# Index Notation

Here we will learn about simple index notation including how to write an expression using index notation and how to simplify expressions written in index form.

There are also index notation worksheets based on Edexcel, AQA and OCR GCSE exam style questions, along with further guidance on where to go next if you’re still stuck.

## What is index notation?

Index notation is a way of representing numbers (constants) and variables (e.g. x and y) that have been multiplied by themselves a number of times.

Below are all examples of expressions involving index notations:

$3^{4} \quad \quad \quad a^{5} \quad \quad \quad 2 x^{7} \quad \quad \quad \frac{1}{2}^{2 x} \quad \quad \quad \left(4 y^{2} x^{4}\right)^{7} \quad \quad \quad z^{-\frac{5}{2}}$

We use index notations, or the plural ‘indices’, to simplify expressions or solve equations involving powers.

E.g.

Simplify 6 × 6 × 6 × 6

6 is being multiplied by itself 4 times.

We can therefore write this as 64

This is said as “6 to the power of 4

E.g

Simplify  x × x × x × x

x is being multiplied by itself 4 times.

We can therefore write this as x 4

This is said as “x to the power of 4

E.g

Simplify 2y × 2y × 2y × 2y

In this question the term 2y is being multiplied by itself 4 times.

Therefore you can write this as 2y4

This is said as “2y to the power of 4

### What is Index Notation ### Key vocabulary for index notation

Term:

A single number (constant) or variable

E.g.
In the expression 4x − 7 both 4x and −7 are terms.

Coefficients:

The number which the variable is being multiplied by

E.g.
In 2x3 the coefficient is 2.

Integer:

A whole number

E.g.
1, 7 and 1003.

Index (also called exponent or powers):

The index number is the amount of times you multiply a number/variable by itself. It is normally shown about the base number

E.g.
The index number in 54 is 4.

Note: the plural of index is indices

Note: you will see index number as a superscript

Base Number:

The number/unknown that is being multiplied by itself an amount of times

E.g.
The base number in 54 is 5 and in 2x3 the base number is x.

Square Numbers:

A number/variable that is ‘squared’ is multiplied by itself

E.g.
4 × 4 can be written as 42 and is spoken as 4 squared” or 4 to the power of 2.

A square number is found when we multiply an integer (whole number) by itself.

E.g.

Cube Number:

A number/variable that is ‘cubed’ is multiplied by itself three times.

E.g.
4 × 4 × 4 can be written as 43 and is spoken as 4 cubed”.

A cube number is found when we multiply an integer (whole number) by itself three times.

E.g.

## How to simplify expressions involving index notation

1. Identify whether the base numbers for each term are the same
In higher tier questions you may need to manipulate the base numbers first
2. Identify the operation/s being undertaken between the terms
3. Follow the rules of index notation to simplifying the expression

### Explain how to simplify expressions involving index notation ## Related lessons on laws of indices

Index notation is part of our series of lessons to support revision on laws of indices. You may find it helpful to start with the main laws of indices lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

## Simplifying index notations examples

### Example 1: finding the value of an expression involving index notation and multiplication

Simplify 32 × 33

1. Identify whether the base numbers for each term are the same

The base number is 3 and is the same in each term.

2 Identify the operation/s being undertaken between the terms

The terms are being multiplied.

3Follow the rules of index notation to simplify each expression

32 is equivalent to 3 × 3

33 is equivalent to 3 × 3 × 3

Therefore:

$\begin{array}{l} 3^{2} \times 3^{3} \\ =3 \times 3 \times 3 \times 3 \times 3 \\ =3^{5} \\ \end{array}$

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

You will notice the index number 5 is the same as the two original indices added together, this is because they share the same base number.
This is a topic we will explore further in lessons on laws of indices.

Step-by-step guide: Multiplying indices

### Example 2: simplifying an expression involving unknowns and multiplication

Simplify 3x2 × 4x3

The base number is x and is the same in each term.

Note: the 3 and 4 are the coefficients of the terms.

The terms are being multiplied.

3x2 is equivalent to 3 × x × x

4x3 is equivalent to 4 × x × x × x

Therefore:

\begin{array}{l} 3 x^{2} \times 4 x^{3} \\ =3 \times x \times x \times 4 \times x \times x \times x \end{array}

We can write this as:

$3 \times 4 \times x \times x \times x \times x \times x$

Therefore the expression can be simplified to:

\begin{array}{l} 12 \times x^{5} \\ =12 x^{5} \end{array}

$3 x^{2} \times 4 x^{3}=12 x^{5}$

Again here you will notice the index number 5 is the same as the two original indices added together, this is because they share the same base number.

### Example 3: finding the value of an expression involving index notation and division

Simplify 65 ÷ 62

The base number is 6 and is the same in each term.

The terms are being divided.

