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Cambridge IGCSE Mathematics 0580 · Algebra and graphs

Indices II

Using the rules of indices with letters: simplifying products, quotients and powers of terms such as $(5 x^{3})^{2}$ , working with zero and negative indices, and solving equations such as $2^{x} = 32$ by matching bases. Extended students also use fractional indices and solve equations such as $5^{x + 1} = 25^{x}$ .

C2.4E2.4Core and ExtendedFree lesson

What the syllabus says

C2.4 (Core)

1 Understand and use indices (positive, zero and negative).
2 Understand and use the rules of indices.

Notes and examples

e.g. $2^{x} = 32$ . Find the value of $x$ .
e.g. simplify:
$(5 x^{3})^{2}$
$12 a^{5} \div 3 a^{- 2}$
$6 x^{7} y^{4} \times 5 x^{- 5} y$ .
Knowledge of logarithms is not required.

E2.4 (Extended)

1 Understand and use indices (positive, zero, negative and fractional).
2 Understand and use the rules of indices.

Notes and examples

e.g. solve:
$32^{x} = 2$
$5^{x + 1} = 25^{x}$ .
e.g. simplify:
$3 x^{- 4} \times \frac{2}{3} x^{\frac{1}{2}}$
$\frac{2}{5} x^{\frac{1}{2}} \div 2 x^{- 2}$
$(\frac{2 x^{5}}{3})^{3}$ .
Knowledge of logarithms is not required.

Quoted from the Cambridge IGCSE Mathematics 0580 syllabus for 2025 to 2027.

Indices with letters

Everything you know about indices with numbers works with letters, because a letter is a number. $x^{4}$ means $x \times x \times x \times x$ , and the three rules still hold whenever the base is the same:

Multiplying: add the indices. $x^{m} \times x^{n} = x^{m + n}$
Dividing: subtract the indices. $x^{m} \div x^{n} = x^{m - n}$
A power of a power: multiply the indices. $(x^{m})^{n} = x^{m n}$

The zero and negative indices follow from rule 2. $x^{3} \div x^{3}$ is 1 (anything divided by itself), and by the rule it is $x^{0}$ . So $x^{0} = 1$ for any $x \neq 0$ . And $x^{2} \div x^{5} = \frac{x \times x}{x \times x \times x \times x \times x} = \frac{1}{x^{3}}$ , while the rule gives $x^{- 3}$ . So

$x^{- n} = \frac{1}{x^{n}}$

Notice where the index sits. In $3 x^{- 2}$ only the $x$ has the index, so $3 x^{- 2} = \frac{3}{x^{2}}$ . In $(3 x)^{- 2}$ the whole bracket has it, so $(3 x)^{- 2} = \frac{1}{9 x^{2}}$ .

Multiplying and dividing terms

A term such as $6 x^{7} y^{4}$ is a number multiplied by powers of letters. When you multiply or divide terms, deal with each part separately:

the numbers (coefficients) are multiplied or divided as ordinary numbers;
the indices of each letter are added (multiplying) or subtracted (dividing), one letter at a time.

A letter with no index written has index 1: $y = y^{1}$ . That is easy to forget when you add indices.

Never multiply the indices when you multiply terms, and never add the coefficients: $2 x^{3} \times 4 x^{2} = 8 x^{5}$ , not $6 x^{5}$ and not $8 x^{6}$ .

Worked example 1

Simplify (a) $6 x^{7} y^{4} \times 5 x^{- 5} y$ (b) $12 a^{5} \div 3 a^{- 2}$ .

(a) Numbers: $6 \times 5 = 30$ .
$x$ : $x^{7} \times x^{- 5} = x^{7 + (- 5)} = x^{2}$ .
$y$ : $y^{4} \times y^{1} = y^{5}$ . So the answer is $30 x^{2} y^{5}$ .
(b) Numbers: $12 \div 3 = 4$ .
$a$ : $a^{5} \div a^{- 2} = a^{5 - (- 2)} = a^{7}$ . Subtracting a negative index adds it.
So the answer is $4 a^{7}$ .

Answer: (a) $30 x^{2} y^{5}$ (b) $4 a^{7}$

Raising a term to a power

$(5 x^{3})^{2}$ means $5 x^{3} \times 5 x^{3}$ . Multiply the numbers, $5 \times 5 = 25$ , and add the indices, $x^{3 + 3} = x^{6}$ . So $(5 x^{3})^{2} = 25 x^{6}$ .

The quick way: every factor inside the bracket is raised to the power. The number is raised to the power as a number ( $5^{2} = 25$ ), and each index is multiplied by the power ( $x^{3 \times 2} = x^{6}$ ).

