Indices II

Syllabus coverage

Understand and use indices (positive, zero and negative)
Understand and use the rules of indices, e.g. simplify $(5 x^{3})^{2}$ , $12 a^{5} \div 3 a^{- 2}$ and $6 x^{7} y^{4} \times 5 x^{- 5} y$
Extended only: understand and use indices that are fractional as well as positive, zero and negative
Extended only: solve equations such as $32^{x} = 2$ and $5^{x + 1} = 25^{x}$ by writing both sides as powers of the same base

The index laws with an example beside each, and the expression (2x²y)³ worked as a flow of arrows.

Letters as bases: what an index is counting

In this topic the base is a letter rather than a number. $x^{4}$ means $x \times x \times x \times x$ , and the index 4 counts how many copies of $x$ are multiplied together. Nothing else changes when the base is a letter: the laws are the reason $x^{4} \times x^{3}$ can be written as $x^{7}$ without expanding anything.

The number in front of a letter is the coefficient, and it is not part of the index. In $6 x^{7} y^{4}$ the coefficient is 6, the base of the first power is $x$ with index 7, and the base of the second is $y$ with index 4. Coefficients are multiplied and divided in the ordinary way; only the indices follow the index laws.

The laws apply to powers of the same base and to nothing else. $x^{3} \times x^{4} = x^{7}$ , but $x^{3} \times y^{4}$ cannot be simplified at all, and neither can $x^{3} + x^{4}$ . Adding two powers is not multiplying them: $x^{3} + x^{4}$ stays as it is, because there is no law that collects it.

Law	In symbols	Example
Multiplying powers of the same base	$a^{m} \times a^{n} = a^{m + n}$	$x^{4} \times x^{3} = x^{7}$
Dividing powers of the same base	$a^{m} \div a^{n} = a^{m - n}$	$x^{7} \div x^{4} = x^{3}$
A power of a power	$(a^{m})^{n} = a^{m n}$	$(x^{3})^{2} = x^{6}$
A power of a product	$(a b)^{n} = a^{n} b^{n}$	$(2 x)^{3} = 8 x^{3}$
Zero index	$a^{0} = 1$	$x^{0} = 1$
Negative index	$a^{- n} = \frac{1}{a^{n}}$	$x^{- 2} = \frac{1}{x^{2}}$
Fractional index (Extended only)	$a^{\frac{1}{n}} = \sqrt[n]{a}$	$x^{\frac{1}{2}} = \sqrt{x}$

Every law in that table needs the same base.

x^{3} \times x^{4} = x^{7}

is a law;

x^{3} \times y^{4}

has no law, and it is already in its simplest form.

Multiplying and dividing powers of the same base

To multiply powers of the same base, add the indices: $a^{m} \times a^{n} = a^{m + n}$ . The reason is only counting, because $x^{4} \times x^{3}$ is four $x$ s multiplied by three more $x$ s, which is seven altogether. The syllabus example is $6 x^{7} y^{4} \times 5 x^{- 5} y$ . Deal with the numbers first, then each letter in turn: $6 \times 5 = 30$ ; for $x$ , $7 + (- 5) = 2$ ; for $y$ , $4 + 1 = 5$ . So the answer is $30 x^{2} y^{5}$ .

To divide powers of the same base, subtract the indices: $a^{m} \div a^{n} = a^{m - n}$ . The syllabus example is $12 a^{5} \div 3 a^{- 2}$ . The numbers divide: $12 \div 3 = 4$ . The indices subtract: $5 - (- 2) = 7$ , so the answer is $4 a^{7}$ . Subtracting a negative index adds it, which is why dividing by a negative power makes the index larger rather than smaller.

Write the answer with the coefficient first and the letters in alphabetical order, as in $30 x^{2} y^{5}$ . A letter with no index written has index 1, so the $y$ at the end of $6 x^{7} y^{4} \times 5 x^{- 5} y$ contributes $y^{4 + 1} = y^{5}$ . Forgetting that hidden 1 is the most common single slip in a multiplication question.

