Exponential Growth & Decay
Exponential growth and decay models appear throughout GCSE mathematics, from population dynamics to compound interest calculations. Students often struggle with the distinction between linear and exponential change, particularly when applying the formula y = a · bˣ to real-world scenarios.
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Why it matters
Exponential functions model countless real-world phenomena that GCSE students encounter daily. Bank accounts earning compound interest at 5% annually demonstrate growth where money multiplies rather than adds linearly. Disease spread during pandemics follows exponential patterns, with cases potentially doubling every 3 days without intervention. Car depreciation typically follows exponential decay, losing 15-20% of value annually. Radioactive decay in physics follows half-life patterns, crucial for medical treatments and carbon dating. Mobile phone battery discharge, population growth in biology, and even social media follower growth all follow exponential patterns. Understanding these models helps students interpret graphs, make predictions, and solve practical problems involving compound changes over time.
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 100 bacteria and doubles every hour. How many bacteria are there after 3 hours?
Answer: 800
- Identify the doubling pattern → 100 × 2³ — The colony doubles 3 times, so multiply by 2³.
- Calculate the power → 2³ = 8 — 2 multiplied by itself 3 times is 8.
- Multiply by the starting amount → 100 × 8 = 800 — There are 800 bacteria after 3 hours.
A town has 10,000 people and grows by 20% per year. How many people live there after 2 years?
Answer: 14400
- Find the growth factor → 1 + 20/100 = 1.2 — A 20% increase means multiplying by 1.2 each year.
- Year 1 → 10000 × 1.2 = 12000 — After year 1 the population is 12000.
- Year 2 → 12000 × 1.2 = 14400 — After year 2 the population is 14400.
- Verify with formula → A = 10000 × 1.2² = 14400 — 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 2 years?
Answer: £162,000.00
- Find the decay factor → 1 − 10/100 = 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.
- Verify with formula → A = 200,000 × 0.9² = 162,000 — Using A = P × (1 − r)ᵗ confirms the answer.
Common mistakes
- Students confuse exponential and linear growth, calculating 500 + 3 × 50 = 650 instead of 500 × 1.1³ = 665.5 for 10% annual growth over 3 years.
- When finding decay factors, students add the percentage instead of subtracting, using 1.15 instead of 0.85 for 15% annual depreciation.
- Students forget to convert percentages to decimals, writing 500 × 20² = 200,000 instead of 500 × 1.2² = 720 for 20% growth over 2 years.
- Many students mix up the base and exponent positions, calculating 3² × 100 = 900 instead of 100 × 2³ = 800 for population doubling 3 times.