65 is equivalent to 6 × 6 × 6 × 6 × 6

62 is equivalent to 6 × 6

You can write 65 ÷ 62 in the form

$\frac{6^{5}}{6^{2}}$

Therefore:

\begin{array}{l} 6^{5} \div 6^{2} \\\\ =\frac{6^{5}}{6^{2}} \\\\ =\frac{6 \times 6 \times 6 \times 6 \times 6 \times 6}{6 \times 6} \\\\ =6 \times 6 \times 6 \\\\ =6^{3}\\\\ \end{array}
$6^{5} \div 6^{2}=6^{3}$

You will notice the index number 3 is the same as the two original indices subtracted from one another, this is because they share the same base number.
This is a topic we will explore further in lessons on laws of indices.

Step-by-step guide: Dividing indices

### Example 4: simplifying an expression involving unknowns and division

Simplify 10y5 ÷ 5y2

The base number is y and is the same in each term.

The terms are being divided.

10y5 is equivalent to 10 × y5

5y2 is equivalent to 5 × y2

Therefore:

$10 y^{5} \div 5 y^{2}=$
$\frac{10 \times y^{5}}{5 \times y^{2}}$

You can simplify the coefficient as 10 ÷ 5 = 2

You can simplify the

$\frac{y^{5}}{y^{2}} \quad \text{to} \quad y^{2} \quad \quad$

(see previous example)

Therefore:

\begin{array}{l} \frac{10 \times y^{5}}{5 \times y^{2}} \\\\ =2 y^{3} \\\\ \end{array}
$10 y^{5} \div 5 y^{2}=2 y^{3}$

You will notice the index number 3 is the same as the two original indices subtracted from one another.
This is a topic we will explore further in lessons on laws of indices.

### Example 5: finding the value of an expression involving index notation and brackets

Simplify

$\left(3^{2}\right)^{3}$

The base number is 3 for the term inside the bracket.

The term 32 is “to the power of 3”.

$\left(3^{2}\right)^{3}$

is equivalent to 32 × 32 × 32 because the term 32 is being cubed.

Therefore:

(see example 1 for extended explanation)

$\left(3^{2}\right)^{3}= 3^{6}$

You will notice the index number 6 is the same as the two original indices multiplied together.
This is a topic we will explore further in lessons on laws of indices.

Step-by-step guide: Brackets with indices

### Example 6: simplifying an expression involving unknowns and brackets

Simplify

$\left(2 z^{3}\right)^{5}$

The base number inside the brackets is z.

The term 2z3 is “to the power of 5”.

$\left(2 z^{3}\right)^{5}$

is equivalent to 2z3 × 2z3 × 2z3 × 2z3 × 2z3

Therefore:

\begin{array}{l} \left(2 z^{3}\right)^{5} \\\\ =2 z^{3} \times 2 z^{3} \times 2 z^{3} \times 2 z^{3} \times 2 z^{3} \\\\ =32 x^{15} \end{array}

You will notice the index number 15 is the same as the two original indices multiplied together whilst 2 has also been multiplied by itself 5 times.

This is a topic we will explore further in lessons on laws of indices.

Step-by-step guide: Laws of indices

### Negative indices, fractional indices, and power of zero

Below are three further examples involving negative indices, fractional indices and the power of zero.

### Example 7: a variable to the power of zero

If we continue the above pattern below we can see that 20 must be equal to 1.

E.g.
Base Number 2

\begin{array}{l} 2^{4}=2 \times 2 \times 2 \times 2=16 \\ 2^{3}=2 \times 2 \times 2=8 \\ 2^{2}=2 \times 2=4 \\ 2^{1}=2=2 \\ 2^{0}=1 \end{array}

E.g.
Base Number 3

\begin{array}{l} 3^{4}=3 \times 3 \times 3 \times 3=81 \\ 3^{3}=3 \times 3 \times 3=27 \\ 3^{2}=3 \times 3=9 \\ 3^{1}=3=3 \\ 3^{0}=1 \end{array}

This pattern is true for any base number therefore a0 is equal to 1.

a0 = 1

Step-by-step guide: Power of 0

### Example 8: a variable to a negative power

We can continue the pattern shown in example 1 for find out a-1

Remember each time we reduce the power by one we are dividing the value by the base number.

E.g.
Base Number 2

\begin{array}{l} 2^{2}=2 \times 2=4 \\ 2^{1}=2=2 \\ 2^{0}=1 \\ 2^{-1}=\frac{1}{2} \end{array}

E.g.
Base Number 3

\begin{array}{l} 3^{2}=3 \times 3=9 \\ 3^{1}=3=3 \\ 3^{0}=1 \\ 3^{-1}=\frac{1}{3} \end{array}

Because this pattern is true for any base number, a-1 is equal to one divided by the base number.

