Numbers: Numbers and number relationships

# Unit 2: Introduction to exponents

Natashia Bearam-Edmunds

### Unit outcomes

By the end of this unit you will be able to:

• Use exponential notation.
• Understand the basic exponential identities $\scriptsize ({{a}^{n}}=a \times a \times a \times a \mbox{ (n times) } ;{{a}^{(-n)}}=\displaystyle \frac{1}{{{a}^{n}}};\displaystyle \frac{1}{{{a}^{-n}}}={{a}^{n}};{{a}^{0}}=1)$.
• Multiply exponents with the same base $\scriptsize ({{a}^{m}}\times {{a}^{n}}={{a}^{(m+n)}})$.
• Divide exponents with the same base $\scriptsize (\displaystyle \frac{{{a}^{m}}}{{{a}^{n}}}={{a}^{(m-n)}})$.

## What you should know

Before you start this unit, make sure you can:

Describe the real number system and explain the difference between natural numbers, whole numbers, integers, rational numbers and irrational numbers. Go back to unit 1 in this subject outcome if you need to revise this.

## Introduction

We are living in exponential technological times. There are over $\scriptsize 100$ billion searches on Google every month, a massive increase from 2006 when that number was $\scriptsize 2.7$ billion. In 1984 the number of connected internet devices was $\scriptsize 1~000$, in 1992 it was $\scriptsize 1~000~000$, in 2008 it was $\scriptsize 1~000~000~000$, around $\scriptsize 17$ billion in 2016 and just over $\scriptsize 26$ billion by the end of 2019.

It is estimated that by the end of 2025 there will be approximately $\scriptsize 76$ billion internet-connected devices worldwide. This form of rapid growth can best be described using exponents. Exponential increases start off slowly but then sharply increase to a tremendous explosion in size.

Are new internet devices increasing at the same rate each year? Or is the rate of new devices also growing exponentially? Think about this.

Can you think of another current situation where we are seeing exponential growth?

Exponents are a powerful way to describe rapid increases or decreases in growth – growth where the rate of change itself is increasing or decreasing. Exponential notation is useful to describe very large and very small numbers.

## Exponential growth

Exponents are useful for describing growth in many different real-life situations. But what exactly are exponents? The best way to start learning what exponents are is to see them in action. In activity 2.1 we will do just that.

### Activity 2.1: Investigate exponential growth using a chessboard and grains of rice

Time required: 5 minutes

What you need:

• a chessboard (if you do not have a chessboard you can make your own $\scriptsize 8\times 8$ grid)
• grains of rice

What to do:

Use grains of rice and a chessboard to discover exponential growth.

You will place grains of rice on the board using patterns that follow either repeated multiplication or repeated addition.

Place one grain of rice on the first square, two grains on the second square, four grains on the third square, eight grains on the fourth square and continue adding rice to the board in this doubling (repeated multiplication by two) pattern.

On the other half of the board, place two grains of rice on the first square, four grains on the second square, six grains on the third square, eight grains on the fourth square and continue adding two more grains of rice to the board in this repeated addition of two pattern.

• How many grains of rice do you place on the sixth square when following repeated multiplication?
• How many grains of rice do you place on the twentieth square when following repeated addition?
• Which method (repeated multiplication or repeated addition) will reach at least $\scriptsize 100$ grains of rice more quickly? Why?
• Which method shows a faster rate of increase?

What did you find?

• Did the doubling grow more quickly than expected?
• Did you see that repeated multiplication results in rapid increases in growth?
• Would you be able to place the correct amount of rice on every square under repeated multiplication and repeated addition?
• If you were able to continue the pattern of doubling all the way up to the last square on the board and add the numbers together there would be over $\scriptsize 18$ quadrillion grains of rice on the board, which is around $\scriptsize 1~000$ times more than the current annual global production of rice. The grains of rice would weigh over $\scriptsize 460$ billion tons, which would be a heap of rice larger than Mount Everest!

### Take note!

The video called “Chess and exponents illustrates” the concept of exponential growth, which you have just investigated in the activity.

Chess and exponents (Duration: 2.01)

We often shorten the names of objects or people as it’s quicker and easier to remember and say shorter names, for example:

• CD instead of Compact Disc
• USB instead of Universal Serial Bus.

In Maths it is no different. We find quicker, easier ways to carry out operations by shortening their processes.

Multiplication is a short way to write repeated addition.

Would you rather write $\scriptsize 2+2+2+2+2$ or $\scriptsize 5\times 2$?

How about $\scriptsize a+a+a+a+a$ or $\scriptsize 5\times a$?

Which expressions are quicker to write and easier to work with?

Hopefully you agree that multiplication is a much quicker and easier way to express repeated addition.

Just as multiplication is a short way to write repeated addition, similarly, exponents are a short way to write repeated multiplication.

Look at the following expressions. Which do you think would be easier to write and work with?

$\scriptsize 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2$ or $\scriptsize {{2}^{10}}$?

Certainly $\scriptsize {{2}^{10}}$ is quicker to write and leaves less room for error.

