Modelling Exponential Growth and Decay

A LevelAQAEdexcelOCR

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.

A LevelAQAEdexcelOCR

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 LevelAQAEdexcelOCR

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

Modelling Exponential Growth and Decay 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

Modelling Exponential Growth and Decay Worksheet and Example Questions

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