Exponential Growth & Decay
Exponential growth and decay describe quantities that change by multiplying by a constant factor over equal time periods. The general form y = a · bˣ represents this relationship, where a is the initial value, b is the growth factor, and x is the number of time periods. When b > 1, the quantity grows exponentially; when 0 < b < 1, it decays exponentially.
Why it matters
Exponential growth and decay appear throughout science, finance, and real-world modelling. Population growth often follows exponential patterns, with human populations doubling roughly every 70 years at current rates. Investment calculations rely on compound interest, where £1000 at 5% annual interest becomes £1276.28 after 5 years. Radioactive decay follows exponential patterns, with carbon-14 having a half-life of 5730 years for archaeological dating. Medical dosages decay exponentially in the body, with caffeine having a half-life of about 6 hours. Car values typically depreciate exponentially, losing 15-20% annually. These concepts form the foundation for logarithms and advanced calculus topics in A-level mathematics, connecting algebra to real scientific applications and financial literacy.
How to solve exponential growth & decay
Exponential Growth
- General form: y = a · bˣ, where a is the starting value and b is the growth factor.
- If b > 1, the quantity grows; if 0 < b < 1, it decays.
- Percent growth of r% means b = 1 + r/100.
- To find y after x periods, substitute and evaluate.
Example: A population of 500 grows 10% per year. After 3 years: y = 500 · 1.10³ ≈ 665.5.
Worked examples
A bacteria colony starts with 50 bacteria and doubles every hour. How many bacteria are there after 5 hours?
Answer: 1600
- Identify the doubling pattern → 50 × 2⁵ — The colony doubles 5 times, so multiply by 2⁵.
- Calculate the power → 2⁵ = 32 — 2 multiplied by itself 5 times is 32.
- Multiply by the starting amount → 50 × 32 = 1600 — There are 1600 bacteria after 5 hours.
A town has 5,000 people and grows by 20% per year. How many people live there after 3 years?
Answer: 8640
- Find the growth factor → 1 + 20100 = 1.2 — A 20% increase means multiplying by 1.2 each year.
- Year 1 → 5000 × 1.2 = 6000 — After year 1 the population is 6000.
- Year 2 → 6000 × 1.2 = 7200 — After year 2 the population is 7200.
- Year 3 → 7200 × 1.2 = 8640 — After year 3 the population is 8640.
- Verify with formula → A = 5000 × 1.2³ = 8640 — Using A = P × (1 + r)ᵗ confirms the answer.
A car worth £200,000.00 loses 10% of its value each year. What is it worth after 3 years?
Answer: £145,800.00
- Find the decay factor → 1 − 10100 = 0.9 — Losing 10% means multiplying by 0.9 each year.
- Year 1 → 200000 × 0.9 = 180000 — After year 1 the value is £180,000.00.
- Year 2 → 180000 × 0.9 = 162000 — After year 2 the value is £162,000.00.
- Year 3 → 162000 × 0.9 = 145800 — After year 3 the value is £145,800.00.
- Verify with formula → A = 200,000 × 0.9³ = 145,800 — Using A = P × (1 − r)ᵗ confirms the answer.
Common mistakes
- Confusing growth factor with percentage rate, such as using 20 instead of 1.2 for 20% growth, giving 50 × 20³ = 400,000 instead of 50 × 1.2³ = 86.4
- Adding percentages instead of multiplying factors, calculating 100 + 10% + 10% + 10% = 130 instead of 100 × 1.1³ = 133.1 for three years of 10% growth
- Using the wrong base for doubling problems, writing 100 × 3⁴ = 8100 instead of 100 × 2⁴ = 1600 when something doubles 4 times