But how did we get from $\scriptsize 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2$ to $\scriptsize 2^{10}$?

Count the number of times that $\scriptsize 2$ has been multiplied by itself. What do you get?

You would have counted that $\scriptsize 2$ is multiplied $\scriptsize 10$ times. It is no coincidence that the number ‘floating above’ the $\scriptsize 2$, which is called the exponent, is also $\scriptsize 10$. Let’s go over each part of $\scriptsize {{2}^{10}}$.

### Take note!

$\scriptsize 2^{10}$ is called exponential form or a power. The base is the number that is repeatedly multiplied. The exponent, which is sometimes also referred to as the ‘index’ or ‘power’, tells us how many times the base is multiplied by itself.

We read $\scriptsize 2^{10}$ as $\scriptsize 2$ to the power of $\scriptsize 10$. This tells us that the base $\scriptsize 2$ has been multiplied by itself $\scriptsize 10$ times.

We can generalise this as:

$\scriptsize {{a}^{n}}=a\times a\times a...\times a\text{ (n times) (}a\in \mathbb{R}\text{ }a\ne 0,n\in \mathbb{N}\text{)}$

To make sure you understand, work through the next two examples.

### Example 2.1

Write $\scriptsize 3\times 3\times 3\times 3\times 3$ in exponential form

Solution

Do you see that the base of $\scriptsize 3$ is multiplied $\scriptsize 5$ times?

Since this is repeated multiplication, writing the expression using exponential form will be a shorter way to rewrite it.

$\scriptsize 3\times 3\times 3\times 3\times 3={{3}^{5}}$

### Example 2.2

Write $\scriptsize {{4}^{3}}$ in expanded form

Solution

Here you have to write the expression without exponents. In other words, you need to show the repeated multiplication (expanded form) of the expression.

The base of $\scriptsize {{4}^{3}}$ is $\scriptsize 4$ and the exponent is $\scriptsize 3$ so $\scriptsize 4$ has been multiplied by itself $\scriptsize 3$ times:

$\scriptsize {{4}^{3}}=4\times 4\times 4$

Now try this exercise on your own.

### Exercise 2.1

1. Rewrite the following expressions in exponential form:
1. $\scriptsize 4\times 4\times 4\times 4\times 4$
2. $\scriptsize 2\times 2\times 2\times 3\times 3$
3. $\scriptsize 2+2+3+3+3$
2. Write the following in expanded form:
1. $\scriptsize {{3}^{4}}$
2. $\scriptsize {{3}^{2}}\times {{2}^{3}}$
3. $\scriptsize {{x}^{3}}{{y}^{2}}$
4. $\scriptsize {{(-2)}^{4}}$
3. Fill in $\scriptsize \gt$, $\scriptsize \lt$ or $\scriptsize =$ to make the statements true:
1. $\scriptsize {{3}^{2}}\_\_\_2\times 2\times 2\times 2$
2. $\scriptsize {{3}^{2}}.2\_\_\_{{2}^{2}}.3$
3. $\scriptsize \displaystyle \frac{1}{2}\times \displaystyle \frac{1}{2}\times \displaystyle \frac{1}{2}\_\_\_{{\left( \displaystyle \frac{1}{2} \right)}^{3}}$
4. Write the following numbers in exponential form:
1. $\scriptsize 25$
2. $\scriptsize 729$
3. $\scriptsize 48$

The full solutions are at the end of the unit.

## Multiplying powers with the same base

Every number or variable (letter) that is not already in exponential form can be written with a power of $\scriptsize 1$.

$\scriptsize 3$ can be written as $\scriptsize {{3}^{1}}$

$\scriptsize x$ can be written as $\scriptsize {{x}^{1}}$

$\scriptsize 2xy$ can be written as $\scriptsize {{2}^{1}}{{x}^{1}}{{y}^{1}}$

You can think of the invisible exponent of $\scriptsize 1$ as a ‘ghost’ floating above the numbers and variables (figure 2).

The exponent of $\scriptsize 1$ is always present, even when it is not written. We will use this fact to show that $\scriptsize 3\times 3\times 3\times 3={{3}^{4}}$.

We know that each $\scriptsize 3$ can be written as $\scriptsize {{3}^{1}}$ so $\scriptsize 3\times 3\times 3\times 3={{3}^{1}}\times {{3}^{1}}\times {{3}^{1}}\times {{3}^{1}}$. You have already seen that $\scriptsize 3\times 3\times 3\times 3$ means $\scriptsize {{3}^{4}}$. So can you see what we have actually done to the exponents?

$\scriptsize {{3}^{1}}\times {{3}^{1}}\times {{3}^{1}}\times {{3}^{1}}={{3}^{4}}={{3}^{1+1+1+1}}$

We have just added the exponents together.

This is a very useful rule especially when we are working with a number of different expressions with the same bases being multiplied. We can apply this rule as long as the bases of the exponents are the same.

