Use of exponential functions
Exponential functions are used to describe exponential growth and decay processes.
Exponential growth and decay have the following function equation:
$t...$ time
$a ...$ initial level
$b ...$ growth factor
$N(t) ...$ value in dependence on $t$
Exponential growth and decay
Hint
 If the growth factor $b$ is greater than 1, it is an exponential growth (increase).
 If the growth factor $b$ is less than 1 and greater than 0, it is an exponential decay (decrease).
EXAMPLE
One person takes 6mg of a drug. It is broken down by $\frac13$ each day. How much mg is still in the body after 7 days?

Determine values from the task
Every day $\frac13$ is broken down, which means $\frac23$ is left. The decrease factor is $b=\frac23$
The initial level $a=6$
The 7th day is of interest. Therefore, the time is $t=7$

Insert values in the formula and calculate
$N(\color{purple}{t})=\color{red}{a}\cdot \color{green}{b}^\color{purple}{t}$
$N(\color{purple}{7})=\color{red}{6}\cdot (\color{green}{\frac23})^\color{purple}{7}\approx0.35$
On the 7th day there is still about 0.35 mg of the drug in the body.
Percentage growth
If there is an increase or decrease with a constant percentage growth factor, it is percentage growth. This is also an exponential growth process and can therefore be described with an exponential function.
When increasing with a percentage growth rate $p$ % growth factor $(1+\frac{p}{100})$ applies and when decreasing with a percentage decrease rate $p$ % the decrease factor $(1\frac{p}{100})$ applies.
Substituting the growth or decrease factor into the above formula for $b$, one obtains with a percentage increase the equation
and at percentage decrease the equation
EXAMPLE
3m² of the water surface of a lake are covered with a type of algae, which grows annually by 50%. Calculate the covered area for the next 5 years.

Determine values from the task
The percentage growth rate is $p=50$
The initial level $a=3$
For the time $t$ , the years 1 to 5 must be used.

Insert values in the formula
This is a percentage increase as there is a percentage growth factor and the area is growing. First, we put the constant values $p$ and $a$ into the percentage increase formula.
$N(\color{purple}{t})=\color{red}{a}\cdot (1+\frac{\color{green}{p}}{100})^\color{purple}{t}$
$N(\color{purple}{t})=\color{red}{3}\cdot (1+\frac{\color{green}{50}}{100})^\color{purple}{t}=\color{red}{3}\cdot1.5^\color{purple}{t}$Now use the values 1 to 5 for $t$ and calculate.
$N(\color{purple}{1})=\color{red}{3}\cdot1.5^\color{purple}{1}=4.5$
$N(\color{purple}{2})=\color{red}{3}\cdot1.5^\color{purple}{2}=6.75$
$N(\color{purple}{3})=\color{red}{3}\cdot1.5^\color{purple}{3}\approx10.13$
$N(\color{purple}{4})=\color{red}{3}\cdot1.5^\color{purple}{4}\approx15.19$
$N(\color{purple}{5})=\color{red}{3}\cdot1.5^\color{purple}{5}\approx22.78$