling
Question
A culture starts with $200$ bacteria. The number doubles every hour. How many bacteria are there after $4$ hours?
Step-by-step solution
1. Step 1
  Doubling each hour means a multiplier of $2$ per hour. After $n$ hours: $A = 200 \times 2^{n}$ .
  $A = 200 \times 2^{4}$
2. Step 2
  Evaluate $2^{4} = 16$ .
  $A = 200 \times 16 = 3200$
Answer
$3{,}200$ bacteria
Examiner tip
On a 'doubles every hour' question, the multiplier is $2$ (not $1 + 0.02$ ). Examiners credit candidates who write the formula explicitly before evaluating.
6Decay over a short period
CoreDirect calculation• decay, context
Question
The amount of a drug in a patient's blood decays by $20\%$ each hour. The initial dose is $400\,\text{mg}$ . Find the amount remaining after $3$ hours.
Step-by-step solution
1. Step 1
  Decay multiplier: $1 - 0.20 = 0.80$ .
2. Step 2
  Apply for $3$ hours.
  $A = 400 \times (0.80)^{3} = 400 \times 0.512$
3. Step 3
  Evaluate.
  $A = 204.8\,\text{mg}$
Answer
$204.8\,\text{mg}$
7Compound interest framed as exponential growth
ExtendedDirect calculation• Adapted from 0580/22 Oct/Nov 2023 Q14• compound interest, growth, context
Question
An investment of $3,200 grows at $2.5\%$ per year, compounded annually. (a) Write down the multiplier for one year. (b) Find the value of the investment after $8$ years.
Step-by-step solution
1. Step 1
  (a) Multiplier $= 1 + 0.025 = 1.025$ .
2. Step 2
  (b) Apply for $n = 8$ years.
  $A = 3200 \times (1.025)^{8}$
3. Step 3
  Evaluate $(1.025)^{8} = 1.2184\dots$
  $A = 3200 \times 1.21840 = 3898.89$
Answer
(a) Multiplier $= 1.025$ (b) $3,898.89
Examiner tip
The 2023 mark scheme awards a method mark for the correct multiplier. Stating $1.025$ explicitly is worth a mark even if the final arithmetic slips.
8Find the decay rate (reverse)
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q4• decay, reverse, rate
Question
A car worth $24,000 when new is worth $15,737.79 after $3$ years. Assuming constant percentage depreciation, find the annual rate of depreciation.
Step-by-step solution
1. Step 1
  Use $A = P(1 - r)^{n}$ and substitute the known values.
  $15737.79 = 24000 \times (1 - r)^{3}$
2. Step 2
  Divide both sides by $24000$ .
  $(1 - r)^{3} = 0.65574\dots$
3. Step 3
  Take the cube root.
  $1 - r = \sqrt[3]{0.65574} = 0.8700$
4. Step 4
  So $r = 1 - 0.87 = 0.13$ , i.e. $13\%$ per year.
Answer
$13\%$ per year
Examiner tip
The 2024 mark scheme awards method marks for explicitly showing the cube root. Candidates who simply quote the answer without working lose 2 of the 3 marks.
9Projected population — predict over a long horizon
ChallengeMulti-step problem• Adapted from 0580/42 Oct/Nov 2024 Q5• growth, population, projection
Question
A village has a population of $1{,}850$ in $2025$ , growing at $2.4\%$ per year. (a) Predict its population in $2040$ . (b) After how many full years will the population first exceed $3{,}000$ ?
Step-by-step solution
1. Step 1
  (a) Years between $2025$ and $2040$ : $n = 15$ .
  $P_{2040} = 1850 \times (1.024)^{15}$
2. Step 2
  Evaluate $(1.024)^{15} = 1.4279\dots$
  $P_{2040} = 1850 \times 1.4279 = 2641.65$
3. Step 3
  Round to the nearest integer: $2{,}642$ people.
4. Step 4
  (b) Solve $1850 \times 1.024^{n} > 3000$ , i.e. $1.024^{n} > 1.6216$ .
5. Step 5
  Try $n = 20$ : $1.024^{20} = 1.6084 < 1.6216$ . Try $n = 21$ : $1.024^{21} = 1.6470 > 1.6216$ . ✓
Answer
(a) Approximately $2{,}642$ people in $2040$ . (b) After $21$ full years (i.e. in $2046$ ).
Examiner tip
The 2024 examiner report notes that candidates often confuse 'full years' with rounded years. The population must exceed the threshold, so always check both the value at $n$ and $n - 1$ before stating the answer.
10Show that a decaying quantity first drops below a threshold at year $n$
ChallengeShow that / prove• Adapted from 0580/42 May/Jun 2023 Q22• show that, decay, threshold, reasoning
Question
A radioactive sample has initial mass $500\,\text{g}$ and decays at $8\%$ per year. Show that the mass first drops below $200\,\text{g}$ during the $11$ th year (i.e. between the end of year $10$ and end of year $11$ ).
Step-by-step solution
1. Step 1
  The mass after $n$ complete years is $A_{n} = 500 \times (0.92)^{n}$ . We must check $A_{10}$ and $A_{11}$ .
2. Step 2
  Compute $A_{10}$ .
  $A_{10} = 500 \times (0.92)^{10} = 500 \times 0.43439 = 217.20$
3. Step 3
  At the end of year $10$ , the mass is $217.20\,\text{g}$ , which is still above $200\,\text{g}$ .
4. Step 4
  Compute $A_{11}$ .
  $A_{11} = 500 \times (0.92)^{11} = 500 \times 0.39964 = 199.82$
5. Step 5
  At the end of year $11$ the mass is $199.82\,\text{g}$ , which is below $200\,\text{g}$ . Since $A_{10} > 200$ and $A_{11} < 200$ , the mass first falls below $200\,\text{g}$ during the $11$ th year. QED.
Answer
$A_{10} = 217.20\,\text{g} > 200$ and $A_{11} = 199.82\,\text{g} < 200$ , so the mass first drops below $200\,\text{g}$ during year $11$ .
Examiner tip
'Show that' questions require both sides of the boundary to be tested. Examiner reports flag that candidates often compute only $A_{11}$ and omit $A_{10}$ , losing the comparison mark.

