Numbers: Numbers and number relationships

# Unit 3: Raising a power to a power, exponent laws

Natashia Bearam-Edmunds

### Unit outcomes

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

• Raise a power to a power.
• Simplify exponential expressions by applying the exponent laws.

## What you should know

Before you start this unit, make sure you can:

Here is a short self-assessment to make sure you have the skills you need to proceed with this unit.

Simplify:

1. $\scriptsize 5{{x}^{0}}\times \displaystyle \frac{2x}{{{x}^{2}}}$
2. $\scriptsize {{2}^{{2x}}}{{.3}^{x}}{{.2}^{{x-1}}}$

Solutions:

1. .
\scriptsize \begin{align*} 5{{x}^{0}}\times \displaystyle \frac{{2x}}{{{{x}^{2}}}} &=5(1)\times 2{{x}^{{1-2}}}\\&=10{{x}^{{-1}}}\\&=\displaystyle \frac{{10}}{x}\end{align*}
2. .
\scriptsize \begin{align*}{{2}^{{2x}}}{{.3}^{x}}{{.2}^{{x-1}}} &={{2}^{{2x+x-1}}}\cdot {{3}^{x}}\\&={{2}^{{3x-1}}}{{3}^{x}}\end{align*}

## Introduction

So far we have worked with the product and quotient exponent rules and learnt the basic identities. You have also seen that exponents are used as a quick way to simplify expressions with repeated multiplication. By now you should be familiar with the following exponent rules and identities.

### Take note!

 Exponent rules and identities Example $\scriptsize {{a}^{n}}=a\times a\times a\ldots \times a~\left[ {\text{n }\!\!~\!\!\text{ times}} \right]$ $\scriptsize 4\times 4\times 4\times 4\times 4={{4}^{5}}$ $\scriptsize {{a}^{m}}\times {{a}^{n}}={{a}^{{m+n}}}$ $\scriptsize {{x}^{2}}{{y}^{3}}\times {{x}^{3}}y={{x}^{5}}{{y}^{4}}$ $\scriptsize \displaystyle \frac{{{{a}^{m}}}}{{{{a}^{n}}}}={{a}^{{m-n}}}$ $\scriptsize \displaystyle \frac{{{{2}^{5}}}}{{{{2}^{2}}}}={{2}^{3}}=8$ $\scriptsize {{a}^{0}}=1$ $\scriptsize {{({{3}^{{-2}}}{{a}^{{-6}}})}^{0}}~=1$ $\scriptsize {{a}^{{-m}}}=\displaystyle \frac{1}{{{{a}^{m}}}}\text{ ; }\displaystyle \frac{1}{{{{a}^{{-m}}}}}={{a}^{m}}$ $\scriptsize \displaystyle \frac{1}{{{{x}^{{-2}}}}}={{x}^{2}}$

## Raising a power to a power

You may have seen a base with an exponent raised to another exponent. This is called raising a power to a power. Look at $\scriptsize {{({{2}^{2}})}^{3}}$. This is an example of a power raised to another power. Think about what the expression means.

It helps to expand the brackets and then write the answer in exponential form.

From the exponential form, we can work out a rule for raising a power to a power. HINT: Look at the exponents you started with and the ones you end up with. What is the relationship between the two?

Let’s work through this together:

In the expression $\scriptsize {{({{2}^{2}})}^{3}}$we see that the power $\scriptsize {{2}^{2}}$ is raised to an exponent of three outside the bracket. This means that $\scriptsize {{2}^{2}}$ is multiplied by itself three times. If we expand $\scriptsize {{({{2}^{2}})}^{3}}$ we get: $\scriptsize {{({{2}^{2}})}^{3}}={{2}^{2}}\times {{2}^{2}}\times {{2}^{2}}$. By using the rule for multiplying powers with the same base and adding the exponents this becomes $\scriptsize {{2}^{2}}\times {{2}^{2}}\times {{2}^{2}}={{2}^{{2+2+2}}}={{2}^{6}}$.

Is there a quick way to get from $\scriptsize {{({{2}^{2}})}^{3}}$  to $\scriptsize {{2}^{6}}$? Yes.

We can see that the exponent six in the answer is simply the product of the exponents in the given expression so $\scriptsize {{({{2}^{2}})}^{3}}={{2}^{{2\times 3}}}={{2}^{6}}$.