To simplify $\scriptsize {{x}^{2}}\times {{x}^{4}}$ we could expand and rewrite as $\scriptsize {{x}^{1}}\cdot {{x}^{1}}\times {{x}^{1}}\cdot {{x}^{1}}\cdot {{x}^{1}}\cdot {{x}^{1}}$ and then add up the exponents to get $\scriptsize {{x}^{6}}$. But a much quicker way to simplify $\scriptsize {{x}^{2}}\times {{x}^{4}}$ is to keep the base $\scriptsize x$ and add the exponent $\scriptsize {{x}^{2}}\times {{x}^{4}}={{x}^{2+4}}={{x}^{6}}$.

Now, try to simplify $\scriptsize {{x}^{20}}\times {{x}^{12}}$.

If you decided to first expand $\scriptsize {{x}^{20}}\times {{x}^{12}}$ it would take a long time to write out all those bases and it could lead to many errors. So, using a rule would not only make things easier but it would save time too. I’m sure you will agree that the quickest way to simplify $\scriptsize {{x}^{20}}\times {{x}^{12}}$ is to keep the base $\scriptsize x$ and add the exponents $\scriptsize {{x}^{20}}\times {{x}^{12}}={{x}^{20+12}}={{x}^{32}}$.

### Example 2.3

Write $\scriptsize {{3}^{2}}\times {{3}^{5}}$ with a single base of $\scriptsize 3$.

Solution

To write $\scriptsize {{3}^{2}}\times {{3}^{5}}$ with a single base of $\scriptsize 3$ you could first write each base of $\scriptsize 3$ in expanded form and then count the total number of threes being multiplied together.

$\scriptsize {{3}^{2}}\times {{3}^{5}}=3\times 3\times 3\times 3\times 3\times 3\times 3={{3}^{7}}$

But, a quicker way to simplify is to keep a single base of $\scriptsize 3$ and add the exponents to get $\scriptsize {{3}^{2}}\times {{3}^{5}}={{3}^{2+5}}={{3}^{7}}$.

### Take note!

What happens when the bases are different? Say, for example, we had $\scriptsize {{3}^{2}}\times {{2}^{3}}$. Can we add the exponents together now? If you answered yes, then which base would get to keep the exponents?

### Example 2.4

In $\scriptsize {{3}^{2}}\times {{2}^{3}}$ we are working with different bases of $\scriptsize 3$ and $\scriptsize 2$, so adding the powers together will not work.

Solution

$\scriptsize {{3}^{2}}\times {{2}^{3}}=9\times 8=72$

So, whatever rule you use it must get you back to this answer.

There are three different ways you may have tried to simplify the expression.

1. Let’s say you kept the base of $\scriptsize 3$ and added the exponents to get $\scriptsize {{3}^{2+3}}\times 2$ what will the answer be? $\scriptsize {{3}^{2+3}}\times 2={{3}^{5}}\times 2=486$. This is not equal to $\scriptsize 72$ so it is not correct to simplify $\scriptsize {{3}^{2}}\times {{2}^{3}}$ by keeping the base $\scriptsize 3$.
2. You may have decided to keep the base of $\scriptsize 2$ and add exponents to get $\scriptsize 3\times {{2}^{5}}$. The answer to $\scriptsize 3\times {{2}^{5}}$ is $\scriptsize 96$. Once again, this is not equal to $\scriptsize 72$ so it is not correct to simplify $\scriptsize {{3}^{2}}\times {{2}^{3}}$ by adding the exponents to base $\scriptsize 2$.
3. Lastly, you may have done something like this: $\scriptsize {{\left( 3\times 2 \right)}^{2+3}}={{\left( 3\times 2 \right)}^{5}}$. If you calculate the answer to $\scriptsize {{\left( 3\times 2 \right)}^{5}}$ you will get $\scriptsize 7~776$. This is very far from $\scriptsize 72$! So it is not correct to simplify $\scriptsize {{3}^{2}}\times {{2}^{3}}$ by multiplying the bases together and adding the powers.

In fact, $\scriptsize {{3}^{2}}\times {{2}^{3}}$ cannot be simplified any further using any exponent rules BUT you can arrive at a rule from this example.

Powers with different bases cannot be combined using any exponent rules.

This leads us to some of the other common mistakes that must be avoided when working with exponents with the same base.

### Take note!

$\scriptsize {{2}^{2}}\times {{2}^{5}}\ne {{2}^{10}}$

$\scriptsize {{3}^{3}}\times {{3}^{4}}\ne {{9}^{7}}$

$\scriptsize {{2}^{2}}+{{2}^{3}}\ne {{2}^{5}}$

### Exercise 2.2

1. Simplify the following where possible:
1. $\scriptsize {{3}^{2}}\times {{3}^{4}}$
2. $\scriptsize {{2}^{3}}\times {{5}^{3}}$
3. $\scriptsize {{x}^{2}}{{y}^{3}}\times 2{{x}^{3}}y$
4. $\scriptsize {{a}^{2}}\times {{b}^{3}}\times {{y}^{2}}$
5. $\scriptsize \left( -{{x}^{2}}{{y}^{4}} \right)\left( -3y{{x}^{3}} \right)$
6. $\scriptsize \left( -2{{a}^{3}}{{b}^{2}} \right)\left( -bxa \right)\left( {{a}^{2}}{{x}^{3}}b \right)$
7. $\scriptsize {{2}^{2x}}\times {{2}^{3x}}$
8. $\scriptsize {{3}^{3a}}\times {{3}^{2a+b}}$

The full solutions are at the end of the unit.

