IGCSEMathsCambridge (CIE)International MathsCoreRevision NotesNumberPowers, Roots & Standard FormLaws of Indices

Laws of Indices (Cambridge (CIE) IGCSE International Maths: Core): Revision Note

Exam code: 0607

Written by: Jamie Wood

Reviewed by: Dan Finlay

Updated on 19 January 2026

Laws of indices

What are the laws of indices?

Index laws are rules you can use when doing operations with powers
- They work with both numbers and algebra

Law	Description	How it works
$a^{1} = a$	Anything to the power of 1 is itself	$6^{1} = 6$
$a^{0} = 1$	Anything to the power of 0 is 1	$8^{0} = 1$
$a^{m} \times a^{n} = a^{m + n}$	To multiply indices with the same base, add their powers	$4^{3} \times 4^{2} = (4 \times 4 \times 4) \times (4 \times 4) = 4^{5}$
$a^{m} \div a^{n} = \frac{a^{m}}{a^{n}} = a^{m - n}$	To divide indices with the same base, subtract their powers	$7^{5} \div 7^{2} = \frac{7 \times 7 \times 7 \times 7 \times 7}{7 \times 7} = 7^{3}$
${(a^{m})}^{n} = a^{m n}$	To raise indices to a new power, multiply their powers	${(14^{3})}^{2} = (14 \times 14 \times 14) \times (14 \times 14 \times 14) = 14^{6}$
${(a b)}^{n} = a^{n} b^{n}$	To raise a product to a power, apply the power to both numbers, and multiply	${(3 \times 4)}^{2} = 3^{2} \times 4^{2}$
${(\frac{a}{b})}^{n} = \frac{a^{n}}{b^{n}}$	To raise a fraction to a power, apply the power to both the numerator and denominator	${(\frac{3}{4})}^{2} = \frac{3^{2}}{4^{2}} = \frac{9}{16}$
$a^{- 1} = \frac{1}{a}$ $a^{- n} = \frac{1}{a^{n}}$	A negative power is the reciprocal	$6^{- 1} = \frac{1}{6}$ $11^{- 3} = \frac{1}{11^{3}}$

How do I deal with different bases?

Index laws only work with terms that have the same base
- $2^{3} \times 5^{2}$ cannot be simplified using index laws
Sometimes expressions involve different base values, but one is related to the other by a power
- e.g. $2^{5} \times 4^{3}$
You can use powers to rewrite one of the bases
- $2^{5} \times 4^{3} = 2^{5} \times {(2^{2})}^{3}$
- This can then be simplified more easily, as the two bases are now the same
- $2^{5} \times {(2^{2})}^{3} = 2^{5} \times 2^{6} = 2^{11}$

Worked Example

(a) Find the value of $x$ when $6^{10} \times 6^{x} = 6^{2}$

Answer:

Using the law of indices $a^{m} \times a^{n} = a^{m + n}$ we can rewrite the left hand side

$6^{10} \times 6^{x} = 6^{10 + x}$

So the equation is now

$6^{10 + x} = 6^{2}$

Comparing both sides, the bases are the same, so we can say that

$10 + x = 2$

Subtract 10 from both sides

$x = - 8$

(b) Find the value of $n$ when $5^{n} \div 5^{4} = 5^{6}$

Answer:

Using the law of indices $a^{m} \div a^{n} = a^{m - n}$ we can rewrite the left hand side

$5^{n} \div 5^{4} = 5^{n - 4}$

So the equation is now

$5^{n - 4} = 5^{6}$

Comparing both sides, the bases are the same, so we can say that

$n - 4 = 6$

Add 4 to both sides

$n = 10$

Answer:

Using the law of indices $a^{- n} = \frac{1}{a^{n}}$ we can rewrite the expression

$2^{- 4} = \frac{1}{2^{4}}$

$2^{4} = 2 \times 2 \times 2 \times 2 = 16$ so we can rewrite the expression

$\frac{1}{2^{4}} = \frac{1}{16}$

$\frac{1}{16}$

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Laws of Indices (Cambridge (CIE) IGCSE International Maths: Core): Revision Note

Laws of indices

What are the laws of indices?

How do I deal with different bases?

Worked Example

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Jamie Wood

Reviewer: Dan Finlay