When we raise a power to a power we multiply the exponents.

This is generalised as the following exponent law.

$\scriptsize {{({{a}^{m}})}^{n}}={{a}^{{m\times n}}}$

This is called the exponent power rule.

### Example 3.1

Simplify: $\scriptsize {{({{x}^{2}}{{y}^{3}})}^{3}}$

Solution

If we expand the expression, we get $\scriptsize {{({{x}^{2}}{{y}^{3}})}^{3}}=({{x}^{2}}{{y}^{3}})\times ({{x}^{2}}{{y}^{3}})\times ({{x}^{2}}{{y}^{3}})={{x}^{2}}\times {{x}^{2}}\times {{x}^{2}}\times {{y}^{3}}\times {{y}^{3}}\times {{y}^{3}}$.

We can apply the product rule to multiply powers with the same base to get $\scriptsize {{x}^{6}}{{y}^{9}}$.

We can see that the exponent on each base is the product of the exponents from the original expression.

$\scriptsize {{({{x}^{2}}{{y}^{3}})}^{3}}={{x}^{{2\times 3}}}\times {{y}^{{3\times 3}}}={{x}^{6}}{{y}^{9}}$

We conclude that when different bases multiplied within a bracket are raised to a power, each base must be raised to the power individually.

The above example is generalised as the following exponent law.

$\scriptsize {{({{a}^{m}}{{b}^{n}})}^{p}}={{a}^{{mp}}}\cdot {{b}^{{np}}}$

This is called the power of a product rule.

Now, try to simplify $\scriptsize {{\left( {\displaystyle \frac{a}{b}} \right)}^{4}}$.

### Example 3.2

Simplify: $\scriptsize {{\left( {\displaystyle \frac{a}{b}} \right)}^{4}}$

Solution

If we expand the expression, we get $\scriptsize {{\left( {\displaystyle \frac{a}{b}} \right)}^{4}}=\displaystyle \frac{a}{b}\times \displaystyle \frac{a}{b}\times \displaystyle \frac{a}{b}\times \displaystyle \frac{a}{b}=\displaystyle \frac{{a\times a\times a\times a}}{{b\times b\times b\times b}}$.

We can write this in exponential form as $\scriptsize \displaystyle \frac{{{{a}^{4}}}}{{{{b}^{4}}}}$.

We can see that the exponent of each base is the product of the exponents from the original expression.

$\scriptsize {{\left( {\displaystyle \frac{a}{b}} \right)}^{4}}={{\left( {\displaystyle \frac{a^1}{b^1}} \right)}^{4}}=\displaystyle \frac{{{{a}^{4}}}}{{{{b}^{4}}}}$

We can conclude that when different bases divided within a bracket are raised to a power, each base must be raised to the power individually.

The above example is generalised as the following exponent law.

$\scriptsize {{\left( {\displaystyle \frac{a}{b}} \right)}^{m}}=\displaystyle \frac{{{{a}^{m}}}}{{{{b}^{m}}}}$

This is called the power of a fraction rule.

The basic exponent laws can be applied to many different examples. Look at the examples below and see how each of them differs from the basic case each time.

### Example 3.3

Simplify:

1. $\scriptsize 2{{({{a}^{2}}b)}^{3}}$
2. $\scriptsize {{(2{{y}^{2}})}^{4}}$
3. $\scriptsize 3{{({{x}^{3}})}^{{-2}}}$

Solutions

1. .
\scriptsize \begin{align*} &2{{({{a}^{2}}b)}^{3}}&&\text{The number }2 \text{ is not raised to the power of }3\\ &&& \text{ since it is not inside the brackets.}\\ &=2{{a}^{6}}{{b}^{3}}\\ \end{align*}
2. .
\scriptsize \begin{align*} &{{(2{{y}^{2}})}^{4}}&&\text{Here the coefficient (number in front of a variable) }2\\ &&& \text{ is also raised to the exponent }4\\ &&& \text{ because it is inside the brackets.}\\ &={{2}^{4}}{{y}^{8}}=16{{y}^{8}}\ \end{align*}
3. .
\scriptsize \begin{align*} &3{{({{x}^{3}})}^{-2}} && \text{The coefficient is not raised to the exponent }-2 \text{.}\\ &=3{{x}^{3}}^{(-2)} && \text{Note that }{{x}^{3}}\text{ is raised to a negative exponent, be careful }\\ &&& \text{with the signs.}\\ &=3{{x}^{-6}}\\ &=\displaystyle \frac{3}{{{x}^{6}}} &&\text{Answers are generally left with positive exponents, so make } \\ &&& \text{the exponent positive by writing }{{x}^{-6}} \text{ as }\displaystyle \frac{1}{{{x}^{6}}} \text{.} \end{align*}