## Dividing powers with the same base

By now, you know that division is the reverse operation of multiplication. So, if we add exponents when multiplying with the same bases, what do you think we do to the exponents when we divide the same bases?

Simplify $\scriptsize \displaystyle \frac{{{a}^{5}}}{{{a}^{2}}}$.

We can rewrite the numerator as $\scriptsize {{a}^{5}}=a\times a\times a\times a\times a$.

We can rewrite the denominator as $\scriptsize {{a}^{2}}=a\times a$.

Then $\scriptsize \displaystyle \frac{{{a}^{5}}}{{{a}^{2}}}=\displaystyle \frac{a\times a\times a\times a\times a}{a\times a}$.

Using simple division and cancelling out we get $\scriptsize \displaystyle \frac{{{a}^{5}}}{{{a}^{2}}}=\displaystyle \frac{\times \times a\times a\times a}{\times }=\displaystyle \frac{a\times a\times a}{1}$.

By multiplying powers with the same base we get $\scriptsize \displaystyle \frac{{{a}^{5}}}{{{a}^{2}}}={{a}^{1}}\times {{a}^{1}}\times {{a}^{1}}={{a}^{3}}$.

Without expanding the powers, is there a quick way to get from $\scriptsize \displaystyle \frac{{{a}^{5}}}{{{a}^{2}}}$ to $\scriptsize {{a}^{3}}$?

Hopefully you can see that the quick way to get from $\scriptsize \displaystyle \frac{{{a}^{5}}}{{{a}^{2}}}$ to $\scriptsize {{a}^{3}}$ is to subtract the exponents with the same base $\scriptsize \displaystyle \frac{{{a}^{5}}}{{{a}^{2}}}={{a}^{5-2}}={{a}^{3}}$. We keep the base ‘$\scriptsize a$’ and subtract the exponent $\scriptsize 2$ from $\scriptsize 5$.

### Take note!

When we divide powers with the same base, we keep the base and subtract the exponents. This can be generalised as:

$\scriptsize \displaystyle \frac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\text{ }$

### Example 2.5

Simplify: $\scriptsize \displaystyle \frac{{{2}^{6}}}{{{2}^{4}}}$ and $\scriptsize \displaystyle \frac{2{{x}^{3}}{{y}^{4}}}{{{y}^{2}}}$

Solution

Let’s start with $\scriptsize \displaystyle \frac{{{2}^{6}}}{{{2}^{4}}}$. The base is the same in the numerator and denominator therefore we can apply the rule for dividing exponents with the same base. We will keep the base of $\scriptsize 2$ in the numerator and subtract the powers:

$\scriptsize \displaystyle \frac{{{2}^{6}}}{{{2}^{4}}}={{2}^{6-2}}={{2}^{4}}=16$.

NOTE: If the final answer is easy to work out without a calculator, then write it out without exponents.

With the expression $\scriptsize \displaystyle \frac{2{{x}^{3}}{{y}^{4}}}{{{y}^{2}}}$, we see more than one base. You must remember the same rule that we arrived at when multiplying powers; you can only apply the rules for exponents to powers with the same base.

We see that there is an exponent with base $\scriptsize y$ in both the numerator and denominator so those are the powers we can simplify using the rule.

$\scriptsize \displaystyle \frac{2{{x}^{3}}{{y}^{4}}}{{{y}^{2}}}=2{{x}^{3}}{{y}^{4-2}}$

Keep the $\scriptsize 2$ and $\scriptsize {{x}^{3}}$ as is without changing their exponents.

$\scriptsize \displaystyle \frac{2{{x}^{3}}{{y}^{4}}}{{{y}^{2}}}=2{{x}^{3}}{{y}^{4-2}}=2{{x}^{3}}{{y}^{2}}$

### Exercise 2.3

1. Simplify the following where possible:
1. $\scriptsize \displaystyle \frac{6{{a}^{3}}b}{a}$
2. $\scriptsize \displaystyle \frac{{{3}^{2}}}{{{2}^{4}}}$
3. $\scriptsize \displaystyle \frac{x{{y}^{2}}\times 2{{x}^{2}}y}{3xy}$

The full solutions are at the end of the unit.

There is an important identity that we use often in exponents. The identity states that any base raised to the power of zero is equal to one or $\scriptsize {{a}^{0}}=1$. We can easily arrive at this identity using the rule for dividing exponents with the same base, as shown in the example below.

### Example 2.6

Use the rule for dividing exponents with the same base to prove that $\scriptsize {{a}^{0}}=1$.

