Numbers: Numbers and number relationships

# Unit 4: Rational exponents

Natashia Bearam-Edmunds

### Unit outcomes

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

• Write roots as rational exponents.

## 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. $\scriptsize {{2}^{2}}\times 3\times {{2}^{-1}}$
2. $\scriptsize {{6}^{a}}\times \displaystyle \frac{{{(2a{{b}^{4}})}^{0}}}{{{2}^{a}}{{3}^{a}}}$
3. $\scriptsize \displaystyle \frac{{{3}^{n}}{{9}^{n-3}}}{{{27}^{n-1}}}$

Solutions

1. .
\scriptsize \begin{align} & {{2}^{2}}\times 3\times {{2}^{-1}} \\ & ={{2}^{2-1}}\cdot 3 \\ & =6 \\ \end{align}
2. .
\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. .
\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}^{2}}^{n-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}

## Introduction

So far, we have worked with exponents that are integers (positive and negative whole numbers) and by now you know that the exponent shows how many times the base is multiplied by itself. However, exponents can be any rational number, which means they can be fractions too. So, what do fractional exponents represent?

We know that $\scriptsize {{2}^{3}}=2\times 2\times 2$. What do you think $\scriptsize {{8}^{\displaystyle \frac{1}{3}}}$ is equal to?

What about $\scriptsize {{9}^{\displaystyle \frac{1}{2}}}$? What do you think it means?

Use what you already know about exponents to simplify these powers with fractional exponents.

You can rewrite $\scriptsize {{8}^{\displaystyle \frac{1}{3}}}$ as $\scriptsize {{({{2}^{3}})}^{\displaystyle \frac{1}{3}}}$ using prime factorisation. Then, using the power rule $\scriptsize {{({{a}^{m}})}^{n}}={{a}^{m\times n}}$, you will get $\scriptsize {{({{2}^{3}})}^{\displaystyle \frac{1}{3}}}=2$. Similarly, you can show that $\scriptsize {{9}^{\displaystyle \frac{1}{2}}}={{({{3}^{2}})}^{\displaystyle \frac{1}{2}}}=3$.

We see that we can work out ‘nice’ answers to some bases raised to fractional exponents but this still does not answer the question ‘what is a fractional exponent?’ Will the answers always be ‘nice’ whole numbers?

## Fractional exponents

### Example 1

Think about this: $\scriptsize \sqrt[3]{8}=2$ and $\scriptsize \sqrt{9}=3$. How do we know this?

$\scriptsize 2\times 2\times 2=8$ , so $\scriptsize 2$ is the cube root of $\scriptsize 8$.

$\scriptsize 3\times 3=9$, so $\scriptsize 3$ is the square root of $\scriptsize 9$.

Let us look at the relationship between $\scriptsize {{8}^{\displaystyle \frac{1}{3}}}$ and $\scriptsize \sqrt[3]{8}$ and the relationship between $\scriptsize {{9}^{\displaystyle \frac{1}{2}}}$ and $\scriptsize \sqrt{9}$.

So far we have seen that $\scriptsize {{(8)}^{\displaystyle \frac{1}{3}}}=2$ and $\scriptsize {{9}^{\displaystyle \frac{1}{2}}}=3$ but you also know that $\scriptsize \sqrt[3]{8}=2$ and $\scriptsize \sqrt{9}=3$.

$\scriptsize {{(8)}^{\displaystyle \frac{1}{3}}}=2=\sqrt[3]{8}$. So these expressions are equivalent, hence $\scriptsize {{(8)}^{\displaystyle \frac{1}{3}}}$ means $\scriptsize \sqrt[3]{8}$.

$\scriptsize {{9}^{\displaystyle \frac{1}{2}}}=3=\sqrt{9}$. So $\scriptsize {{9}^{\displaystyle \frac{1}{2}}}$ is the same as $\scriptsize \sqrt{9}$.

It seems that fractional exponents describe the roots of numbers. But let us test this idea with a few more examples.

From what you’ve learnt, you can see that:

$\scriptsize \sqrt[3]{8}={{(8)}^{\displaystyle \frac{1}{3}}}$

$\scriptsize \sqrt{9}={{(9)}^{\displaystyle \frac{1}{2}}}$

$\scriptsize \sqrt[3]{27}=\text{ }{{(27)}^{\displaystyle \frac{1}{3}}}$

$\scriptsize \sqrt[4]{16}={{(16)}^{\displaystyle \frac{1}{4}}}$

In fact, by definition, a fractional exponent is equivalent to finding some root of a number.