### Exercise 3.1

Simplify each expression as far as possible and write answers with positive exponents:

1. $\scriptsize {{(xy)}^{{-2}}}$

2. $\scriptsize {{\left( {-3x{{y}^{2}}} \right)}^{2}}{{(-3x)}^{{-2}}}$

3. $\scriptsize {{\left( {\displaystyle \frac{a}{{{{b}^{{-2}}}c}}} \right)}^{3}}$

4. $\scriptsize 2{{\left( {-{{a}^{2}}{{b}^{4}}} \right)}^{2}}\left( {-4{{a}^{3}}{{b}^{6}}} \right){{\left( {-16{{a}^{6}}{{b}^{8}}} \right)}^{{-1}}}$

5. $\scriptsize {{\left( {{{2}^{{-2}}}} \right)}^{{2p+1}}}$

6. $\scriptsize {{\left( {\displaystyle \frac{{3{{q}^{{2x}}}}}{{{{q}^{{3x}}}{{y}^{{-2a-1}}}}}} \right)}^{2}}$

The full solutions are at the end of the unit.

### Note

Do you need more practise raising a power to a power? Try this online activity called Powers of powers. You will need an internet connection.

## Prime factorising and exponents

The word ‘prime’ in ‘prime factorising’ tells us that this type of factorising has something to do with prime numbers. Remember that a prime number is a number that has only two factors; the number one and itself. What are some examples of prime numbers?

The numbers $\scriptsize 2;3;5;17$ are examples of prime numbers. Prime factorisation is the process of breaking a number up into its prime factors.

For example, $\scriptsize 12$  broken down into its prime factors is $\scriptsize 2 \times 2 \times 3$. Did you notice that there is repeated multiplication? That means we can use exponents to simplify the expression further: $\scriptsize 2^2 \cdot 3$. This is a method you will often rely on to simplify powers.

### Did you know?

Is $\scriptsize 1$ a prime number? Watch the video called “1 and Prime Numbers” about prime factorisation. It explains why $\scriptsize 1$ is not a prime number using the Fundamental Theorem of Arithmetic. It also explains more about prime numbers.

1 and Prime Numbers (Duration: 5.21)

### Did you know?

You can use your calculator to prime factorise any number. The video called “How to do Prime Factorisation on a Casio FX-83GT PLUS” shows the calculator steps.

Have a look at this next example of prime factorisation.

### Example 3.4

Use prime factorisation to simplify: $\scriptsize {{\left( 18 \right)}^{2x}}$

Solution

Start by breaking $\scriptsize 18$ up into its prime factors.

$\scriptsize 18=9\times 2={{3}^{2}}\times 2$

So $\scriptsize {{18}^{2x}}={{({{3}^{2}}\times 2)}^{2x}}$

Next, raise each of the prime factors to the power of $\scriptsize 2x$.

$\scriptsize {{({{3}^{2}}\times 2)}^{2x}}={{3}^{2\times 2x}}\times {{2}^{1\times 2x}}$

$\scriptsize {{3}^{2\times 2x}}\times {{2}^{1\times 2x}}={{3}^{4x}}\cdot {{2}^{2x}}$

Hence $\scriptsize {{18}^{2x}}={{3}^{4x}}\cdot {{2}^{2x}}$

### Take note!

You can use a calculator to find the prime factors of any number. Here are the steps for the Casio fx-82ZA PLUS calculator, shown below. If you have a different calculator check your manual for the steps.

To find the prime factors of $\scriptsize 72$:

Step 1: Enter $\scriptsize 72$ and press $\scriptsize =$

Step 2: Press SHIFT and FACT (the key that shows degrees and minutes o’’’)

The calculator will display the prime factors as $\scriptsize {{2}^{2}}\times {{3}^{2}}$

Use your calculator to factorise: $\scriptsize 56;112;725$

When working with exponents, all the laws of operation for algebra apply.