$6 x^{7} y^{4} \times 5 x^{- 5} y = 30 x^{2} y^{5}$
$12 a^{5} \div 3 a^{- 2} = 4 a^{7}$
$4 x^{3} y^{5} \times 3 x^{- 1} y^{2} = 12 x^{2} y^{7}$
$15 x^{8} \div 5 x^{3} = 3 x^{5}$

Divide the numbers and subtract the indices. Mixing the two up, by dividing the indices or subtracting the coefficients, is the commonest error in the whole topic.

Brackets: powers of powers and powers of products

When a power is raised to another power, multiply the indices: $(a^{m})^{n} = a^{m n}$ . The syllabus example is $(5 x^{3})^{2}$ . The 5 is squared as well, because the bracket multiplies everything inside it by itself: $5^{2} = 25$ and $3 \times 2 = 6$ , so $(5 x^{3})^{2} = 25 x^{6}$ .

The same rule handles a product inside the bracket. $(2 x^{2} y)^{3} = 2^{3} \times x^{2 \times 3} \times y^{1 \times 3} = 8 x^{6} y^{3}$ . Every factor in the bracket takes the outside power, the number included, and a letter with no index still has an index of 1 to multiply.

A negative sign inside the bracket disappears when the outside power is even and survives when it is odd: $(- 3 x^{2})^{2} = 9 x^{4}$ , but $(- 2 x)^{3} = - 8 x^{3}$ . Work the sign out separately from the arithmetic, because it is easy to lose it halfway through.

$(5 x^{3})^{2} = 25 x^{6}$
$(2 x^{2} y)^{3} = 8 x^{6} y^{3}$
$(- 3 x^{2})^{2} = 9 x^{4}$
$(x^{4})^{3} = x^{12}$

Everything inside the bracket is raised to the outside power, so the coefficient changes too.

(5 x^{3})^{2} = 25 x^{6}

, never

5 x^{6}

The zero index and negative indices

$a^{0} = 1$ for every non-zero base, letters included: $x^{0} = 1$ and $(3 x^{2})^{0} = 1$ . The division law explains it, because $x^{4} \div x^{4} = x^{4 - 4} = x^{0}$ , and anything divided by itself is 1.

A negative index means a reciprocal: $a^{- n} = \frac{1}{a^{n}}$ . So $x^{- 2} = \frac{1}{x^{2}}$ , and $3 x^{- 2} = \frac{3}{x^{2}}$ , because the coefficient 3 stays on top and only the power moves down. A negative index never makes an answer negative.

The laws work on negative indices in exactly the same way. $x^{- 3} \div x^{- 5} = x^{- 3 - (- 5)} = x^{2}$ , and $x^{- 2} \times x^{5} = x^{3}$ . A division can leave you with a negative index, so finish the job: $x^{3} \div x^{5} = x^{- 2} = \frac{1}{x^{2}}$ . Moving a factor across the fraction line just changes the sign of its index, which tidies an answer in one step: $\frac{4 x^{- 3}}{2 x^{- 5}} = 2 x^{- 3 + 5} = 2 x^{2}$ .

$x^{0} = 1$ , whatever the value of $x$
$x^{- 2} = \frac{1}{x^{2}}$ and $3 x^{- 2} = \frac{3}{x^{2}}$
$\frac{x^{- 3}}{x^{- 5}} = x^{- 3 + 5} = x^{2}$
$\frac{4 x^{- 3}}{2 x^{- 5}} = 2 x^{2}$

A negative index moves a power to the bottom of a fraction. If your final answer still has a negative index in it, it is not in its simplest form.

Fractional indices (Extended only)

A fractional index is a root in disguise: $a^{\frac{1}{n}} = \sqrt[n]{a}$ . So $x^{\frac{1}{2}} = \sqrt{x}$ , $x^{\frac{1}{3}} = \sqrt[3]{x}$ , and $8^{\frac{1}{3}} = 2$ . A half power is a square root, a third power is a cube root, and a quarter power is a fourth root.

When the numerator is not 1, the index does two jobs: the denominator is the root and the numerator is the power. $x^{\frac{3}{2}} = (\sqrt{x})^{3} = \sqrt{x^{3}}$ . Take the root first so the numbers stay small: $27^{\frac{2}{3}} = (\sqrt[3]{27})^{2} = 3^{2} = 9$ .