Solution

We know that any number or variable, other than $\scriptsize 0$, divided by itself is equal to $\scriptsize 1$.

$\scriptsize \displaystyle \frac{2}{2}=1\text{ ; }\displaystyle \frac{3}{3}=1\text{ };\text{ }\displaystyle \frac{a}{a}=1$

Using the exponent law for dividing:

\scriptsize \begin{align} \displaystyle \frac{2}{2}& ={{2}^{1-1}}={{2}^{0}}=1\text{ } \\ \displaystyle \frac{3}{3}& ={{3}^{1-1}}={{3}^{0}}=1\text{ } \\ \displaystyle \frac{a}{a}& ={{a}^{1-1}}={{a}^{0}}=1 \\ \end{align}

$\scriptsize \displaystyle \frac{2}{2}={{2}^{1-1}}={{2}^{0}}=1\text{ }$

$\scriptsize \displaystyle \frac{3}{3}={{3}^{1-1}}={{3}^{0}}=1\text{ }$

$\scriptsize \displaystyle \frac{a}{a}={{a}^{1-1}}={{a}^{0}}=1$

So we can see that an exponent of $\scriptsize 0$ means that the base has been divided by itself and will be equal to $\scriptsize 1$.

Therefore, any base raised to the power of zero is equal to one or $\scriptsize {{a}^{0}}=1$.

So far, we have worked with powers where the exponent in the numerator is greater than the exponent in the denominator but what happens in a case like this $\scriptsize \displaystyle \frac{{{3}^{3}}}{{{3}^{5}}}$?

### Example 2.7

Simplify: $\scriptsize \displaystyle \frac{{{3}^{3}}}{{{3}^{5}}}$

Solution

We see the bases are the same and that the exponent in the denominator is greater than the exponent in the numerator: $\scriptsize 5 \gt 3$.

Using basic expansion, we see that $\scriptsize \displaystyle \frac{{{3}^{3}}}{{{3}^{5}}}=\displaystyle \frac{3\times 3\times 3}{3\times 3\times 3\times 3\times 3}=\displaystyle \frac{1}{{{3}^{2}}}$

Using the law for dividing exponents, keep the base of $\scriptsize 3$ in the numerator and subtract the powers.

$\scriptsize \displaystyle \frac{{{3}^{3}}}{{{3}^{5}}}={{3}^{3-5}}$

Can you see what happens to the sign of the exponent now?

$\scriptsize {{3}^{3-5}}={{3}^{-2}}$

You have already seen by expansion that $\scriptsize \displaystyle \frac{{{3}^{3}}}{{{3}^{5}}}=\displaystyle \frac{1}{{{3}^{2}}}$ but we now also know that $\scriptsize {{3}^{3-5}}={{3}^{-2}}$. So we can conclude that $\scriptsize {{3}^{-2}}=\displaystyle \frac{1}{{{3}^{2}}}$.

To write $\scriptsize {{3}^{-2}}$ with a positive exponent you simply find the positive reciprocal. Remember that a reciprocal is a fraction where the numerator and denominator switch places.

So $\scriptsize {{3}^{-2}}$ can be written as $\scriptsize \displaystyle \frac{{{3}^{-2}}}{1}$ and its positive reciprocal will be $\scriptsize \displaystyle \frac{1}{{{3}^{+2}}}$. You will notice that the sign of the exponent has changed when its position changed.

$\scriptsize {{3}^{-2}}=\displaystyle \frac{1}{{{3}^{2}}}=\displaystyle \frac{1}{9}$.

This example leads us to an important concept – negative exponents. Negative exponents produce fractions. You will use the following identity when dealing with negative exponents.

Similarly:

The table below will help you understand how to change from negative to positive exponents.

 Negative exponent Rewrite as positive exponent Answer $\scriptsize {{3}^{-2}}$ $\scriptsize {{3}^{-2}}=\displaystyle \frac{1}{{{3}^{+2}}}$ $\scriptsize {{3}^{-2}}=\displaystyle \frac{1}{{{3}^{+2}}}=\displaystyle \frac{1}{9}$ $\scriptsize {{(2x)}^{-2}}$ $\scriptsize {{(2x)}^{-2}}=\displaystyle \frac{1}{{{(2x)}^{2}}}$ $\scriptsize {{(2x)}^{-2}}=\displaystyle \frac{1}{{{(2x)}^{2}}}=\displaystyle \frac{1}{4{{x}^{2}}}$ $\scriptsize \displaystyle \frac{1}{{{2}^{-3}}}$ $\scriptsize \displaystyle \frac{1}{{{2}^{-3}}}={{2}^{+3}}$ $\scriptsize \displaystyle \frac{1}{{{2}^{-3}}}={{2}^{+3}}=8$ $\scriptsize \displaystyle \frac{2}{{{x}^{-3}}}$ $\scriptsize \displaystyle \frac{2}{{{x}^{-3}}}=2{{x}^{+3}}$ $\scriptsize \begin{array}{l}\displaystyle \frac{2}{{{{x}^{{-3}}}}}&=\displaystyle \frac{2}{1}\times \displaystyle \frac{1}{{{{x}^{{-3}}}}}\\&=2\times {{x}^{{+3}}}\\&=2{{x}^{3}}\end{array}$

We can also use patterns to visually understand what negative exponents represent. Have a look at figure 3 and the explanation in the ‘Did you know?’ box.

