Modelling Exponential Growth and Decay

Modelling Exponential Growth and Decay

A LevelAQAEdexcelOCREdexcel 2022OCR 2022

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.

A Level AQA Edexcel OCR

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.

A LevelAQAEdexcelOCR

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.

A LevelAQAEdexcelOCR

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}

A LevelAQAEdexcelOCR
A Level AQA Edexcel OCR

Example 1: Sketching an Exponential Graph

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

[3 marks]

A LevelAQAEdexcelOCR

Example 2: Sketching a Logarithmic Graph

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

[3 marks]

A LevelAQAEdexcelOCR

Example Questions

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.

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}

Additional Resources

MME

Exam Tips Cheat Sheet

A Level
MME

Formula Booklet

A Level

Worksheet and Example Questions

You May Also Like...

A Level Maths Revision Cards

The best A level maths revision cards for AQA, Edexcel, OCR, MEI and WJEC. Maths Made Easy is here to help you prepare effectively for your A Level maths exams.

£14.99
View Product

A Level Maths – Cards & Paper Bundle

A level maths revision cards and exam papers for Edexcel. Includes 2022 predicted papers based on the advance information released in February 2022! MME is here to help you study from home with our revision cards and practise papers.

From: £22.99
View Product

Transition Maths Cards

The transition maths cards are a perfect way to cover the higher level topics from GCSE whilst being introduced to new A level maths topics to help you prepare for year 12. Your ideal guide to getting started with A level maths!

£8.99
View Product