Population Growth Models

Model growing populations of bacteria, cities and animals with exponential functions, doubling time and the rule of 70.

Advanced30 min18 problems
Practice Problems

Definition

A population model is a function that predicts how many individuals (people, bacteria, animals) there will be after a time tt. When a population grows by the same percentage every year, it follows an exponential model:
P(t)=P0(1+r)tP(t) = P_0 (1 + r)^t
  • P0P_0 = initial population (at t=0t = 0)
  • rr = yearly growth rate as a decimal (3% means r=0.03r = 0.03)
  • 1+r1 + r = growth factor
  • tt = time in years
When we know the doubling time dd (how long it takes the population to double), we can write the same kind of model as:
P(t)=P0⋅2t/dP(t) = P_0 \cdot 2^{t/d}
The exponent td\frac{t}{d} counts how many doublings have happened. A quick estimate links the two forms, the rule of 70:
d≈70growth rate in percentd \approx \frac{70}{\text{growth rate in percent}}
It is only an approximation, but it is very good for small rates.
The bar chart shows a population of 1000010000 that grows 10%10\% per year: 1000010000, 1100011000, 1210012100, 1331013310, 1464114641. Each bar is 1.11.1 times the one before it, so the jumps get bigger every year.

Try it now

A town of 20002000 people grows by 4%4\% per year. Which function models its population after tt years?

Worked Examples

A city has 5000050000 inhabitants and grows by 2%2\% per year. Estimate its population after 5 years.
1

Identify P0P_0, rr and tt

P0=50000P_0 = 50000, r=0.02r = 0.02, t=5t = 5 → Values identified
2

Find the growth factor

1+r=1+0.02=1.021 + r = 1 + 0.02 = 1.02 → Growth factor 1.021.02
3

Write the model

P(t)=50000⋅1.02tP(t) = 50000 \cdot 1.02^t → Model ready
4

Evaluate at t=5t = 5

P(5)=50000⋅1.025≈50000⋅1.10408≈55204P(5) = 50000 \cdot 1.02^5 \approx 50000 \cdot 1.10408 \approx 55204 → P(5)≈55204P(5) \approx 55204

Common Mistakes

Using the percent itself in the growth factor: 50000⋅(1+2)t50000 \cdot (1 + 2)^t

Why it's wrong: A rate of 2%2\% is 0.020.02 as a decimal. Writing 1+2=31 + 2 = 3 would mean the city triples every year.

Correct: Convert the percent first: 2%=0.022\% = 0.02, so the growth factor is 1.021.02 and P(t)=50000⋅1.02tP(t) = 50000 \cdot 1.02^t.

Forgetting to divide by the doubling time: 300⋅2180300 \cdot 2^{180}

Why it's wrong: The exponent must count doublings, not minutes. In 180 minutes there are only 18030=6\frac{180}{30} = 6 doublings.

Correct: Use P(t)=P0⋅2t/dP(t) = P_0 \cdot 2^{t/d}: 300⋅2180/30=300⋅26=19200300 \cdot 2^{180/30} = 300 \cdot 2^6 = 19200.

Using the decimal rate in the rule of 70: 700.035=2000\frac{70}{0.035} = 2000 years

Why it's wrong: The rule of 70 is built for the rate written as a percent number. With the decimal you get an answer 100 times too large.

Correct: Divide 70 by the percent number: 703.5=20\frac{70}{3.5} = 20 years, and remember it is an estimate.

Dividing by the later count when finding the rate: 1201320≈9.1%\frac{120}{1320} \approx 9.1\%

Why it's wrong: The growth rate is measured relative to where the population started, not where it ended.

Correct: Divide the new count by the old count: 13201200=1.1\frac{1320}{1200} = 1.1, so r=10%r = 10\%.

Interactive Visual

Sequence Explorer

300,
600,
1200,
2400,
4800,
9600,
19,200
, ...

Formula

a_n = 300 × 2^(n-1)
1234567
a_1

300

a_5

4800

a_10

153,600

a_20

157,286,400

Bar Chart

t = 0(10,000)
t = 1(11,000)
t = 2(12,100)
t = 3(13,310)
t = 4(14,641)

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Practice Problems

18 problems
Problem 1 of 18
Easy
A town of 20002000 people grows by 4%4\% per year. Which function models its population after tt years?

Why It Matters

Population models help people plan ahead:
  • Cities use growth forecasts to decide how many schools, homes and buses they will need in 10 or 20 years
  • Biologists and doctors predict how fast bacteria multiply in a sample or in an infection
  • Wildlife managers track endangered species and invasive animals to see whether a population is recovering or exploding
  • Governments compare linear and exponential forecasts to plan water, energy and food supplies
Knowing the difference between a straight line and a curve that keeps doubling can change a plan completely: an exponential forecast can be thousands of people higher after only a few decades. Real populations cannot grow exponentially forever, so a good forecaster also knows when the model stops working.

Real World Applications

City Planning

City planners forecast how many residents a city will have so they can build enough schools, apartments and public transport in time.

Example:

A city of 250000250000 that grows 1%1\% per year would have about 250000⋅1.0110≈276156250000 \cdot 1.01^{10} \approx 276156 residents after 10 years.

