Exponential Growth & Decay
Exponential growth and decay problems challenge students to think beyond linear patterns, where quantities multiply rather than add by fixed amounts. In LK20 Trinn 10, students work with population models that double every 3 hours or car values that depreciate 15% annually, building critical thinking skills for real-world scenarios.
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Why it matters
Exponential models appear everywhere in science and finance. Bacteria colonies double every 20 minutes during optimal conditions, requiring healthcare workers to predict infection spread. Investment portfolios growing at 7% annually turn $10,000 into $19,672 after 10 years through compound interest. Radioactive materials with 8-year half-lives help doctors calculate safe radiation doses, while car values depreciating 12% yearly affect insurance calculations. Population growth at 2.1% annually means cities must plan infrastructure for 50% more residents every 33 years. Students who master these concepts understand pandemic modeling, retirement planning, and environmental decay rates that directly impact their future careers and personal decisions.
How to solve exponential growth & decay
Exponential Growth
- General form: y = a Β· bΛ£, 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 Β· 1.10Β³ β 665.5.
Worked examples
A bacteria colony starts with 500 bacteria and doubles every hour. How many bacteria are there after 5 hours?
Answer: 16000
- Identify the doubling pattern β 500 Γ 2β΅ β The colony doubles 5 times, so multiply by 2β΅.
- Calculate the power β 2β΅ = 32 β 2 multiplied by itself 5 times is 32.
- Multiply by the starting amount β 500 Γ 32 = 16000 β There are 16000 bacteria after 5 hours.
A town has 8,000 people and grows by 10% per year. How many people live there after 2 years?
Answer: 9680
- Find the growth factor β 1 + 10/100 = 1.1 β A 10% increase means multiplying by 1.1 each year.
- Year 1 β 8000 Γ 1.1 = 8800 β After year 1 the population is 8800.
- Year 2 β 8800 Γ 1.1 = 9680 β After year 2 the population is 9680.
- Verify with formula β A = 8000 Γ 1.1Β² = 9680 β Using A = P Γ (1 + r)α΅ confirms the answer.
A radioactive sample of 120 g has a half-life of 5 years. How much remains after 10 years?
Answer: 30 g
- Find number of half-lives β 10 Γ· 5 = 2 β In 10 years, the sample halves 2 times.
- Halving 1 β 120 Γ· 2 = 60 β After 5 years: 60 g remaining.
- Halving 2 β 60 Γ· 2 = 30 β After 10 years: 30 g remaining.
- Verify with formula β 120 Γ (1/2)Β² = 30 β Using the half-life formula confirms the answer.
Common mistakes
- βStudents often add the growth rate instead of multiplying: for 500 bacteria doubling 3 times, they calculate 500 + 500 + 500 + 500 = 2000 instead of 500 Γ 2Β³ = 4000.
- βConfusing growth factor with growth rate: for 15% growth, students use b = 0.15 instead of b = 1.15, calculating 1000 Γ 0.15Β² = 22.5 instead of 1000 Γ 1.15Β² = 1322.5.
- βMixing up half-life periods: with 80g decaying over 12 years (4-year half-life), students calculate 80 Γ· 12 = 6.67g instead of 80 Γ (1/2)Β³ = 10g.
- βUsing wrong base for time conversion: when population doubles every 3 hours, students calculate 6 hours as 2Β² instead of 2βΆ/Β³ = 2Β² = 4 times growth.
Practice on your own
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