# 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    ## 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 Level    ## 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 Level   ## 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 Level   A Level   ## Example 1: Sketching an Exponential Graph

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

[3 marks] A Level   ## Example 2: Sketching a Logarithmic Graph

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

[3 marks] A Level   ## 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}

A Level

A Level

A Level

## 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 ### 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 ### 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