### Did you know?

If we start at $\scriptsize {{2}^{0}}$ and move up the left hand side of the diagram we add $\scriptsize 1$ to the exponent of $\scriptsize 0$ and get $\scriptsize {{2}^{1}}$. This is the same as $\scriptsize {{2}^{1}}=1\times 2=2$. We multiplied by $\scriptsize 2$ to move up the diagram on the right hand side. Similarly, if we add to the exponent of $\scriptsize {{2}^{1}}$ we get $\scriptsize {{2}^{2}}$, this is the same as $\scriptsize {{2}^{2}}=1\times 2\times 2=4$. So as we move up the diagram, we add exponents and the answers get larger.

Now, start at $\scriptsize {{2}^{2}}$ and move down the left hand side of the diagram by subtracting $\scriptsize 1$ from the exponent. You will see that each time we decrease the exponent by $\scriptsize 1$ we divide the previous answer on the right hand side of the diagram by $\scriptsize 2$. So as we move down the diagram, we subtract exponents and the answers get smaller.

When we get to $\scriptsize {{2}^{0}}$ and subtract one from the exponent we will get $\scriptsize {{2}^{-1}}$ so to work out the answer for $\scriptsize {{2}^{-1}}$ we divide the previous answer on the right $\scriptsize {{2}^{0}}=1$ by $\scriptsize 2$ and get that $\scriptsize {{2}^{-1}}=\displaystyle \frac{1}{2}$. Similarly, for $\scriptsize {{2}^{-2}}$ we divide the previous answer $\scriptsize {{2}^{-1}}=\displaystyle \frac{1}{2}$ by $\scriptsize 2$ and get $\scriptsize {{2}^{-2}}=1\div 2\div 2=\displaystyle \frac{1}{{{2}^{2}}}$.

Hopefully, you see that positive, zero and negative exponents form a simple pattern as you move up and down the diagram.

### Exercise 2.4

1. Simplify without using a calculator and write answers with positive exponents:
1. $\scriptsize {{3}^{12}}\div {{3}^{9}}$
2. $\scriptsize \displaystyle \frac{15{{x}^{12}}{{y}^{5}}}{12{{x}^{9}}{{y}^{3}}}$
3. $\scriptsize \displaystyle \frac{{{2}^{x+3}}}{{{2}^{2+x}}}$
4. $\scriptsize \displaystyle \frac{3}{{{2}^{-2}}}$
5. $\scriptsize \displaystyle \frac{{{a}^{2}}{{x}^{-2}}}{{{a}^{-1}}{{x}^{3}}}$

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

• How to use exponential form to write a number as $\scriptsize {{a}^{n}}$ where $\scriptsize n$ is any natural number and $\scriptsize a$ is any real number $\scriptsize \ne 0$.
• The law for multiplying exponents with the same base $\scriptsize {{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\text{ }$.
• The law for dividing exponents with the same base $\scriptsize \displaystyle \frac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\text{ }$.
• The identity $\scriptsize {{a}^{0}}=1$.
• The identity for negative exponents: $\scriptsize {{a}^{-m}}=\displaystyle \frac{1}{{{a}^{m}}}\text{ ; }\displaystyle \frac{1}{{{a}^{-m}}}={{a}^{m}}$.

# Unit 2: Assessment

#### Suggested time to complete: 10 minutes

1. Simplify as far as possible and write your answers with positive exponents:
1. $\scriptsize {{({{3}^{-2}}{{a}^{-6}})}^{0}}~\times {{3}^{-2}}$
2. $\scriptsize -8{{x}^{2}}{{y}^{3}}\div 6{{x}^{3}}y$
3. $\scriptsize \displaystyle \frac{(2{{x}^{2}}y)(-4x{{y}^{2}})}{-2xy}$
4. $\scriptsize \displaystyle \frac{{{2}^{a-2}}{{3}^{a+3}}}{{{2}^{-2+a}}}$
2. Andy and Alex find different answers to the same question. Look at each method below and state who is correct. Give a reason for your answer.

Andy’s solution

\scriptsize \begin{align} & 3\times {{2}^{2}}+5 \\ & ={{6}^{2}}+5 \\ & =41 \\ \end{align}.

Alex’s solution

\scriptsize \begin{align} & 3\times {{2}^{2}}+5 \\ & =3\times 4+5 \\ & =17 \\ \end{align}

The full solutions are at the end of the unit.

### Note

Think about what you learnt in this unit. Can you:

• Rewrite expressions with repeated addition in exponential form?
• Expand expressions written in exponential form?
• Multiply exponents with the same base?
• Divide exponents with the same base?