The two classic wrong answers are $5 x^{6}$ (forgetting to square the 5) and $10 x^{6}$ (doubling the 5 instead of squaring it). The same goes for negative powers: $(2 x)^{- 3} = 2^{- 3} x^{- 3} = \frac{1}{8 x^{3}}$ .

Worked example 2

Simplify $\frac{(2 a^{4} b)^{3}}{4 a^{5} b^{2}}$ .

Top first: $(2 a^{4} b)^{3} = 2^{3} \times a^{4 \times 3} \times b^{1 \times 3} = 8 a^{12} b^{3}$ .
Divide the numbers: $8 \div 4 = 2$ .
$a$ : $a^{12} \div a^{5} = a^{7}$ . $b$ : $b^{3} \div b^{2} = b^{1} = b$ .
Answer: $2 a^{7} b$ .

Answer: $2 a^{7} b$

Check yourself

Simplify $(4 y^{5})^{2}$ .

Answer it in the portal, where your answer is checked as you type →

Solving equations with indices

In an equation such as $2^{x} = 32$ the unknown is the index. Logarithms are not needed (and are not on the syllabus). Instead, write both sides as powers of the same base, then the indices must be equal:

$2^{x} = 32 = 2^{5}, so x = 5$

This works because powers of 2 are all different: $2^{4} = 16$ and $2^{6} = 64$ , so only $x = 5$ gives 32.

Fractions and decimals give negative indices. $3^{x} = \frac{1}{9}$ : since $9 = 3^{2}$ , $\frac{1}{9} = 3^{- 2}$ and $x = - 2$ . In the same way $10^{x} = 0.001 = \frac{1}{1000} = 10^{- 3}$ , so $x = - 3$ . And $7^{x} = 1$ means $x = 0$ .

It helps to know these powers by heart: the powers of 2 up to $2^{10} = 1024$ , of 3 up to $3^{5} = 243$ , of 5 up to $5^{4} = 625$ , and the powers of 10.

Worked example 3

Find the value of $x$ in each equation. (a) $2^{x} = 32$ (b) $5^{x} = \frac{1}{125}$ (c) $4^{x} \times 4^{3} = 4^{10}$

(a) $32 = 2^{5}$ , so $x = 5$ .
(b) $125 = 5^{3}$ , so $\frac{1}{125} = 5^{- 3}$ and $x = - 3$ .
(c) The left side is $4^{x + 3}$ , so $x + 3 = 10$ and $x = 7$ .

Answer: (a) $x = 5$ (b) $x = - 3$ (c) $x = 7$

Check yourself

$3^{x} = 81$ . Find the value of $x$ .

Answer it in the portal, where your answer is checked as you type →

Indices on a calculator

On a calculator paper you may need the value of an expression such as $4.2 x^{- 3}$ for a decimal $x$ . Use the power key ( $x^{y}$ or ^) and put a negative index in brackets: for $x = 1.5$ , type $4.2 \times 1.5$ ^ $(- 3)$ . The calculator does the power first, as it should.

If you prefer, rewrite first: $4.2 x^{- 3} = \frac{4.2}{x^{3}}$ , then $\frac{4.2}{3.375} = 1.244 \dots$ Either way, do a rough check: ${1.5}^{3}$ is a little over 3, so the answer should be a little over 1.

Fractional indices with letters Extended

Extended candidates also use fractional indices. As with numbers, $x^{\frac{1}{2}} = \sqrt{x}$ , $x^{\frac{1}{3}} = \sqrt[3]{x}$ and $x^{\frac{m}{n}} = (\sqrt[n]{x})^{m}$ . The three rules apply unchanged, so you add, subtract and multiply fractions:

$3 x^{- 4} \times \frac{2}{3} x^{\frac{1}{2}} = 2 x^{- 4 + \frac{1}{2}} = 2 x^{- \frac{7}{2}}$

$\frac{2}{5} x^{\frac{1}{2}} \div 2 x^{- 2} = \frac{1}{5} x^{\frac{1}{2} - (- 2)} = \frac{1}{5} x^{\frac{5}{2}}$

$(\frac{2 x^{5}}{3})^{3} = \frac{2^{3} x^{15}}{3^{3}} = \frac{8 x^{15}}{27}$

In the last one, the bracket includes the 3 on the bottom, so it is cubed as well.

A term such as $(16 x^{8})^{\frac{3}{4}}$ needs a fractional power of a number and of a letter. Do the number as in 1.7 (fourth root of 16 is 2, cubed is 8) and multiply the index of the letter, $8 \times \frac{3}{4} = 6$ , to get $8 x^{6}$ .

Worked example 4 Extended

Simplify (a) $(16 x^{8})^{\frac{3}{4}}$ (b) $\frac{4 x^{\frac{3}{2}}}{8 x^{- \frac{1}{2}}}$ .

