Exponential Growth

We will examine situations where quantities are increasing exponentially. This situation is known as $\textit{exponential growth modelling}$, and frequently occurs in real life around us.

The population of species, people, bacteria and investment usually $\textit{growth}$ in an exponential way.

Growth is exponential when the quantity present is multiplied by a constant for each unit time interval. this constant is called the $\textit{growth}$ or $\textit{compounding factor}$ is greater than one.

Considering a situation where the changes to a certain population of bacteria over time is a good example of $\textit{exponential growth}$.

It can be seen that the population is increasing but also that the growth rate is increasing; that is, the graph is getting steeper.

Consider a population of $10$ bacteria which, under favourable conditions, increase by $20\%$ each day.
To increase a quantity by $20\%$, it is known to multiply it by $1.2$.

If $B_n$ is the population of bacteria after $n$ days, then:

\( \begin{align} \displaystyle
B_0 &= 10 &\textit{the original population} \\
B_1 &= B_0 \times 1.2 = 10 \times 1.2 \\
B_2 &= B_1 \times 1.2 = 10 \times 1.2 \times 1.2 = 10 \times 1.2^2 \\
B_3 &= B_2 \times 1.2 = 10 \times 1.2^2 \times 1.2 = 10 \times 1.2^3 \\
\end{align} \)
and so on.

The population is $1.2$ times every day, so the $\textit{growth}$ or $\textit{compounding factor}$ is $2$.
The pattern above shows that $B_n = 10 \times 1.2^n$.

In general, the exponential growth function has an equation of the form: $y=Aa^{kx}$

  • $A$, $a$ and $k$ are constants.
  • $a>1$ and $k>0$ are the growth or compounding factor.
  • $A$ is the initial value of $y$ (when $x=0$)

Example 1

The weight $B_n$ of bacteria in a colony $n$ hours after establishment is given by $B_n = 100 \times 5^{0.2n}$ grams.

(a)   Find the initial weight.

\( \begin{align} \displaystyle
B_0 &= 100 \times 5^{0.2 \times 0} \\
&= 100 \times 5^0 \\
&= 100 \times 1 \\
&= 100
\end{align} \)

(b)   Find the weight after $5$ hours.

\( \begin{align} \displaystyle
B_5 &= 100 \times 5^{0.2 \times 5} \\
&= 100 \times 5^1 \\
&= 100 \times 5 \\
&= 500
\end{align} \)

(c)   Find the percentage increase from $n=10$ to $n=20$.

\( \begin{align} \displaystyle
\dfrac{B_{20}-B_{10}}{B_{10}} \times 100\% &= \dfrac{100 \times 5^{0.2 \times 20}-100 \times 5^{0.2 \times 10}}{100 \times 5^{0.2 \times 10}} \times 100 \% \\
&= \dfrac{5^{0.2 \times 20}-5^{0.2 \times 10}}{5^{0.2 \times 10}} \times 100 \% \\
&= \dfrac{5^4-5^2}{5^2} \times 100 \% \\
&= \dfrac{5^2(5^2-1)}{5^2} \times 100 \% \\
&= (5^2-1) \times 100 \% \\
&= 2400 \%
\end{align} \)

Example 2

The speed $V_t$ is given by $V_t = 8 \times 2^{0.5t}$, where $t$ is the temperature in $^{\circ}C$. Find the temperature when the speed is $24$, correcting to three significant figures.

\( \begin{align} \displaystyle
8 \times 2^{0.5t} &= 24 \\
2^{0.5t} &= 3 \\
0.5t &= \log_2{3} \\
t &= \dfrac{1}{0.5} \log_2{3} \\
&= 3.1699 \cdots \\
&= 3.17 ^{\circ}C
\end{align} \)

 

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