1Try It Yourself

A fast-growing city has 8000080000 residents and grows by 5%5\% per year.

How many residents will it have after 3 years?

Step 1: Write the mathematical expression

Calculate 80000⋅1.05380000 \cdot 1.05^3:

Bacteria in the Lab

Microbiologists use doubling times to predict when a culture will be large enough to study, or how fast an infection can spread before treatment.

Example:

Some bacteria double about every 20 minutes in ideal conditions, which is 3 doublings, or a factor of 23=82^3 = 8, every hour.

2Try It Yourself

A culture starts with 10001000 bacteria that double every 2020 minutes.

How many bacteria are there after 1 hour and 40 minutes?

Step 1: Write the mathematical expression

Calculate 1000⋅2100/201000 \cdot 2^{100/20}:

Wildlife Recovery

Conservation teams count animals every year, build a growth model from the counts and check whether a protected species is recovering. They also know that a reserve can only feed a limited number of animals, its carrying capacity, so growth will slow down later.

Example:

If a herd grows from 500500 to 540540 animals in a year, the growth factor is 540500=1.08\frac{540}{500} = 1.08, so the herd grows 8%8\% per year while food is plentiful.

3Try It Yourself

A wolf pack in a reserve grew from 6464 wolves to 8080 wolves in one year.

If the same yearly percentage continues, how many wolves will there be 2 years after the count of 80?

Step 1: Write the mathematical expression

Calculate 80⋅(80/64)280 \cdot (80/64)^2:

Key Takeaways

  • 1A population that grows by the same percentage each year follows P(t)=P0(1+r)tP(t) = P_0 (1 + r)^t, with rr as a decimal
  • 2With a doubling time dd, the model is P(t)=P0⋅2t/dP(t) = P_0 \cdot 2^{t/d}, where td\frac{t}{d} counts the doublings
  • 3The rule of 70 estimates the doubling time: d≈70percent rated \approx \frac{70}{\text{percent rate}}; it is an approximation, not an exact formula
  • 4From two yearly counts, the growth factor is P1P0\frac{P_1}{P_0} and the rate is r=P1P0−1r = \frac{P_1}{P_0} - 1
  • 5Linear forecasts add the same number each year; exponential forecasts multiply, so they pull ahead over time
  • 6Real populations slow down as they approach the carrying capacity of their environment, so exponential models work best for early growth

Frequently Asked Questions

The exact doubling time is d=ln⁡2ln⁡(1+r)d = \frac{\ln 2}{\ln(1 + r)}, and for small rates ln⁡(1+r)\ln(1 + r) is very close to rr. Since ln⁡2≈0.693\ln 2 \approx 0.693, the doubling time is about 0.693r=69.3percent\frac{0.693}{r} = \frac{69.3}{\text{percent}}, rounded to 70. You will meet logarithms in a later lesson; for now, use the rule as a quick estimate.
The exact doubling time is d=ln⁡2ln⁡(1+r)d = \frac{\ln 2}{\ln(1 + r)}, and for small rates ln⁡(1+r)\ln(1 + r) is very close to rr. Since ln⁡2≈0.693\ln 2 \approx 0.693, the doubling time is about 0.693r=69.3percent\frac{0.693}{r} = \frac{69.3}{\text{percent}}, rounded to 70. You will meet logarithms in a later lesson; for now, use the rule as a quick estimate.
For rates up to about 10%10\% it is usually within a year. At 2%2\% it gives 3535 years and the exact value is about 35.035.0 years. At 7%7\% it gives 1010 years and the exact value is about 10.210.2 years. For large rates the error grows.
The carrying capacity is the largest population an environment can support with its food, water and space. As a population gets close to it, growth slows down and the curve flattens into an S shape. This is called logistic growth; in this lesson we only describe it, without a formula.
Look at the data. If the population increases by about the same number each year, use a linear model. If it increases by about the same percentage each year (the ratios P1P0\frac{P_1}{P_0}, P2P1\frac{P_2}{P_1} are about equal), use an exponential model.

Glossary

Population model
A function P(t)P(t) that predicts the size of a population after time tt
Initial population
The population P0P_0 at the start, when t=0t = 0
Growth rate
The percentage by which the population grows each period, written as a decimal rr in the formula
Doubling time
The time dd it takes a population to double in size
Rule of 70
The estimate d≈70percent rated \approx \frac{70}{\text{percent rate}} for the doubling time
Carrying capacity
The largest population an environment can support for a long time

Formula Card

Population model with a yearly rate

P(t)=P0(1+r)tP(t) = P_0 (1 + r)^t

$P_0$ = initial population, $r$ = yearly growth rate as a decimal, $t$ = years.

Doubling-time model

P(t)=P0⋅2t/dP(t) = P_0 \cdot 2^{t/d}

$d$ = doubling time, in the same unit as $t$. The exponent $\frac{t}{d}$ is the number of doublings.

Rule of 70 (estimate)

d≈70percent rated \approx \frac{70}{\text{percent rate}}

Approximate doubling time. Use the rate as a percent number, for example $\frac{70}{2} = 35$ years for $2\%$.

Growth factor from two counts

1+r=P1P01 + r = \frac{P_1}{P_0}

$P_0$ and $P_1$ are counts taken one period apart.

Today's challengeOne problem, three tries

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