# Unit 2: Solutions

### Exercise 2.1

1. $\scriptsize 4\times 4\times 4\times 4\times 4={{4}^{5}}$. The base $\scriptsize 4$ has been multiplied $\scriptsize 5$ times so the exponent is $\scriptsize 5$.
2. You need to be careful with this expression since there is a combination of numbers being multiplied.
$\scriptsize 2\times 2\times 2={{2}^{3}}$ and $\scriptsize 3\times 3={{3}^{2}}$. So $\scriptsize 2\times 2\times 2\times 3\times 3={{2}^{3}}\times {{3}^{2}}$.
3. Did you notice that the numbers are being added in this question so you cannot use exponents?
$\scriptsize 2+2+3+3+3=2\times 2+3\times 3$. Use multiplication to write the expression.
1. $\scriptsize {{3}^{4}}=3\times 3\times 3\times 3$. The base $\scriptsize 3$ has been multiplied $\scriptsize 4$ times.
2. $\scriptsize {{3}^{2}}\times {{2}^{3}}=3\times 3\times 2\times 2\times 2$. There are two different bases so each base must be treated separately.
3. $\scriptsize {{x}^{3}}{{y}^{2}}=x\times x\times x\times y\times y$. The variables (letters that represent unknown numbers) must be multiplied separately, the $\scriptsize x$ is multiplied $\scriptsize 3$ times and $\scriptsize y$ is multiplied twice.
4. $\scriptsize {{(-2)}^{4}}=(-2)\times (-2)\times (-2)\times (-2)=16$. Use brackets to help you multiply negative numbers.
1. .
\scriptsize \begin{align} {{3}^{2}}& =9 \\ 2\times 2\times 2\times 2& =16 \\ & 9 \lt 16 \end{align}
$\scriptsize \therefore {{3}^{2}} \lt 2\times 2\times 2\times 2$
2. .
\scriptsize \begin{align} {{3}^{2}}.2& =9\times 2=18 \\ {{2}^{2}}.3& =4\times 3=12 \\ & 18 \gt 12 \end{align}
$\scriptsize \therefore {{3}^{2}}.2 \gt {{2}^{2}}.3$
3. $\scriptsize \displaystyle \frac{1}{2}\times \displaystyle \frac{1}{2}\times \displaystyle \frac{1}{2}={{\left( \displaystyle \frac{1}{2} \right)}^{3}}$. The base of $\scriptsize \displaystyle \frac{1}{2}$ is multiplied 3 times hence the expressions are equal.
1. $\scriptsize 25={{5}^{2}}$
2. $\scriptsize 729={{3}^{6}}$
3. $\scriptsize 48={{2}^{4}}\times 3$

Back to exercise 2.1

### Exercise 2.2

1. .
\scriptsize \begin{align*} {{3}^{2}}\times {{3}^{4}}&={{3}^{2+4}}\\ & ={{3}^{7}} \end{align*}
2. .
$\scriptsize \begin{array}{l}{{2}^{3}}\times {{5}^{3}}&={{(2\times 5)}^{3}}\\&={{10}^{3}}\\&=1000\end{array}$
3. .
\scriptsize \begin{align*} {{x}^{2}}{{y}^{3}}\times 2{{x}^{3}}y&=2{{x}^{2+3}}{{y}^{3+1}}\\ &=2{{x}^{5}}{{y}^{4}} &&\text{ The exponents with the same bases are added and the coefficient of 2 stays the same. } \end{align*}
4. .
\scriptsize \begin{align*} {{a}^{2}}\times {{b}^{3}}\times {{y}^{2}}={{a}^{2}}{{b}^{3}}{{y}^{2}} &&\text{ This cannot be simplified because all the bases are different. All that has been done is to rewrite the expression without the explicit multiplication signs. } \end{align*}
5. .
\scriptsize \begin{align*} \left( -{{x}^{2}}{{y}^{4}} \right)\left( -3y{{x}^{3}} \right)&=3{{x}^{2+3}}{{y}^{4+1}} \\ &=3{{x}^{5}}{{y}^{5}} &&\text{ Be careful with the signs. } \end{align*}
6. .
\scriptsize \begin{align*} & \left( -2{{a}^{3}}{{b}^{2}} \right)\left( -bxa \right)\left( {{a}^{2}}{{x}^{3}}b \right) \\ & =2{{a}^{3+1+2}}{{b}^{2+1+1}}{{x}^{1+3}} \\ & =2{{a}^{6}}{{b}^{4}}{{x}^{4}} &&\text{ The order of the variables does not matter. Just make sure to allocate the exponents correctly. } \end{align*}
7. .
\scriptsize \begin{align*} {{2}^{2x}}\times {{2}^{3x}}& ={{2}^{2x+3x}} \\ & ={{2}^{5x}} &&\text{ Even though variables appear as exponents, the same rule applies. Keep the base and add the exponents when multiplying powers with the same base. } \end{align*}
8. .
\scriptsize \begin{align*} {{3}^{3a}}\times {{3}^{2a+b}}&={{3}^{3a+2a+b}}\\ &={{3}^{5a+b}}&&\text{ The rules of algebra apply here, you cannot add unlike variables. } \end{align*}