### Exercise 3.2

Simplify as far as possible:

1. $\scriptsize {{3}^{n}}\cdot {{(9)}^{2n}}$
2. $\scriptsize \displaystyle \frac{{{4}^{2x}}}{{{2}^{3x}}}$
3. $\scriptsize {{({{8}^{n}})}^{2}}$
4. $\scriptsize {{25}^{x}}\cdot 5$

The full solutions are at the end of the unit

## Summary

In this unit you have learnt the following:

• The power rule.
• The power of a product rule.
• The power of a fraction rule.
$\scriptsize {{({{a}^{m}})}^{n}}={{a}^{m\times n}};{{({{a}^{m}}{{b}^{n}})}^{p}}={{a}^{mp}}\cdot {{b}^{np}};{{\left( \displaystyle \frac{a}{b} \right)}^{m}}=\displaystyle \frac{{{a}^{m}}}{{{b}^{m}}}$
• How to raise a power to a negative exponent.
• How to use prime factorisation to simplify powers.

# Unit 3: Assessment

#### Suggested time to complete: 20 minutes

1. Arrange the following from least to greatest:
$\scriptsize {{2}^{3}}-{{2}^{{-1}}};{{\left( 2 \right)}^{3}}-{{\left( 2 \right)}^{2}};{{2}^{2}}{{3}^{0}}\div 2$
2. Calculate the following without using a calculator:
1. $\scriptsize {{2}^{2}}\times 3\times {{2}^{{-1}}}$
2. $\scriptsize {{9}^{{-1}}}\times {{(3)}^{2}}\times {{2}^{{-2}}}$
3. $\scriptsize {{6}^{a}}\times \displaystyle \frac{{{{{(2a{{b}^{4}})}}^{0}}}}{{{{2}^{a}}{{3}^{a}}}}$
3. Simplify:
1. $\scriptsize {{({{x}^{2}})}^{{-2}}}\times 2{{({{x}^{3}})}^{2}}$
2. $\scriptsize \displaystyle \frac{{9{{{({{a}^{2}})}}^{{-2}}}}}{{3{{a}^{{-5}}}}}$
3. $\scriptsize \displaystyle \frac{{{{{\left( {{{2}^{{x+2}}}{{3}^{x}}} \right)}}^{2}}}}{{{{6}^{x}}}}$
4. $\scriptsize \displaystyle \frac{{{{3}^{n}}{{9}^{{n-3}}}}}{{{{{27}}^{{n-1}}}}}$

The full solutions are at the end of the unit.

