# Modelling Exponential Growth and Decay

A LevelAQAEdexcelOCR

## 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

@mmerevise

## 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

Gold Standard Education

Gold Standard Education

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 $2031$.

Gold Standard Education

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}

Gold Standard Education

A Level

A Level

## Modelling Exponential Growth and Decay Worksheet and Example Questions

### Logarithms

A Level
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