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 a period of time is a good example of $\textit{exponential growth}$.

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

Consider a population of $10$ bacteria which under favourable conditions it increasingly by $20\%$ each day.

To increase a quantity by $20\%$, it is know that 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 $2$. From the pattern above, we see 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.

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

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