# Modelling Exponential Growth and Decay

## Modelling Exponential Growth and Decay Revision

**Modelling Exponential Functions and the Natural Logarithm**

It is important to know how to use e^{x} and \ln(x) in **real life**. This means you will need to be able to **sketch graphs** of e^{ax+b}+c and \ln(ax+b), and **interpret word-based problems** about **growth** and **decay**.

**Exponential Function Graphs**

You need to know how to **sketch** y=e^{ax+b}+c. The best way to do this is **step by step**.

**Step 1: **Find where the **graph** **intercepts** **the axes**.

**y-axis** is at x=0, so **y-intercept** is at y=e^{b}+c

**x-axis** is at y=0, so **x-intercept** is at e^{ax+b}+c=0, which is an equation **we know how to solve**. Note that if c>0 the **graph** **does not cross the x-axis**.

**Step 2: **Find the **asymptotes** by looking at the **behaviour** of the **graph** as x\rightarrow\pm\infty.

As x\rightarrow\infty, y=e^{ax+b}+c\rightarrow\infty if a>0

and y=e^{ax+b}+c\rightarrow c if a<0

As x\rightarrow -\infty, y=e^{ax+b}+c\rightarrow c if a>0

and y=e^{ax+b}+c\rightarrow\infty if a<0

**Step 3: **Mark all of this information on a **graph** and use it to **plot the graph**.

**Natural Logarithm Graphs**

You need to know how to **sketch** y=\ln(ax+b). The best way to do this is **step by step**.

**Step 1: **Find where the **graph** **intercepts the axes**.

**y-axis** is at x=0, so **y-intercept** is at y=\ln(b)

**x-axis** is at y=0, so **x-intercept** is at ax+b=1, which is

x=\dfrac{1-b}{a}

**Step 2: **Find the** asymptotes** by looking at the **behaviour** of the **graph** as x\rightarrow\pm\infty.

If a>0, as x\rightarrow\infty, y=\ln(ax+b)\rightarrow\infty and x can decrease until ax+b=0, so there is an **asymptote** at x=\dfrac{-b}{a} where

y\rightarrow -\infty

If a<0, as x\rightarrow -\infty, y=\ln(ax+b)\rightarrow \infty and x can increase until ax+b=0, so there is an **asymptote** at x=\dfrac{-b}{a} where

y\rightarrow -\infty

**Step 3: **Mark all of this information on a **graph** and use it to **plot the graph**.

**Modelling Exponential Growth and Decay**

Most real world problems about **exponential functions** involve either something **increasing (growth)** or **decreasing (decay)**. The maths in these problems is not new, but it is **not trivial to interpret what the questions are asking**.

**Example: **A radioactive particle **decays** according to R=Ae^{-0.05t}, where R is radioactivity and t is time. It starts with a radioactivity of 500.

i) Find A

ii) Find the **half-life** of the particle.

**[6 marks]**

i) At t=0, R=500

500=Ae^{-0.05\times0}

500=Ae^{0}

500=A\times1

A=500

ii) The **half-life** is when the radioactivity has **halved**, so at R=\dfrac{500}{2}=250

250=500e^{-0.05t}

\dfrac{250}{500}=e^{-0.05t}

e^{-0.05t}=\dfrac{1}{2}

-0.05t=\ln\left(\dfrac{1}{2}\right)

\begin{aligned}t&=\dfrac{-\ln\left(\dfrac{1}{2}\right)}{0.05}\\[1.2em]&=13.9\end{aligned}

**Example 1: Sketching an Exponential Graph**

**Sketch** y=e^{3x+2}-11, marking any **asymptotes** and **intersections with the axes**.

**[3 marks]**

**Example 2: Sketching a Logarithmic Graph**

**Sketch** y=\ln(6x+5), marking any **asymptotes** and **intersections with the axes**.

**[3 marks]**

## Modelling Exponential Growth and Decay Example Questions

**Question 1: **Plot the graph y=e^{0.3x-0.8}-1, labelling where the graph crosses the axes and any asymptotes.

**[3 marks]**

**Question 2: **Plot the graph y=\ln(22x+35), labelling where the graph crosses the axes and any asymptotes.

**[3 marks]**

**Question 3: **The population of a small city grows according to P=10000e^{0.1(t-2008)} where P is the population and t is the current year.

a) In which year was the city’s population first measured, and what was the population in this year?

b) What is the city’s population in 2021?

c) How many years will it take for the city’s population to reach 100000?

**[7 marks]**

a) From the equation, we can see that the population was first measured in 2008.

The population in 2008 was:

\begin{aligned}P&=10000e^{0.1(2008-2008)}\\[1.2em]&=10000e^{0.1\times0}\\[1.2em]&=10000e^{0}\\[1.2em]&=10000\times1\\[1.2em]&=10000\end{aligned}

b) t=2021

\begin{aligned}P&=10000e^{0.1(2021-2008)}\\[1.2em]&=10000e^{0.1\times13}\\[1.2em]&=10000e^{1.3}\\[1.2em]&=36700\end{aligned}

c) P = 100000

\begin{aligned} 100000 &=10000e^{0.1(t-2008)} \\[1.2em] \dfrac{100000}{10000}&=e^{0.1(t-2008)} \\[1.2em] 10 &=e^{0.1(t-2008)} \\[1.2em] 0.1(t-2008)&=\ln(10) \\[1.2em] t-2008 & =10\ln(10) \\[1.2em] t &=2008+10\ln(10) \\[1.2em] &=2031\end{aligned}

The city’s population will reach 100000 in 2031.

**Question 4: **A colony of bacteria grow very quickly according to y=Ae^{kt} where y is the population and t is time in hours. At time 0 there is only 1 bacterium. At time 10 there are 1000 bacteria.

i) Find A and t.

ii) The container in which the bacteria are in can only support 72000 of them. How long until it is at capacity?

**[5 marks]**

i) At t=0, y=1

Ae^{k\times0}=1

Ae^{0}=1

A\times1=1

A=1

At t=10, y=1000

e^{10k}=1000

10k=\ln(1000)

\begin{aligned}k&=\dfrac{\ln(1000)}{10}\\&=0.691\end{aligned}

ii) y=e^{0.691t}

72000=e^{0.691t}

0.691t=\ln(72000)

\begin{aligned}t&=\dfrac{\ln(72000)}{0.691}\\&=16.2\text{ hours}\end{aligned}

## Modelling Exponential Growth and Decay Worksheet and Example Questions

### Logarithms

A Level## MME Premium Membership

£19.99

/monthLearn an entire GCSE course for maths, English and science on the most comprehensive online learning platform. With revision explainer videos & notes, practice questions, topic tests and full mock exams for each topic on every course, it’s easy to Learn and Revise with the MME Learning Portal.

Sign Up Now