$\scriptsize {{x}^{\displaystyle \frac{1}{n}}}=\text{n-th root of }x$

We generalise this as:

$\scriptsize {{x}^{\displaystyle \frac{1}{n}}}=\sqrt[n]{x}\text{ where }n\in \mathbb{N}\text{ and }x\in \mathbb{R}$

The root symbol has a special name; it is called a radical. Notice that the denominator of the fractional exponent is the same number that appears in the ‘tail’ of the radical symbol. Each part of a radical has its own name.

### Take note!

The radical that you are most familiar with is the square root. With a square root there is no need to write the index of $\scriptsize 2$. It is understood that $\scriptsize \sqrt{a}$ means $\scriptsize \sqrt[2]{a}$.

Try the following examples to better understand how to convert between radical and exponential form. Remember that the term ‘radicand’ refers to the number inside the root sign.

### Example 2

Convert the fractional exponents to radical form and then rewrite from radical form to exponential form:

1. $\scriptsize {{36}^{\displaystyle \frac{1}{2}}}$
2. $\scriptsize {{\left( 125x \right)}^{\displaystyle \frac{1}{5}}}$
3. $\scriptsize \sqrt{15x}$
4. $\scriptsize \sqrt[4]{{{a}^{2}}+{{b}^{2}}}$
5. $\scriptsize \sqrt[3]{{{x}^{2}}}$

Solutions

1. $\scriptsize {{36}^{\displaystyle \frac{1}{2}}}=\sqrt{36}$ You do not need to write the numerator of the fraction ‘$\scriptsize 1$’ in the radicand, and you do not write the denominator ‘$\scriptsize 2$’ as the index of the radical.
2. $\scriptsize {{\left( 125x \right)}^{\displaystyle \frac{1}{5}}}$. The brackets are important as they show us that the entire expression $\scriptsize \left( 125x \right)$ makes up the radicand.
$\scriptsize {{\left( 125x \right)}^{\displaystyle \frac{1}{5}}}=\sqrt[5]{125x}$
3. $\scriptsize \sqrt{15x}$ It is important to use brackets to rewrite the radicand in exponential form.
$\scriptsize \sqrt{(15x})={{\left( 15x \right)}^{\displaystyle \frac{1}{2}}}$
4. $\scriptsize \sqrt[4]{{{a}^{2}}+{{b}^{2}}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{\displaystyle \frac{1}{4}}}$ Use brackets again to indicate that the entire sum makes up the radicand.
5. $\scriptsize \sqrt[3]{{{x}^{2}}}$. Here the radicand has an exponent. In fact, all the radicands in the previous examples have had exponents of $\scriptsize 1$. We have just not written them down.
You know that $\scriptsize {{x}^{\displaystyle \frac{1}{n}}}=\sqrt[n]{{{x}^{1}}}$. This shows us that the exponent of the radicand ‘$\scriptsize 1$’ is the same as the numerator of the fractional exponent, and the index of the root is the same as the denominator.
So, in $\scriptsize \sqrt[3]{{{x}^{2}}}$ the exponent of the radicand must be the same as the numerator of the fractional exponent and the index or root is the same as the denominator.
$\scriptsize \sqrt[3]{{{x}^{2}}}={{x}^{\displaystyle \frac{2}{3}}}$

### Take note!

Example 4.2 shows that the number in the numerator of the fractional exponent becomes the exponent of the radicand and the number in the denominator of the fractional exponent becomes the root.

We can also prove the above by using the exponent law for raising a power to a power. $\scriptsize {{a}^{\displaystyle \frac{m}{n}}}={{a}^{(m\times \displaystyle \frac{1}{n})}}={{\left( {{a}^{m}} \right)}^{\displaystyle \frac{1}{n}}}=\sqrt[n]{{{a}^{m}}}$.

Here is another example for you to work through.

### Example 3

Find the value of $\scriptsize {{4}^{\displaystyle \frac{3}{2}}}$ by converting to radical form. Without any further working out, find the value of $\scriptsize {{4}^{-\displaystyle \frac{3}{2}}}$.

Solution

$\scriptsize {{4}^{\displaystyle \frac{3}{2}}}$ in radical form is $\scriptsize \sqrt{{{4}^{3}}}=\sqrt{4\times 4\times 4}=\sqrt{64}=8$.

$\scriptsize {{4}^{-\displaystyle \frac{3}{2}}}=\displaystyle \frac{1}{8}$

This is because you can rewrite the negative exponent as a positive exponent $\scriptsize {{4}^{-\displaystyle \frac{3}{2}}}=\displaystyle \frac{1}{{{4}^{\displaystyle \frac{3}{2}}}}$.