$a^{-1}=\frac{1}{a}$

Step-by-step guide: Negative indices

### Example 9: a variable to a fractional power

Look at the example below, see if you can spot the pattern before moving onto the next one.

\begin{array}{ll} \left(4^{\frac{1}{2}}\right)^{2}=4^{\frac{1}{2}} \times 4^{\frac{1}{2}}=4^{\frac{1}{2}+\frac{1}{2}}=4^{1}=4 & \therefore\left(4^{\frac{1}{2}}\right)^{2}=4 \\ \left(3^{\frac{1}{2}}\right)^{2}=3^{\frac{1}{2}} \times 3^{\frac{1}{2}}=3^{\frac{1}{2}+\frac{1}{2}}=3^{1}=3 & \therefore\left(3^{\frac{1}{2}}\right)^{2}=3 \\ \left(2^{\frac{1}{2}}\right)^{2}=2^{\frac{1}{2}} \times 2^{\frac{1}{2}}=2^{\frac{1}{2}+\frac{1}{2}}=2^{1}=2 & \therefore\left(2^{\frac{1}{2}}\right)^{2}=2 \end{array}

When you multiply a number by itself it is called ‘squaring a number’.
The inverse of this is called square rooting.

Therefore from the above examples:

\begin{array}{l} \sqrt{4}=4^{\frac{1}{2}} \\ \sqrt{3}=3^{\frac{1}{2}} \\ \sqrt{2}=2^{\frac{1}{2}} \end{array}

Because this pattern is true for any base number, a½ is equal to the square root of a.

$a^{\frac{1}{2}}=\sqrt{a}$

Step-by-step guide: Fractional indices

### Common misconceptions

• Base Numbers

A common error is to incorrectly simplify expressions using the index rules where they do not share a base number.

• Index power of 1

If no index power is shown you need to remember it is being multiplied by itself ‘one time’ e.g 2 = 21. This helps with index notation questions.

• Operations

A common error is to attempt to simplify an expression using the incorrect operation between the base numbers.

• Coefficients and brackets

A common error is to not include the coefficients when expanding brackets.

E.g

$(2 x)^{3}=2 x^{3}$

This should be

$(2x)^{3}=2x\times 2x\times 2x=8x^{3}$

• Fractions and decimals

We can write fractions and decimals using index notation.

E.g.

\begin{array}{l} 0.5^{3}=0.5 \times 0.5 \times 0.5=0.125=\frac{1}{8} \\\\ (\frac{1}{2})^{3}=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8} \end{array}

### Practice index notation questions

1. Write 8 \times 8 \times 8 \times 8 using index notation

8^{4} 32 32^{4} 4096^{4} We are multiplying 8 by itself 4 times which can be written as 8^{4} .

2. Write 3 a \times 3 a \times 3 a \times 3 a using index notation

3a^{4} (3a)^{4} 12a 12a^{4} We are multiplying 3a by itself 4 times which can be written as (3a)^{4} .

3. Simplify  5 x^{5} \times 3 x^{6}

8x^{11} 15x^{30} 15x^{11} 8x^{30} \begin{aligned} 5x^{5} \times 3x^{6} &= 5 \times x \times x \times x \times x \times x \times 3 \times x \times x \times x \times x \times x \times x\\\\ &=15x^{11} \end{aligned}

4. Simplify  z^{5} \div z^{3}

z^{\frac{5}{3}} z^{-2} 2 z^{2} \begin{aligned} z^{5} \div z^{3} &= \frac{z \times z \times z \times z \times z}{z \times z \times z}\\\\ &=z^{2} \end{aligned}

5. Simplify  (3c^{4})^{2}

3c^{8} 6c^{8} 9c^{8} 6c^{6} \begin{aligned} (3c^{4})^{2} &= 3c^{4} \times 3c^{4}\\\\ &=3 \times c \times c \times c \times c \times 3 \times c \times c \times c \times c\\\\ &=9c^{8} \end{aligned}

6. Write g^{-1} in another form

0 \frac{1}{g} \sqrt{g} -g Raising a term to the power of negative 1 is the same as writing 1 over that term. g^{-1}=\frac{1}{g}

7. Find the positive value of  64^{\frac{1}{2}}

8 32 16 \frac{1}{64} The power of \frac{1}{2} means the square root

\sqrt{64}=8

### Index notationGCSE questions

1. Work out the value of:

a) 3^{3}

b) 3 \times 2^{3}

(2 Marks)

a) 3 \times 3 \times 3

= 27

(1)

b) 3 \times 8

= 24

(1)

2. Simplify fully:

a) x^{3} \times x^{3}

b) (2 x)^{4}

(2 Marks)

a) x^{6}

(1)

b) 16 x^{4}

(1)

3. Simplify fully:

a) 3 x^{6} \times 2 x^{4} \times x

b) 12 x^{5} \div 3 x^{2}

c) \left(3 x^{2}\right)^{4}

d) 4^{0}

(7 Marks)

a) 6x^{11}

Correct coefficient

(1)

Correct index number for x

(1)

b) 4x^{3}

Correct coefficient

(1)

Correct index number for x

(1)

c) 81x^{8}

Correct coefficient

(1)

Correct index number for x

(1)

d) 1

(1)

## Learning checklist

You have now learned how to:

• Calculate with roots, and with integer indices
• Simplifying expressions involving sums, products and powers
• Simplify and manipulate algebraic expressions involving indices
• Simplify and manipulate algebraic expressions involving simple negative and fractional indices

## Still stuck?

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Practice paper packs based on the advanced information for the Summer 2022 exam series from Edexcel, AQA and OCR.

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