Exponential Growth and Decay • Topic 3 of 4

Population Models

Populations that grow by a steady percentage each year are a classic application of A = P(1 + r)^t, with P the starting population, r the annual growth rate as a decimal, and t the number of years. A town of 10,000 growing 20% per year has 10,000(1.2)² = 14,400 after 2 years. The model assumes a constant percentage rate, so each year’s increase is larger than the last (it compounds). Set up (1 + r)^t exactly as with general growth, reading “per year” as the period. The SAT uses small year counts so the result is exact, and may ask you to interpret the base or rate in context.

Population growth curve rising and curving upward more steeply each yearPopulation growthyearspopulationPopulation grows by a fixed percent each year

✅ Solved examples

1. A town of 10,000 grows 20% per year. Population after 2 years?
10000(1.2)² = 14,400.
2. A town of 5,000 grows 10% per year. After 2 years?
5000(1.1)² = 6,050.
3. A colony of 2,000 grows 50% per year. After 2 years?
2000(1.5)² = 4,500.
4. In P = 8000(1.1)^t, what is the starting population?
8,000 (the value at t = 0).

✏️ Practice — try these, take hints as needed

1. A town of 4,000 grows 10% per year. Population after 2 years?
4000(1.1)².
4000 × 1.21.
—
4,840.
2. A town of 6,000 grows 20% per year. After 2 years?
6000 × 1.44.
—
—
8,640.
3. A town of 2,000 grows 10% per year. After 2 years?
2000 × 1.21.
—
—
2,420.
4. A colony of 1,000 grows 50% per year. After 2 years?
1000 × 2.25.
—
—
2,250.
5. In P = 12000(1.2)^t, what is the starting population?
Value at t = 0.
—
—
12,000.

📝 Topic test — 8 questions

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