Fractional indices obey the same laws as whole-number ones, which is the point of writing them this way. $x^{\frac{1}{2}} \times x^{\frac{1}{2}} = x^{1} = x$ , and $x^{\frac{1}{2}} \times x^{\frac{3}{2}} = x^{\frac{1}{2} + \frac{3}{2}} = x^{2}$ . Add the fractions exactly as you would add any fractions, over a common denominator.

Fractional and negative indices combine. $x^{- \frac{1}{2}} = \frac{1}{\sqrt{x}}$ , and $9^{- \frac{1}{2}} = \frac{1}{3}$ . Find the root first, then take the reciprocal if the index is negative.

$x^{\frac{1}{2}} = \sqrt{x}$ , $x^{\frac{1}{3}} = \sqrt[3]{x}$
$27^{\frac{2}{3}} = (\sqrt[3]{27})^{2} = 9$
$x^{\frac{1}{2}} \times x^{\frac{3}{2}} = x^{2}$
$9^{- \frac{1}{2}} = \frac{1}{3}$

The denominator of a fractional index is a root, not a divisor.

27^{\frac{2}{3}}

is 9, not 18.

Index equations: making the bases the same

A question such as $2^{x} = 32$ is not asking for logarithms, and the syllabus says so. It is asking you to write both sides as powers of the same base: $32 = 2^{5}$ , so $x = 5$ .

The same idea solves an equation where the unknown sits in the index on both sides. For $5^{x + 1} = 25^{x}$ , write 25 as $5^{2}$ : $5^{x + 1} = 5^{2 x}$ . The bases now match, so the indices must be equal, and $x + 1 = 2 x$ gives $x = 1$ . Check it: $5^{1 + 1} = 25$ and $25^{1} = 25$ .

The answer can be a fraction, which is where the Extended work on fractional indices pays off. $9^{x} = 27$ becomes $3^{2 x} = 3^{3}$ , so $2 x = 3$ and $x = \frac{3}{2}$ . $32^{x} = 2$ becomes $2^{5 x} = 2^{1}$ , so $5 x = 1$ and $x = \frac{1}{5}$ .

Two habits keep these reliable. Write the base you are converting to on both sides before you equate the indices, and substitute your answer back: $2^{5 \times \frac{1}{5}} = 2^{1} = 2$ , as required.

$2^{x} = 32$ : $32 = 2^{5}$ , so $x = 5$
$5^{x + 1} = 25^{x}$ : $x + 1 = 2 x$ , so $x = 1$
$9^{x} = 27$ : $3^{2 x} = 3^{3}$ , so $x = \frac{3}{2}$
$32^{x} = 2$ : $2^{5 x} = 2^{1}$ , so $x = \frac{1}{5}$

Once the bases match, the indices are equal. That single step is the whole method, and it is usually worth a mark on its own.

Key terms you must be able to define

Base: The quantity being raised to a power. In $x^{7}$ the base is $x$ ; in $6 x^{7} y^{4}$ there are two bases, $x$ and $y$ .

Index (power): The number of copies of the base that are multiplied together. In $x^{7}$ the index is 7.

Coefficient: The number multiplying the letter, which is not affected by the index laws. In $6 x^{7}$ the coefficient is 6.

Index law: A rule for combining powers of the same base, such as $a^{m} \times a^{n} = a^{m + n}$ . The laws need matching bases.

Multiplication law: $a^{m} \times a^{n} = a^{m + n}$ . Multiply powers of the same base by adding the indices: $x^{4} \times x^{3} = x^{7}$ .

Division law: $a^{m} \div a^{n} = a^{m - n}$ . Divide powers of the same base by subtracting the indices: $x^{7} \div x^{4} = x^{3}$ .

Power of a power: $(a^{m})^{n} = a^{m n}$ . A bracket multiplies the indices: $(x^{3})^{2} = x^{6}$ .

Power of a product: $(a b)^{n} = a^{n} b^{n}$ . Every factor inside the bracket takes the outside power, so $(2 x)^{3} = 8 x^{3}$ .