(a) Number: $16^{\frac{3}{4}} = (\sqrt[4]{16})^{3} = 2^{3} = 8$ .
Letter: $(x^{8})^{\frac{3}{4}} = x^{8 \times \frac{3}{4}} = x^{6}$ . Answer: $8 x^{6}$ .
(b) Numbers: $\frac{4}{8} = \frac{1}{2}$ .
Letter: $x^{\frac{3}{2} - (- \frac{1}{2})} = x^{\frac{4}{2}} = x^{2}$ .
Answer: $\frac{1}{2} x^{2}$ , which may also be written $\frac{x^{2}}{2}$ .

Answer: (a) $8 x^{6}$ (b) $\frac{1}{2} x^{2}$

Check yourself Extended

Simplify $(27 x^{6})^{\frac{1}{3}}$ .

Answer it in the portal, where your answer is checked as you type →

Solving when the bases are different Extended

Extended questions use bases that are powers of a common number. Write both sides as powers of that number, use the rules to make each side a single power, then equate the indices.

$32^{x} = 2$ : since $32 = 2^{5}$ , the left side is $(2^{5})^{x} = 2^{5 x}$ . So $2^{5 x} = 2^{1}$ , giving $5 x = 1$ and $x = \frac{1}{5}$ .

$5^{x + 1} = 25^{x}$ : since $25 = 5^{2}$ , the right side is $(5^{2})^{x} = 5^{2 x}$ . So $x + 1 = 2 x$ , giving $x = 1$ . Check: $5^{2} = 25^{1}$ .

Keep brackets round an index that is an expression: $(2^{2})^{x - 1} = 2^{2 (x - 1)} = 2^{2 x - 2}$ , not $2^{2 x - 1}$ .

Worked example 5 Extended

Solve $4^{x - 1} = 8^{x}$ .

Both 4 and 8 are powers of 2: $4 = 2^{2}$ and $8 = 2^{3}$ .
Left: $(2^{2})^{x - 1} = 2^{2 (x - 1)} = 2^{2 x - 2}$ .
Right: $(2^{3})^{x} = 2^{3 x}$ .
Equate the indices: $2 x - 2 = 3 x$ , so $x = - 2$ .
Check: $4^{- 3} = \frac{1}{64}$ and $8^{- 2} = \frac{1}{64}$ .

Answer: $x = - 2$

Check yourself Extended

Solve $9^{x} = 27$ .

Answer it in the portal, where your answer is checked as you type →

Key points

Same base: multiply means add the indices, divide means subtract, a power of a power means multiply.
Multiply or divide the numbers in front as ordinary numbers; deal with each letter's index separately. A letter on its own has index 1.
In $(5 x^{3})^{2}$ everything in the bracket is raised to the power: $25 x^{6}$ .
$x^{0} = 1$ and $x^{- n} = \frac{1}{x^{n}}$ . Watch whether a coefficient is inside the bracket.
To solve $a^{x} = b$ , write $b$ as a power of $a$ and compare indices. No logarithms needed.
Extended: $x^{\frac{m}{n}} = (\sqrt[n]{x})^{m}$ , and with different bases write both sides as powers of one number before comparing indices.

Where marks are lost

Not raising the number in a bracket: $(5 x^{3})^{2}$ is $25 x^{6}$ , not $5 x^{6}$ or $10 x^{6}$ .
Adding coefficients when multiplying terms: $4 x^{5} \times 7 x^{- 2} = 28 x^{3}$ , not $11 x^{3}$ .
Forgetting that a letter on its own has index 1: $y^{2} \times y = y^{3}$ .
Sign slips with negative indices: $a^{6} \div a^{- 3} = a^{9}$ , because $6 - (- 3) = 9$ .
In $2^{x} = 32$ , answering 16 (dividing 32 by 2) instead of finding the index.
Extended: losing the bracket round an index, as in $(2^{2})^{x - 1} = 2^{2 x - 1}$ . It is $2^{2 x - 2}$ .
Extended: forgetting to cube the denominator in $(\frac{2 x^{5}}{3})^{3}$ .

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An original lesson written to the Cambridge IGCSE Mathematics 0580 syllabus (2025 to 2027). The worked examples and questions are our own, not Cambridge questions. Back to every IGCSE Mathematics 0580 lesson.

Indices II

Indices with letters

Multiplying and dividing terms

Worked example 1

Raising a term to a power

Worked example 2

Check yourself

Solving equations with indices

Worked example 3

Check yourself

Indices on a calculator

Fractional indices with letters Extended

Worked example 4 Extended

Check yourself Extended

Solving when the bases are different Extended

Worked example 5 Extended

Check yourself Extended

Key points

Where marks are lost

Related IGCSE Mathematics 0580 lessons

Equations

Algebraic fractions

Inequalities

Algebraic manipulation