## Irrational numbers

Some radicals or roots can be written as rational numbers (positive or negative whole numbers or fractions), for example $\scriptsize \sqrt{64}=8$ or $\scriptsize \sqrt[3]{\displaystyle \frac{8}{27}}=\displaystyle \frac{2}{3}$.

But what about $\scriptsize \sqrt[3]{\displaystyle \frac{15}{45}}$ or $\scriptsize \sqrt{55}$ ?

$\scriptsize \sqrt[3]{\displaystyle \frac{15}{45}}=\sqrt[3]{\displaystyle \frac{1}{3}}$. There is no ‘nice’ answer to this cube root.

If you use your calculator to work out the answer, you will get a non-repeating decimal value otherwise known as an irrational number.

Similarly, $\scriptsize \sqrt{55}$ can only be estimated using a calculator.

Some radicals or roots cannot be written as rational numbers and it is best to leave them in radical form, for example $\scriptsize \sqrt{55}$ or $\scriptsize \sqrt[3]{\displaystyle \frac{1}{3}}$.

A calculator can only work out a rough approximation of their value. We call these irrational roots, surds.

### Note

You will learn more about surds in an upcoming unit but if you would like to learn about the basics of surds now, you can watch the video called “What are surds?”.

What are surds? (Duration: 4.20)

### Exercise 4.1

Simplify:

1. $\scriptsize 2{{a}^{\displaystyle \frac{1}{3}}}\times 3{{a}^{-\displaystyle \frac{1}{3}}}$
2. $\scriptsize {{(0.027)}^{\displaystyle \frac{1}{3}}}$
3. $\scriptsize {{(27)}^{-\displaystyle \frac{1}{3}}}$
4. $\scriptsize {{({{(-2)}^{2}}{{a}^{4}}{{b}^{-6}})}^{\displaystyle \frac{1}{2}}}$
5. $\scriptsize {{(5{{x}^{2}})}^{\displaystyle \frac{1}{2}}}\times {{(5{{x}^{4}})}^{\displaystyle \frac{1}{2}}}$
6. $\scriptsize \sqrt[3]{{{x}^{2}}y}\times {{x}^{\displaystyle \frac{1}{3}}}{{y}^{\displaystyle \frac{2}{3}}}$
7. $\scriptsize 6{{({{a}^{6}}{{b}^{12}})}^{\displaystyle \frac{1}{3}}}\times {{(64{{a}^{4}}{{b}^{8}})}^{\displaystyle \frac{1}{2}}}$

The full solutions are at the end of the unit.

### Note

If you would like more practise working with rational exponents, then click on this link.

Here is a useful list of the exponential laws and examples of them.

 LAW EXAMPLE 1. $\scriptsize {{a}^{m}}\cdot {{a}^{n}}={{a}^{m+n}}$ $\scriptsize {{x}^{6}}\cdot {{x}^{2}}={{x}^{6+2}}={{x}^{8}}$ 2. $\scriptsize \displaystyle \frac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ $\scriptsize \displaystyle \frac{{{x}^{6}}}{{{x}^{2}}}={{x}^{6-2}}={{x}^{4}}$ 3. $\scriptsize {{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ $\scriptsize {{\left( {{x}^{6}} \right)}^{5}}={{x}^{6\times 5}}={{x}^{30}}$ 4. $\scriptsize {{\left( {{a}^{m}}\cdot {{b}^{n}} \right)}^{p}}={{a}^{mp}}\cdot {{b}^{np}}$ $\scriptsize {{\left( {{x}^{2}}y \right)}^{7}}={{x}^{14}}\cdot {{y}^{7}}$ 5. $\scriptsize {{\left( \displaystyle \frac{a}{b} \right)}^{m}}=\displaystyle \frac{{{a}^{m}}}{{{b}^{m}}}$ $\scriptsize {{\left( \displaystyle \frac{x}{y} \right)}^{4}}=\displaystyle \frac{{{x}^{4}}}{{{y}^{4}}}$ 6. $\scriptsize {{a}^{\displaystyle \frac{m}{n}}}=\sqrt[n]{{{a}^{m}}}$ $\scriptsize {{x}^{\displaystyle \frac{2}{3}}}=\sqrt[3]{{{x}^{2}}}$

## Summary

In this unit you have learnt the following:

• The definition of a fractional exponent.
• How to convert from a fractional exponent to root form, also called radical form.
• How to convert from radical form to a fractional exponent.
• By now you should be able to understand and apply all of the following exponent rules:

$\scriptsize {{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$

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

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

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

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

$\scriptsize {{x}^{\displaystyle \frac{1}{n}}}=\sqrt[n]{x}\text{ }$

# Unit 4: Assessment

#### Suggested time to complete: 20 minutes

Simplify:

1. $\scriptsize \displaystyle \frac{{{3}^{2x}}}{9}$
2. $\scriptsize \displaystyle \frac{{{(-1)}^{4}}}{{{(-2)}^{-3}}}$
3. $\scriptsize {{m}^{-2t}}\times {{(3{{m}^{t}})}^{3}}$
4. $\scriptsize {{\left( {{({{a}^{36}})}^{\displaystyle \frac{1}{2}}} \right)}^{\displaystyle \frac{1}{3}}}$
5. $\scriptsize {{({{3}^{-1}}+{{2}^{-1}})}^{\displaystyle \frac{1}{2}}}$ Write the answer in radical form.
6. $\scriptsize 12{{({{a}^{10}}{{b}^{20}})}^{\displaystyle \frac{1}{5}}}\times {{(125{{a}^{12}}{{b}^{15}})}^{\displaystyle \frac{1}{3}}}$
7. $\scriptsize \sqrt[3]{\displaystyle \frac{27}{1000}}$
8. $\scriptsize \sqrt[8]{\sqrt[4]{{{a}^{32}}{{b}^{64}}}}$

The full solutions are at the end of the unit.

# Unit 4: Solutions

### Exercise 4.1

1. $\scriptsize 2{{a}^{\displaystyle \frac{1}{3}}}\times 3{{a}^{-\displaystyle \frac{1}{3}}}=6{{a}^{0}}=6$
2. .
\scriptsize \begin{align} & {{(0.027)}^{\displaystyle \frac{1}{3}}}\\ & 0.027=\displaystyle \frac{27}{1000} & & \text{Convert the base from a decimal to a fraction first.}\\ & \therefore {{(0.027)}^{\displaystyle \frac{1}{3}}}={{\left( \displaystyle \frac{27}{1000} \right)}^{\displaystyle \frac{1}{3}}} \\ & ={{\left( \displaystyle \frac{{{3}^{3}}}{{{10}^{3}}} \right)}^{\displaystyle \frac{1}{3}}} \\ & =\displaystyle \frac{3}{10}=0.3 \\ \end{align}
3. .
\scriptsize \begin{align} & {{(27)}^{\displaystyle \frac{-1}{3}}} \\ & ={{({{3}^{3}})}^{\displaystyle \frac{-1}{3}}} \\ & ={{3}^{-1}} \\ & =\displaystyle \frac{1}{3} \\ \end{align}
4. .
\scriptsize \begin{align} & {{({{(-2)}^{2}}{{a}^{4}}{{b}^{-6}})}^{\displaystyle \frac{1}{2}}} \\ & ={{(-2)}^{2\times \displaystyle \frac{1}{2}}}{{a}^{4\times \displaystyle \frac{1}{2}}}{{b}^{-6\times \displaystyle \frac{1}{2}}} \\ & =\displaystyle \frac{-2{{a}^{2}}}{{{b}^{3}}} \\ \end{align}
5. .
\scriptsize \begin{align} & {{(5{{x}^{2}})}^{\displaystyle \frac{1}{2}}}\times {{(5{{x}^{4}})}^{\displaystyle \frac{1}{2}}} \\ & ={{5}^{\displaystyle \frac{1}{2}}}x\times {{5}^{\displaystyle \frac{1}{2}}}{{x}^{2}} \\ & =5{{x}^{3}} \\ \end{align}
6. .
\scriptsize \begin{align} & \sqrt[3]{{{x}^{2}}y}\times {{x}^{\displaystyle \frac{1}{3}}}{{y}^{\displaystyle \frac{2}{3}}}\\ & ={{({{x}^{2}}y)}^{\displaystyle \frac{1}{3}}}\times {{x}^{\displaystyle \frac{1}{3}}}{{y}^{\displaystyle \frac{2}{3}}} & & \text{Rewrite without the cube root in exponential form.}\\ & ={{x}^{^{\displaystyle \frac{2}{3}}}}{{y}^{\displaystyle \frac{1}{3}}}\times {{x}^{\displaystyle \frac{1}{3}}}{{y}^{\displaystyle \frac{2}{3}}} \\ & =xy \\ \end{align}
7. .
\scriptsize \begin{align} & 6{{({{a}^{6}}{{b}^{12}})}^{\displaystyle \frac{1}{3}}}\times {{(64{{a}^{4}}{{b}^{8}})}^{\displaystyle \frac{1}{2}}} \\ & =6{{({{a}^{6}})}^{\displaystyle \frac{1}{3}}}{{({{b}^{12}})}^{\displaystyle \frac{1}{3}}}\times {{({{2}^{6}})}^{^{\displaystyle \frac{1}{2}}}}{{({{a}^{4}})}^{\displaystyle \frac{1}{2}}}{{({{b}^{8}})}^{\displaystyle \frac{1}{2}}} \\ & =6{{a}^{2}}{{b}^{4}}\times {{2}^{3}}{{a}^{2}}{{b}^{4}} \\ & =48{{a}^{4}}{{b}^{8}} \\ \end{align}