Zero index: $a^{0} = 1$ for any non-zero base, including letters: $x^{0} = 1$ .

Negative index: $a^{- n} = \frac{1}{a^{n}}$ . It means a reciprocal, not a negative value: $x^{- 2} = \frac{1}{x^{2}}$ .

Reciprocal: 1 divided by the quantity. The reciprocal of $x^{2}$ is $\frac{1}{x^{2}}$ , which is why $x^{- 2} = \frac{1}{x^{2}}$ .

Fractional index: $a^{\frac{1}{n}} = \sqrt[n]{a}$ : the denominator is the root and the numerator is the power. Extended only.

Simplest form: An answer with the coefficients multiplied out, the letters in order, and no zero or negative indices left: $4 a^{7}$ , not $4 a^{- 2} \times a^{9}$ .

In context

Cambridge sets 0580 in real situations - a recipe, a bank statement, a map, a garden, a graph of a car's journey. These show how the mathematics appears in a question, then you write your own working for the numbers you are given.

Multiplying powers of letters

Simplify $6 x^{7} y^{4} \times 5 x^{- 5} y$ . The numbers multiply: $6 \times 5 = 30$ . Then each letter in turn: $x^{7 + (- 5)} = x^{2}$ and $y^{4 + 1} = y^{5}$ . The answer is $30 x^{2} y^{5}$ .

Dividing powers of letters

Simplify $12 a^{5} \div 3 a^{- 2}$ . Divide the numbers: $12 \div 3 = 4$ . Subtract the indices: $5 - (- 2) = 7$ . So $12 a^{5} \div 3 a^{- 2} = 4 a^{7}$ . The minus sign in the index has made the answer's index larger, not smaller.

A bracket

Simplify $(5 x^{3})^{2}$ and $(2 x^{2} y)^{3}$ . In the first, everything inside is squared: $5^{2} = 25$ and $3 \times 2 = 6$ , giving $25 x^{6}$ . In the second, cube each factor: $2^{3} = 8$ , $x^{2 \times 3} = x^{6}$ and $y^{1 \times 3} = y^{3}$ , giving $8 x^{6} y^{3}$ .

Zero, negative and fractional indices

Write $3 x^{- 2}$ without a negative index, simplify $\frac{4 x^{- 3}}{2 x^{- 5}}$ , and work out $27^{\frac{2}{3}}$ . $3 x^{- 2} = \frac{3}{x^{2}}$ ; $\frac{4 x^{- 3}}{2 x^{- 5}} = 2 x^{- 3 + 5} = 2 x^{2}$ ; $27^{\frac{2}{3}} = (\sqrt[3]{27})^{2} = 3^{2} = 9$ .

Flashcards

Click a card to flip it. Mark honestly — the tracker uses this.

Q1Simplify

x^{4} \times x^{3}

x^{7}

. The bases match, so add the indices: 4 + 3 = 7.

Q2Simplify

x^{7} \div x^{4}

x^{3}

. The bases match, so subtract the indices: 7 - 4 = 3.

Q3Simplify

(5 x^{3})^{2}

25 x^{6}

. Square the 5 as well as the letter, and multiply the indices: 3 x 2 = 6.

Q4Simplify

(2 x^{2} y)^{3}

8 x^{6} y^{3}

. Every factor in the bracket is cubed, the coefficient 2 included.

Q5What is

x^{0}

1, for any non-zero

x

. The division law shows why:

x^{4} \div x^{4} = x^{0}

, and a quantity divided by itself is 1.

Q6Write

3 x^{- 2}

without a negative index.

\frac{3}{x^{2}}

. The coefficient stays on top; only the power moves to the bottom.

Q7Simplify

\frac{x^{- 3}}{x^{- 5}}

x^{2}

. Subtract the indices:

- 3 - (- 5) = 2

Q8Simplify

4 x^{3} y^{5} \times 3 x^{- 1} y^{2}

12 x^{2} y^{7}

. Numbers:

4 \times 3 = 12

. Indices:

3 + (- 1) = 2

for

x

, and

5 + 2 = 7

for

y

Q9What does

x^{\frac{1}{2}}

mean?