Key Formulae — Exponential Growth and Decay

The formulae you need to memorise for exponential growth and decay on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Exponential growth
$A = P(1 + r)^{n}$
$P$
starting amount
$r$
growth rate per period (decimal: $5\% = 0.05$ )
$n$
number of periods
$A$
final amount after $n$ periods
When to use
Populations, savings, any quantity that grows by a fixed percentage each period.
Exponential decay
$A = P(1 - r)^{n}$
$r$
decay rate per period (decimal)
When to use
Depreciation of cars, radioactive decay, drug clearance — any percentage decrease per period.

Key Definitions and Keywords — Exponential Growth and Decay

Definitions to memorise and the exact keywords mark schemes credit for exponential growth and decay answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Exponential growth
Examiner keyword
An increase by a fixed percentage of the current value at each step.
Exponential decay
Examiner keyword
A decrease by a fixed percentage of the current value at each step.
Depreciation
The loss in value of an asset over time, usually modelled as exponential decay.
Multiplier
Examiner keyword
$1 + r$ for growth, $1 - r$ for decay. Multiplying by this each period gives the new value.

Common Mistakes and Misconceptions — Exponential Growth and Decay

The traps other students keep falling into on exponential growth and decay questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Using $(1 + r)^{n}$ for decay (or vice versa)
0580/42 — every recent series
Why it happens
Students forget to subtract for decay.
How to avoid it
Decay → $(1 - r)$ . Growth → $(1 + r)$ . Always check whether the quantity is increasing or decreasing.
✕Using the rate as a whole number ( $r = 5$ instead of $0.05$ )
Why it happens
Students paste percentage values directly into the formula.
How to avoid it
Always divide by $100$ first.
✕Multiplying by the rate instead of using the power
Why it happens
Treating exponential growth like simple proportion: $5000 \times 1.04 \times 5$ is wrong.
How to avoid it
Power, not multiplication: $(1.04)^{5}$ .
✕Rounding mid-calculation
Why it happens
Students round $(1.04)^{5}$ to $1.16$ early.
How to avoid it
Keep at least 4 d.p. in the multiplier; round only at the end.

Exponential Growth and Decay — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Exponential Growth and Decay

Short Notes - Exponential Growth and Decay

Detailed Study Notes

At a glance

What you’ll learn

Exponential growth — the formula

Exponential decay

Half-life and doubling time

Spotting an exponential question

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Exponential Growth and Decay

1Calculate exponential growth (population)

2Calculate exponential decay (depreciation)

3Find the time required (trial-and-improvement)

4Apply exponential decay to half-life

5Bacterial doubling — short calculation

6Decay over a short period

7Compound interest framed as exponential growth

8Find the decay rate (reverse)

9Projected population — predict over a long horizon

10Show that a decaying quantity first drops below a threshold at year nnn

Exponential growth

Exponential decay

Exponential growth

Exponential decay

Depreciation

Multiplier

✕Using (1+r)n(1 + r)^{n}(1+r)n for decay (or vice versa)

✕Using the rate as a whole number (r=5r = 5r=5 instead of 0.050.050.05)

✕Multiplying by the rate instead of using the power

✕Rounding mid-calculation

Exponential Growth and Decay — frequently asked questions

Exponential Growth and Decay

Short Notes - Exponential Growth and Decay

Detailed Study Notes

At a glance

What you’ll learn

Exponential growth — the formula

Exponential decay

Half-life and doubling time

Spotting an exponential question

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Exponential Growth and Decay

1Calculate exponential growth (population)

2Calculate exponential decay (depreciation)

3Find the time required (trial-and-improvement)

4Apply exponential decay to half-life

5Bacterial doubling — short calculation

6Decay over a short period

7Compound interest framed as exponential growth

8Find the decay rate (reverse)

9Projected population — predict over a long horizon

10Show that a decaying quantity first drops below a threshold at year nnn

Exponential growth

Exponential decay

Exponential growth

Exponential decay

Depreciation

Multiplier

✕Using (1+r)n(1 + r)^{n}(1+r)n for decay (or vice versa)

✕Using the rate as a whole number (r=5r = 5r=5 instead of 0.050.050.05)

✕Multiplying by the rate instead of using the power

✕Rounding mid-calculation

Exponential Growth and Decay — frequently asked questions

10Show that a decaying quantity first drops below a threshold at year $n$

✕Using $(1 + r)^{n}$ for decay (or vice versa)

✕Using the rate as a whole number ( $r = 5$ instead of $0.05$ )

10Show that a decaying quantity first drops below a threshold at year $n$

✕Using $(1 + r)^{n}$ for decay (or vice versa)

✕Using the rate as a whole number ( $r = 5$ instead of $0.05$ )