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Exponential Growth and Decay

GCSELevel 6-7Cambridge iGCSE

Exponential Growth and Decay Revision

Exponential Growth and Decay

Exponential growth and decay describe the change of a current value.

Exponential growth is when the growth rate increases in proportion to the growing total or number.

Exponential decay is the opposite, where the decay rate decreases in proportion to the current total or number.


Make sure you are happy with the following topics before continuing:

Exponential Growth

When a value is growing exponentially, the greater the quantity, the faster the value grows. We often see exponential growth in population models.


A population of ducks is growing exponentially. Originally, there were \textcolor{blue}{6} ducks. The number of ducks doubles each year.

As the number of ducks doubles each year, the population of ducks will always be multiplied by a power of 2. We can model this exponentially with the equation:

\textcolor{purple}{D} = \textcolor{blue}{A}\times 2^\textcolor{orange}{x}

\textcolor{purple}{D} = number of ducks after \textcolor{orange}{x}\text{ years}

\textcolor{blue}{A} = initial number of ducks

For example, we could work out how many ducks there would be after 3 years by inputting \textcolor{orange}{x}=3

\textcolor{purple}{D} = \textcolor{blue}{6}\times 2^\textcolor{orange}{3}

\textcolor{purple}{D} = \textcolor{blue}{6}\times 2^\textcolor{orange}{3}

\textcolor{purple}{D} = 48

Level 6-7GCSECambridge iGCSE

Exponential Decay

When a value is decaying exponentially, the lower the quantity, the slower the value decreases. We may also see exponential decay in population models for endangered species.


A population of elephants is decaying exponentially. There are currently \textcolor{blue}{72900} elephants. The number of elephants divides by 3 each year.

As the number of elephants divides by 3, the population size will be multiplied by a power of \dfrac{1}{3}

\textcolor{purple}{E} = \textcolor{blue}{A}\times \dfrac{1}{3}^\textcolor{orange}{x}

This may also be written as:

\textcolor{purple}{E} = \textcolor{blue}{A}\times 3^\textcolor{orange}{-x}

\textcolor{purple}{E} = number of elephants after \textcolor{orange}{x}\text{ years}

\textcolor{blue}{A} = initial number of elephants

For example, we could work out how many elephants there would be after 5 years by inputting \textcolor{orange}{x}=-5

\textcolor{purple}{E} = \textcolor{blue}{72900}\times 3^\textcolor{orange}{-5}

\textcolor{purple}{E} = \textcolor{blue}{72900}\times 3^\textcolor{orange}{-5}

\textcolor{purple}{E} =300

Level 6-7GCSECambridge iGCSE

Example 1: Exponential Growth

A colony of bacteria increases exponentially each hour. The colony had 500 bacteria initially, and increased to 32,000 after 3 hours. Calculate by how much the population increases per hour.

[3 marks]

We can use the equation we found earlier:

\textcolor{purple}{32,000} = \textcolor{blue}{500}\times r^\textcolor{orange}{3}

Where r is the population increase per year.

64 = r^\textcolor{orange}{3}

Cube root each side:

\sqrt[3]{64} = r\\

r = 4

Level 6-7GCSECambridge iGCSE

Example 2: Exponential Decay

Radioactive substances become less toxic over time, and the time taken for them to become half as toxic is called their half-life. Carbon has a half life of 5730\text{ years}. If a sample contains 1\text{ g} of carbon, work out how many grams of carbon it would have after 22920\text{ years}.

[2 marks]

The unit of time used in this exponential decay is 5730\text{ years}, so we can work out how many of these units we are working out the exponential decay for:

22920\div 5730 =4 units of time.

We can write this as an equation: 

\textcolor{purple}{A} = \textcolor{blue}{1}\times \dfrac{1}{2}^\textcolor{orange}{4}

\textcolor{purple}{A} = \textcolor{blue}{1}\times 2^\textcolor{orange}{-4}

Where A is the amount of Carbon.

\textcolor{purple}{A} = 2^\textcolor{orange}{-4}\\

\textcolor{purple}{A} = \dfrac{1}{16}\text{ g}

Level 6-7GCSECambridge iGCSE

Exponential Growth and Decay Example Questions

We know with exponential growth, the greater the quantity, the faster it grows, so the biggest difference will be between 8 and 9 (d).

This means, the population grows most between 8 and 9

We can form an equation:

583200=A\times 3^6\\ A=\dfrac{583200}{3^6}\\

A=800 bears initially.

We can form an equation:

W = 100,000\times 2^{-5}\\ W =3125

Therefore, as the population would be higher than 3000, the killer whales will not be critically endangered.

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