\sqrt{x}

. A half power is a square root, because the denominator of the index is the root. Extended only.

Q10Work out

27^{\frac{2}{3}}

9. Cube root first:

\sqrt[3]{27} = 3

. Then the power:

3^{2} = 9

Q11Simplify

x^{\frac{1}{2}} \times x^{\frac{3}{2}}

x^{2}

, because

\frac{1}{2} + \frac{3}{2} = 2

. The index laws work on fractional indices too.

Q12Solve

5^{x + 1} = 25^{x}

x = 1

. Write

25 = 5^{2}

, so the indices give

x + 1 = 2 x

Q13What is the coefficient in

6 x^{7} y^{4}

, and what happens to it in

(6 x^{7} y^{4})^{2}

The coefficient is 6, and it is squared like everything else in the bracket:

36 x^{14} y^{8}

0 of 13 marked as known

Exam technique for this topic

Write the law you are using on the line above your answer, for instance $a^{m} \times a^{n} = a^{m + n}$ . The method mark is for the law, so it survives a slip in the arithmetic.
Keep the coefficients out of the index work. Multiply or divide the numbers first, then add or subtract the indices of each letter separately: two steps, and both are marked.
Never leave a zero or negative index in a final answer. $x^{- 2}$ is $\frac{1}{x^{2}}$ and $x^{0}$ is 1; writing $x^{- 2} = - x^{2}$ or $3 x^{0} = 3$ scores nothing.
For an index equation, show both sides as powers of the same base before you equate the indices. $9^{x} = 27$ needs the line $3^{2 x} = 3^{3}$ for the method mark, even when you can see the answer.
Extended only: for a fractional index take the root before the power, and add fractional indices over a common denominator. $x^{\frac{1}{2}} \times x^{\frac{1}{3}} = x^{\frac{5}{6}}$ , not $x^{\frac{1}{6}}$ .

Why students lose marks here

A student wrote

x^{3} \times x^{4} = x^{12}

Why it loses marksMultiplying powers of the same base adds the indices. Multiplying them is the rule for a power of a power,

(x^{3})^{4}

, which is a different question.

The answer that scores

x^{3} \times x^{4} = x^{3 + 4} = x^{7}

A student wrote

(5 x^{3})^{2} = 5 x^{6}

Why it loses marksThe outside power applies to everything inside the bracket, so the coefficient 5 is squared as well. Squaring only the letter gives an answer five times too small.

The answer that scores

(5 x^{3})^{2} = 5^{2} \times x^{3 \times 2} = 25 x^{6}

A student wrote

x^{- 2} = - x^{2}

Why it loses marksA negative index means a reciprocal, not a negative value. Nothing in the answer is negative: the power moves to the bottom of a fraction.

The answer that scores

x^{- 2} = \frac{1}{x^{2}}

, and

3 x^{- 2} = \frac{3}{x^{2}}

A student wrote

12 a^{5} \div 3 a^{- 2} = 4 a^{3}

Why it loses marksThe indices are subtracted, and subtracting a negative adds:

5 - (- 2) = 7

, not 3. Dividing by

a^{- 2}

is the same as multiplying by

a^{2}

, so the index grows.

The answer that scores

12 a^{5} \div 3 a^{- 2} = 4 a^{5 - (- 2)} = 4 a^{7}

Where this fits in IGCSE Maths

This page is one topic of Cambridge IGCSE Maths 0580. IGCSE Maths is the course home: every area and every topic in Cambridge's own order, the papers you sit and what each one tests, and the exam-practice tools. The full topic list, with each topic linked to its revision page, is on the IGCSE Maths syllabus page.

Syllabus coverage

Letters as bases: what an index is counting

Multiplying and dividing powers of the same base

Brackets: powers of powers and powers of products

The zero index and negative indices

Fractional indices (Extended only)

Index equations: making the bases the same

Key terms you must be able to define

In context

Flashcards

Exam-style questions

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Exam technique for this topic

Why students lose marks here

Where this fits in IGCSE Maths