How accurate is the rule of 70?

For rates up to about $10\%$ it is usually within a year. At $2\%$ it gives $35$ years and the exact value is about $35.0$ years. At $7\%$ it gives $10$ years and the exact value is about $10.2$ years. For large rates the error grows.

What is carrying capacity?

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.

Should I use a linear or an exponential model?

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 $\frac{P_1}{P_0}$, $\frac{P_2}{P_1}$ are about equal), use an exponential model.

Teacher Resources

Population Growth Models

Q: Why does the rule of 70 work?

The exact doubling time is $d = \frac{\ln 2}{\ln(1 + r)}$, and for small rates $\ln(1 + r)$ is very close to $r$. Since $\ln 2 \approx 0.693$, the doubling time is about $\frac{0.693}{r} = \frac{69.3}{\text{percent}}$, rounded to 70. You will meet logarithms in a later lesson; for now, use the rule as a quick estimate.

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

Advanced30 minLesson

Printable worksheet

Definition

A population model is a function that predicts how many individuals (people, bacteria, animals) there will be after a time

t

. When a population grows by the same percentage every year, it follows an exponential model:

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

$P_0$ = initial population (at $t = 0$ )
$r$ = yearly growth rate as a decimal (3% means $r = 0.03$ )
$1 + r$ = growth factor
$t$ = time in years

When we know the doubling time

d

(how long it takes the population to double), we can write the same kind of model as:

P(t) = P_0 \cdot 2^{t/d}

The exponent

\frac{t}{d}

counts how many doublings have happened. A quick estimate links the two forms, the rule of 70:

d \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

10000

that grows

10\%

per year:

10000

11000

12100

13310

14641

. Each bar is

1.1

times the one before it, so the jumps get bigger every year.

Try it now

A town of

2000

people grows by

4\%

per year. Which function models its population after

t

years?

Worked Examples

A city has $50000$ inhabitants and grows by $2\%$ per year. Estimate its population after 5 years.

Identify $P_0$ , $r$ and $t$

$P_0 = 50000$ , $r = 0.02$ , $t = 5$ → Values identified

Find the growth factor

$1 + r = 1 + 0.02 = 1.02$ → Growth factor $1.02$

Write the model

$P(t) = 50000 \cdot 1.02^t$ → Model ready

Evaluate at $t = 5$

$P(5) = 50000 \cdot 1.02^5 \approx 50000 \cdot 1.10408 \approx 55204$ → $P(5) \approx 55204$

Answer: After 5 years the city will have about 55204 inhabitants.

Common Mistakes

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

Why it's wrong: A rate of $2\%$ is $0.02$ as a decimal. Writing $1 + 2 = 3$ would mean the city triples every year.

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

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

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

Correct: Use $P(t) = P_0 \cdot 2^{t/d}$ : $300 \cdot 2^{180/30} = 300 \cdot 2^6 = 19200$ .

Using the decimal rate in the rule of 70: $\frac{70}{0.035} = 2000$ 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: $\frac{70}{3.5} = 20$ years, and remember it is an estimate.

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

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: $\frac{1320}{1200} = 1.1$ , so $r = 10\%$ .

Interactive Visual

Sequence Explorer

First Term (a₁ = 300)

Common Ratio (r = 2)

300,

600,

1200,

2400,

4800,

9600,

19,200

, ...

Formula

a_n = 300 × 2^(n-1)

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)

Interactive Sandbox

Expression Calculator

Try these:

History

No calculations yet

Practice Problems

18 problems

Problem 1 of 18

Easy

A town of $2000$ people grows by $4\%$ per year. Which function models its population after $t$ 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 $250000$ that grows $1\%$ per year would have about $250000 \cdot 1.01^{10} \approx 276156$ residents after 10 years.

1Try It Yourself

A fast-growing city has $80000$ residents and grows by $5\%$ per year.

How many residents will it have after 3 years?

Step 1: Write the mathematical expression

Calculate $80000 \cdot 1.05^3$ :

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 $2^3 = 8$ , every hour.

2Try It Yourself

A culture starts with $1000$ bacteria that double every $20$ minutes.

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

Step 1: Write the mathematical expression

Calculate $1000 \cdot 2^{100/20}$ :

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

3Try It Yourself

A wolf pack in a reserve grew from $64$ wolves to $80$ 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 $80 \cdot (80/64)^2$ :

Key Takeaways

1A population that grows by the same percentage each year follows $P(t) = P_0 (1 + r)^t$ , with $r$ as a decimal
2With a doubling time $d$ , the model is $P(t) = P_0 \cdot 2^{t/d}$ , where $\frac{t}{d}$ counts the doublings
3The rule of 70 estimates the doubling time: $d \approx \frac{70}{\text{percent rate}}$ ; it is an approximation, not an exact formula
4From two yearly counts, the growth factor is $\frac{P_1}{P_0}$ and the rate is $r = \frac{P_1}{P_0} - 1$
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 = \frac{\ln 2}{\ln(1 + r)}

, and for small rates

\ln(1 + r)

is very close to

r

. Since

\ln 2 \approx 0.693

, the doubling time is about

\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.

Glossary

Population model: A function $P(t)$ that predicts the size of a population after time $t$
Initial population: The population $P_0$ at the start, when $t = 0$
Growth rate: The percentage by which the population grows each period, written as a decimal $r$ in the formula
Doubling time: The time $d$ it takes a population to double in size
Rule of 70: The estimate $d \approx \frac{70}{\text{percent rate}}$ 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) = P_0 (1 + r)^t

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

Doubling-time model

P(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 \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 = \frac{P_1}{P_0}

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

Population Growth Models

Definition

Worked Examples

Common Mistakes

Interactive Visual

Sequence Explorer

Formula

Bar Chart

Interactive Sandbox

Expression Calculator

History

Practice Problems

Why It Matters

Real World Applications

City Planning

Bacteria in the Lab

Wildlife Recovery

Key Takeaways

Frequently Asked Questions

Glossary

Formula Card

Population model with a yearly rate

Doubling-time model

Rule of 70 (estimate)

Growth factor from two counts

More in This Topic

Introduction to Exponential Functions

Exponential Growth

Graphing Exponential Functions

Exponential Decay