# Unit 3: Solutions

### Exercise 3.1

1. .
$\scriptsize {{(xy)}^{-2}}=\displaystyle \frac{1}{{{(xy)}^{2}}}=\displaystyle \frac{1}{{{x}^{2}}{{y}^{2}}}$
2. .
\scriptsize \begin{align*} &{{\left( {-3x{{y}^{2}}} \right)}^{2}}{{(-3x)}^{{-2}}} &&\text{Move }{{(-3x)}^{{-2}}}\text{ to the denominator to make the exponent positive.}\\ &=\displaystyle \frac{{{{{\left( {-3x{{y}^{2}}} \right)}}^{2}}}}{{{{{(-3x)}}^{{+2}}}}} &&\text{Make the exponent of the second bracket positive.}\\ &=\displaystyle \frac{{{{{(-3)}}^{2}}{{x}^{2}}{{y}^{4}}}}{{{{{(-3)}}^{2}}{{x}^{2}}}} &&\text{Apply the power of a product rule. The } 3 \text{ must also be raised to the exponent of }2 \text{.}\\ &=\displaystyle \frac{{9{{x}^{2}}{{y}^{4}}}}{{9{{x}^{2}}}}={{y}^{4}}\end{align*}
3. .
\scriptsize \begin{align*} &{{\left( {\displaystyle \frac{a}{{{{b}^{{-2}}}c}}} \right)}^{3}}\\ &={{\left( {\displaystyle \frac{{a{{b}^{2}}}}{c}} \right)}^{3}} &&\text{Make the negative exponent positive by moving the power to the numerator.}\\ &=\displaystyle \frac{{{{a}^{3}}{{b}^{6}}}}{{{{c}^{3}}}}\end{align*}
4. .
\scriptsize \begin{align*} &2{{\left( {-{{a}^{2}}{{b}^{4}}} \right)}^{2}}\left( {-4{{a}^{3}}{{b}^{6}}} \right){{\left( {-16{{a}^{6}}{{b}^{8}}} \right)}^{{-1}}}\\ &=\displaystyle \frac{{2{{{\left( {-{{a}^{2}}{{b}^{4}}} \right)}}^{2}}\left( {-4{{a}^{3}}{{b}^{6}}} \right)}}{{{{{\left( {-16{{a}^{6}}{{b}^{8}}} \right)}}^{{+1}}}}} &&\text{Move the bracket with the negative exponent to the denominator.}\\ &=\displaystyle \frac{{2({{a}^{4}}{{b}^{8}})(-4{{a}^{3}}{{b}^{6}})}}{{-16{{a}^{6}}{{b}^{8}}}} &&\text{Raise the first bracket to the power of 2 before removing brackets and simplifying further.}\\ &=\displaystyle \frac{{-8{{a}^{7}}{{b}^{{14}}}}}{{-16{{a}^{6}}{{b}^{8}}}}\\ &=\displaystyle \frac{{-8{{a}^{{7-6}}}{{b}^{{14-8}}}}}{{-16}}\\ &=\displaystyle \frac{{a{{b}^{6}}}}{2}\end{align*}
5. .
\scriptsize \begin{align*} &{{\left( {{{2}^{{-2}}}} \right)}^{{2p+1}}}\\ &={{\left( {\displaystyle \frac{1}{{{{2}^{2}}}}} \right)}^{{2p+1}}}&&\text{Variables are also multiplied if they form part of the exponent.}\\ &=\displaystyle \frac{1}{{{{2}^{{4p+2}}}}} \end{align*}
6. .
\scriptsize \begin{align*} &{{\left( {\displaystyle \frac{{3{{q}^{{2x}}}}}{{{{q}^{{3x}}}{{y}^{{-2a-1}}}}}} \right)}^{2}}\\ &={({3q}^{2x-3x}{y}^{2a+1})}^{2}\\ &={3}^{2}{q}^{-2x}{y^{4a+2}}&&\text{Each part of each exponents must be raised to the power of 2.}\\ &=\displaystyle \frac{9{y}^{4a+2}}{{q}^{2x}} \end{align*}

Back to exercise 3.1

### Exercise 3.2

1. .
\scriptsize \begin{align*} &{{3}^{n}}\cdot {{(9)}^{{2n}}}\\ &={{3}^{n}}\cdot {{({{3}^{2}})}^{{2n}}} &&\text{Prime factorise 9.}\\ &={{3}^{n}}\cdot {{3}^{{4n}}}\\ &={{3}^{{5n}}} \end{align*}
2. .
\scriptsize \begin{align*} &\displaystyle \frac{{{{4}^{{2x}}}}}{{{{2}^{{3x}}}}}\\ &=\displaystyle \frac{{{{{({{2}^{2}})}}^{{2x}}}}}{{{{2}^{{3x}}}}}\\ &=\displaystyle \frac{{{{2}^{{4x}}}}}{{{{2}^{{3x}}}}}\\ &={{2}^{{4x-3x}}}\\ &={{2}^{x}} \end{align*}
3. .
\scriptsize \begin{align*} &{{({{8}^{n}})}^{2}}\\ &={{({{({{2}^{3}})}^{n}})}^{2}} &&\text{Work from within the brackets and prime factorise }8 \text { to } {2}^{3}\\ &={{({{2}^{3}}^{n})}^{2}}\\ &={{2}^{{6n}}} \end{align*}
4. .
\scriptsize \begin{align*} &{{25}^{x}}\cdot 5\\ &={{({{5}^{2}})}^{x}}\cdot 5\\ &={{5}^{{2x}}}\cdot {{5}^{1}}\\ &={{5}^{{2x+1}}}\\ \end{align*}