Back to Exercise 4.1

### Unit 4: Assessment

1. $\scriptsize \displaystyle \frac{{{3}^{2x}}}{9}=\displaystyle \frac{{{3}^{2x}}}{{{3}^{2}}}={{3}^{2x-2}}$
2. .
\scriptsize \begin{align} & \displaystyle \frac{{{(-1)}^{4}}}{{{(-2)}^{-3}}}={{(-1)}^{4}}{{(-2)}^{3}} \\ & =(1)(-8) \\ & =-8 \\ \end{align}
3. .
\scriptsize \begin{align} & {{m}^{-2t}}\times {{(3{{m}^{t}})}^{3}} \\ & ={{m}^{-2t}}\times {{3}^{3}}{{m}^{3t}} \\ & =27{{m}^{t}} \\ \end{align}
4. .
\scriptsize \begin{align} & {{\left( {{({{a}^{36}})}^{\displaystyle \frac{1}{2}}} \right)}^{\displaystyle \frac{1}{3}}} \\ & ={{\left( {{a}^{18}} \right)}^{\displaystyle \frac{1}{3}}} \\ & ={{a}^{6}} \\ \end{align}
5. .
\scriptsize \begin{align} & {{({{3}^{-1}}+{{2}^{-1}})}^{\displaystyle \frac{1}{2}}} \\ & ={{(\displaystyle \frac{1}{3}+\displaystyle \frac{1}{2})}^{\displaystyle \frac{1}{2}}} \\ & ={{(\displaystyle \frac{1\times 2+1\times 3}{6})}^{^{\displaystyle \frac{1}{2}}}} \\ & ={{(\displaystyle \frac{5}{6})}^{^{\displaystyle \frac{1}{2}}}} \\ & =\sqrt{\displaystyle \frac{5}{6}} \\ \end{align}
6. .
\scriptsize \begin{align} & 12{{({{a}^{10}}{{b}^{20}})}^{\displaystyle \frac{1}{5}}}\times {{(125{{a}^{12}}{{b}^{15}})}^{\displaystyle \frac{1}{3}}} \\ & =12{{({{a}^{10}})}^{^{\displaystyle \frac{1}{5}}}}{{({{b}^{20}})}^{\displaystyle \frac{1}{5}}}\times {{(125)}^{^{\displaystyle \frac{1}{3}}}}{{({{a}^{12}})}^{^{\displaystyle \frac{1}{3}}}}{{({{b}^{15}})}^{\displaystyle \frac{1}{3}}} \\ & =12{{a}^{2}}{{b}^{4}}\times 5{{a}^{4}}{{b}^{5}} \\ & =60{{a}^{6}}{{b}^{9}} \\ \end{align}
7. .
\scriptsize \begin{align} & \sqrt[3]{\displaystyle \frac{27}{1000}} \\ & =\sqrt[3]{\displaystyle \frac{{{3}^{3}}}{{{10}^{3}}}} & & \text{Rewrite the fraction in exponential form.} \\ & =\displaystyle \frac{{{({{3}^{3}})}^{\displaystyle \frac{1}{3}}}}{{{({{10}^{3}})}^{\displaystyle \frac{1}{3}}}} & & \text{Convert from root to exponential form.}\\ & =\displaystyle \frac{3}{10} \\ \end{align}
8. .
\scriptsize \begin{align} & \sqrt[8]{\sqrt[4]{{{a}^{32}}{{b}^{64}}}} \\ & =\sqrt[8]{{{({{a}^{32}})}^{\displaystyle \frac{1}{4}}}{{({{b}^{64}})}^{\displaystyle \frac{1}{4}}}} \\ & =\sqrt[8]{{{(a)}^{\displaystyle \frac{32}{4}}}{{(b)}^{\displaystyle \frac{64}{4}}}} \\ & =\sqrt[8]{{{a}^{8}}{{b}^{16}}} \\ & ={{a}^{\displaystyle \frac{8}{8}}}{{b}^{\displaystyle \frac{16}{8}}} \\ & =a{{b}^{2}} \\ \end{align}

Back to Unit 4: Assessment