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>Types of Functions>Exponential Functions with Euler's Number

Exponential Functions with Euler's Number

When something increases or decreases by the same percentage in each period, you have exponential (percentage) growth. The exponential growth may be negative, meaning the graph decreases to the right rather than ascending upwards as it otherwise would.

In exponential functions, you can either use an arbitrary number $b$ as the base or $e$ as the base.

Theory

Exponential Functions

Exponential functions can have $e$ as the base or an arbitrary number $b$ as the base. In both cases, $a$ is a constant. They look like this:

f (x) = a \cdot e^{k x} f (x) = a \cdot b^{x}

Note! These functions are reformulations of each other, so they have identical graphs $(b = e^{k})$ . Note that the variable $x$ is now in the exponent! The symbols $a$ , $b$ and $k$ are numbers.

When the value of $a$ in the function is positive, the graph looks like one of the two graphs below.

Rule

The Graph of an Exponential Function

The graphs of an increasing and a decreasing exponential function

$a$ is the $y$ value when $x = 0$ , $b = e^{k}$ is the growth factor.

$0 < b < 1$ blue graph, $b > 1$ red graph.

Note! You are expected to be able to convert from $b = e^{k}$ and from $e^{k} = b$ .

Rule

Conversion from $b = e^{k}$ and from $e^{k} = b$

\begin{array}{l} b^{x} & = {(e^{\ln b})}^{x} = e^{k x}, where k = \ln b \\ e^{k x} & = {(e^{k})}^{x} = b^{x}, where b = e^{k} \end{array}

The following is an overview of how the function behaves for different values of $a$ and $b > 0$ .

$a > 0$ and $b > 1$ :: The graph runs along the $x$ -axis and rises sharply to the right.
$a > 0$ and $0 < b < 1$ :: The graph decreases sharply to the right and flattens along the $x$ -axis.
$a \in ℝ$ and $b = 1$ :: The graph is a horizontal line through $a$ .
$a < 0$ and $b > 1$ :: The graph runs along the $x$ -axis and decreases sharply below the $x$ -axis.
$a < 0$ and $0 < b < 1$ :: The graph rises sharply and flattens along the $x$ -axis.

In general, $b > 1$ gives you a fixed percentage increase, $0 < b < 1$ gives you a fixed percentage reduction, and $b = 1$ gives you no change.

The number $b = e^{k}$ acts as a growth factor. The value of $a$ affects the sign of the functional values.

Example 1

You have the function $f (x) = 3 \cdot 2^{x}$ . This intersects the $y$ -axis at $y = 3$ and grows exponentially. This form of growth is very powerful. References to this type of graph are also often used in everyday speech, when someone says something is experiencing “exponential” growth.

Example of increasing exponential function

Example 2

You have the function $f (x) = 3 \cdot 0 . 5^{x}$ . This intersects the $y$ -axis at $y = 3$ and decreases exponentially. This form of reduction is also very powerful. It’s like the downward version of the previous example.

Example of decreasing exponential function

Example 3

In a particular chemical reaction, the concentration of a substance is given by

f (t) = 2.50 - 2.50 \cdot e^{- 0.012 t},

where $t$ is the time measured in seconds, and $f (t)$ is measured in mmol/L.
You are given the following tasks:

1.
What is the concentration after 15 seconds? How long does it take before the concentration is $2.00 mmol/L$ ?
2.
Draw the graph of $f$ . What value will the concentration approach if the reaction lasts for a very long time?
3.
What is the reaction rate when the concentration is $2.00 mmol/L$ ?

1.

To find the concentration after 15 seconds, you have to insert

t = 15

f (t)

. You then get

\begin{array}{l} f (15) & = 2.50 - 2.50 \cdot e^{- 0.012 \cdot 15} \\ \approx 0.412 \end{array}

\begin{array}{l} f (15) = 2.50 - 2.50 \cdot e^{- 0.012 \cdot 15} \approx 0.412 \end{array}

The concentration after 15 seconds is

0.412

mmol/L.

Now you need to find out how long it takes before the concentration is $2.00$ mmol/L. Here you have to solve the equation $f (t) = 2.00$ . You set it up and get

2.50 - 2.50 \cdot e^{- 0.012 t} = 2.00

You subtract $2.50$ on both sides, and get

- 2.50 \cdot e^{- 0.012 t} = - 0.50

Then you divide by $- 2.50$ , and the equation becomes

e^{- 0.012 t} = 0.2

You now have an equation in the form $e^{a} = b$ . You apply $\ln$ on both sides, and you get

- 0.012 t = \ln 0.2 = - 1.61

Finally, you divide by $- 0.012$ and find the solution

t = \frac{- 1.61}{- 0.012} \approx 134

This means that it takes 134 seconds before the concentration is $2.00$ mmol/L

2.

First you draw the graph of

f

. It looks like this:

Example of decreasing exponential with Euler’s number

To find out which value the concentration is approaching if the reaction lasts for a very long time, you can either look at the graph and see that it goes towards $2.50$ , or insert a properly large $t$ and get $f (100 000) \approx 2.50$ . This means that if the reaction takes a very long time, the concentration approaches $2.50$ mmol/L Another way you can say this is to say that $y = 2.5$ is the horizontal asymptote.

3.