# Formulas, graphs & relations » Exponential relation

## Contents

1. General2. Formula

3. Table

4. Graph

5. Making formulas

## 1. General

An exponential relation is a relation in which one of the variables is an exponent. You use these relations when working for example with growth factors or interest after interest.

## 2. Formula

As said one of the variables needs to be an exponent.

The formulas are always of the type: `h` = `b` · `g ^{t}`

Examples are

`y`= 50 · 0.73

^{x}and

`h`= 12 · 1.03

^{t}.

## 3. Table

Below you will see a table that corresponds to a exponential relation.

t |
0 | 1 | 2 | 3 | 4 | 5 | ||||||

h |
600 | 630 | 661.5 | 694.575 | 729.303... | 765.769... | ||||||

×1.05 |
×1.05 |
×1.05 |
×1.05 |
×1.05 |
||||||||

When there is a constant increase in the top row of the table, you have to be able to multiply with the same number in the bottom row. When all the factors at the arrows are slightly different but rounded off always the same number, you may also speak of an exponential relation. The deviation is then because the numbers in the table are rounded numbers.

## 4. Graph

The graph of an exponential relation is:

- an increasingly steeper rising line (growth factor greater than 1).

- an increasingly slower falling falling line (growth factor between 0 and 1).

## 5. Making formulas

The formula is always of the type `h` = `b` · `g ^{t}`

Where

`b`is the initial amount (

`t`= 0) and

`g`is the growth factor.

See the table underneath for a couple of examples of growth factors.

In- or decrease in % | growth factor |

increase 5% | 1.05 |

increase 8.5% | 1.085 |

decrease 12% | 0.88 |

decrease 3.5% | 0.965 |

*Example 1*

In a savings account is 1500 euros and every year 3.5% interest is received.

The formula you can calculate the amount `a` on the savings account after `t` years is:

`a` = 1500 · 1.035^{t}

*Example 2*

A tree is 8 m high. Every year the tree grows with 6%.

The formula with which you can calculate the height `h` of the tree after `t` years is:

`h` = 8 · 1.06^{t}

*Example 3*

A car has a new price of 35 000 euros. Every year the car loses 10% of its worth.

The formula with which you can calculate the value `v` of the car after `t` years is:

`v` = 35 000 · 0.9^{t}

#### Example 4: Making formula for a graph.

Let's take the graph above. We can clearly read off points (0, 3) and (7, 8).

When we fill in 3 as the initial amount we will get `y` = 3 · `g ^{x}`

Now we will fill in the other point (

`y`= 7 and

`x`= 8 ).

We will get an equation which we can solve.

^{ }8 | = 3 · g^{7} |

g^{7} | = 8 : 3^{ } |

^{ }g | = = (8 : 3)^{(1 : 7)} = 1.15 |

The formula will therefore be `y` = 3 · 1.15^{x}