Back to exercise 2.2

### Exercise 2.3

1. .
\scriptsize \begin{align*} \displaystyle \frac{6{{a}^{3}}b}{a}& =6{{a}^{3-1}}b\\ & =6{{a}^{2}}b \end{align*}
2. .
\scriptsize \begin{align*} \displaystyle \frac{{{3}^{2}}}{{{2}^{4}}} &&\text{ Bases are different so you cannot simplify any further using exponent rules. }\\ = \displaystyle \frac{9}{16} &&\text{ You can write answer as a common fraction. } \end{align*}
3. .
\scriptsize \begin{align*} \displaystyle \frac{x{{y}^{2}}\times 2{{x}^{2}}y}{3xy}& = \displaystyle \frac{2{{x}^{1+2}}{{y}^{2+1}}}{3xy}\\ &= \displaystyle \frac{2{{x}^{3}}{{y}^{3}}}{3xy}\\ &= \displaystyle \frac{2{{x}^{3-1}}{{y}^{3-1}}}{3}\\ &= \displaystyle \frac{2}{3}{{x}^{2}}{{y}^{2}} \end{align*}

Back to exercise 2.3

### Exercise 2.4

1. .
\scriptsize \begin{align*} {{3}^{12}}\div {{3}^{9}} & ={{3}^{12-9}}\\ & ={{3}^{3}}\\ & =27 \end{align*}
2. .
\scriptsize \begin{align*} \displaystyle \frac{15{{x}^{12}}{{y}^{5}}}{12{{x}^{9}}{{y}^{3}}}& =\displaystyle \frac{5{{x}^{12-9}}{{y}^{5-3}}}{4}\\ & =\displaystyle \frac{5{{x}^{3}}{{y}^{2}}}{4} \end{align*}
3.  .
\scriptsize \begin{align*} \displaystyle \frac{{{2}^{x+3}}}{{{2}^{2+x}}} & ={{2}^{x+3-(2+x)}} &&\text{ Note: apply the same rule even though there are variables in the exponent. }\\ & ={{2}^{x+3-2-x}} &&\text{ Use brackets when subtracting the exponent so you remember to change the signs. }\\ & ={{2}^{x+3-2-x}} \\ & = {{2}^{1}}\\ &=2 \end{align*}
4. .
\scriptsize \begin{align*} \displaystyle \frac{3}{{{2}^{-2}}} & =3\cdot {{2}^{2}} &&\text{ The sign of the exponent must change to a positive when you move it to the numerator. }\\ & =12 \end{align*}
5. .
\scriptsize \begin{align*} \displaystyle \frac{{{a}^{2}}{{x}^{-2}}}{{{a}^{-1}}{{x}^{3}}}& = \displaystyle \frac{{{a}^{2-(-1)}}{{x}^{-2-(3)}}}{1}\\ & ={{a}^{2+1}}{{x}^{-2-(3)}}\\ & ={{a}^{3}}{{x}^{-5}}\\ & =\displaystyle \frac{{{a}^{3}}}{{{x}^{5}}} \end{align*}

Back to exercise 2.4

### Unit 2: Assessment

1. .
\scriptsize \begin{align*} {({3}^{-2}{a}^{-6})}^{0}\times{3}^{-2}& =1\times\displaystyle \frac{1}{{3}^{2}} &&\text{ Remember: }{a}^{0}=1 \text{ and } {a}^{-m}=\displaystyle \frac{1}{{a}^{m}}\\ &=\displaystyle \frac{1}{9} \end{align*}
2. .
\scriptsize \begin{align*} -8{{x}^{2}}{{y}^{3}}\div 6{{x}^{3}}y & =\displaystyle \frac{-8{{x}^{2}}{{y}^{3}}}{6{{x}^{3}}y} \\ & =\displaystyle \frac{-8{{x}^{2-3}}{{y}^{3-1}}}{6}\\ & =\displaystyle \frac{-4x{{y}^{2}}}{3} \end{align*}
3. .
\scriptsize \begin{align*} \displaystyle \frac{(2{{x}^{2}}y)(-4x{{y}^{2}})}{-2xy} & =\displaystyle \frac{({{x}^{2}}y)(-4x{{y}^{2}})}{-xy}\\ & =\displaystyle \frac{-4{{x}^{2+1}}{{y}^{1+2}}}{-xy}\\ & =4{{x}^{2+1-1}}{{y}^{1+2-1}}\\ & =4{{x}^{2}}{{y}^{2}} \end{align*}
4. .
\scriptsize \begin{align*} \displaystyle \frac{{{2}^{a-2}}{{3}^{a+3}}}{{{2}^{-2+a}}} & ={{2}^{a-2-(-2+a)}}{{3}^{a+3}}\\ & ={{2}^{0}}{{3}^{a+3}}\\ & ={{3}^{a+3}} \end{align*}
1. Alex is correct. Andy made the mistake of multiplying different bases together.