Back to exercise 3.2

### Unit 3: Assessment

1. .
\scriptsize \begin{align*} &{{2}^{3}}-{{2}^{{-1}}}=8-\displaystyle \frac{1}{2}=7\displaystyle \frac{1}{2}\\ &{{\left( 2 \right)}^{3}}-{{\left( 2 \right)}^{2}}=8-4=4\\ &{{2}^{2}}{{3}^{0}}\div 2=\displaystyle \frac{{4\times 1}}{2}=2\\ &\text{From least to greatest: } {{2}^{2}}{{3}^{0}}\div 2 ; {{\left( 2 \right)}^{3}}-{{\left( 2 \right)}^{2}} ; {{2}^{3}}-{{2}^{{-1}}}\\ \end{align*}
2. .
1. .
\scriptsize \begin{align*} &{{2}^{2}}\times 3\times {{2}^{{-1}}}\\ &={{2}^{{2-1}}}\cdot 3\\ &=6\\ \end{align*}
2. .
\scriptsize \begin{align*} &{{9}^{{-1}}}\times {{(3)}^{2}}\times {{2}^{{-2}}}\\ &=\displaystyle \frac{{{{{\left( {{{3}^{2}}} \right)}}^{{-1}}}\times {{{(3)}}^{2}}}}{{{{2}^{2}}}}\\ &=\displaystyle \frac{{{{3}^{{-2+2}}}}}{{{{2}^{2}}}}\\ &=\displaystyle \frac{{{{3}^{0}}}}{4}=\displaystyle \frac{1}{4}\\ \end{align*}
3. .
\scriptsize \begin{align*} &{{6}^{a}}\times \displaystyle \frac{{{{{(2a{{b}^{4}})}}^{0}}}}{{{{2}^{a}}{{3}^{a}}}}\\ &={{(3\cdot 2)}^{a}}\times \displaystyle \frac{1}{{{{2}^{a}}{{3}^{a}}}}\\ &=\displaystyle \frac{{{{3}^{a}}{{2}^{a}}}}{{{{2}^{a}}{{3}^{a}}}}\\ &=1\\ \end{align*}
3. .
1. .
\scriptsize \begin{align*} &{{({{x}^{2}})}^{{-2}}}\times 2{{({{x}^{3}})}^{2}}\\ &={{x}^{{-4}}}\times 2{{x}^{6}}\\ &=2{{x}^{{-4+6}}}\\ &=2{{x}^{2}}\\ \end{align*}
2. .
\scriptsize \begin{align*} &\displaystyle \frac{{9{{{({{a}^{2}})}}^{{-2}}}}}{{3{{a}^{{-5}}}}}\\ &=\displaystyle \frac{{9{{a}^{{-4}}}}}{{3{{a}^{{-5}}}}}\\ &=3{{a}^{{-4-(-5)}}}\\ &=3a\\ \end{align*}
3. .
\scriptsize \begin{align*} &\displaystyle \frac{{{{{\left( {{{2}^{{x+2}}}{{3}^{x}}} \right)}}^{2}}}}{{{{6}^{x}}}}\\ &=\displaystyle \frac{{{{2}^{{2(x+2)}}}{{3}^{{2x}}}}}{{{{{(2\cdot 3)}}^{x}}}}\\ &=\displaystyle \frac{{{{2}^{{2x+4}}}{{3}^{{2x}}}}}{{{{2}^{x}}{{3}^{x}}}}\\ &={{2}^{{2x+4-x}}}{{3}^{{2x-x}}}\\ &={{2}^{{x+4}}}{{3}^{x}}\\ \end{align*}
4. .
\scriptsize \begin{align*} &\displaystyle \frac{{{{3}^{n}}{{9}^{{n-3}}}}}{{{{{27}}^{{n-1}}}}}\\ &=\displaystyle \frac{{{{3}^{n}}{{{({{3}^{2}})}}^{{n-3}}}}}{{{{{\left( {{{3}^{3}}} \right)}}^{{n-1}}}}}\\ &=\displaystyle \frac{{{{3}^{n}}{{3}^{2n-6}}}}{{{{3}^{{3n-3}}}}}\\ &={{3}^{{n+2n-6-(3n-3)}}}\\ &={{3}^{{3n-6-3n+3}}}\\ &={{3}^{{-3}}}=\displaystyle \frac{1}{{{{3}^{3}}}}=\displaystyle \frac{1}{{27}}\\ \end{align*}

Back to the assessment

## License

National Curriculum (Vocational) Mathematics Level 2 by Natashia Bearam-Edmunds is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.