National 5MathsSQARevision NotesNumerical SkillsLaws of IndicesLaws of Indices

Laws of Indices (SQA National 5 Maths): Revision Note

Exam code: X847 75

Written by: Roger B

Reviewed by: Dan Finlay

Updated on 5 November 2025

Simplifying expressions using the 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	$x^{1} = x$
$a^{0} = 1$	Anything to the power of 0 is 1	$b^{0} = 1$
$a^{m} \times a^{n} = a^{m + n}$	To multiply indices with the same base, add their powers	$c^{3} \times c^{2} = (c \times c \times c) \times (c \times c) = c^{5}$
$a^{m} \div a^{n} = \frac{a^{m}}{a^{n}} = a^{m - n}$	To divide indices with the same base, subtract their powers	$d^{5} \div d^{2} = \frac{d \times d \times d \times d \times d}{d \times d} = d^{3}$
${(a^{m})}^{n} = a^{m n}$	To raise indices to a new power, multiply their powers	${(e^{3})}^{2} = (e \times e \times e) \times (e \times e \times e) = e^{6}$
${(a b)}^{m} = a^{m} b^{m}$	To raise a product to a power, apply the power to both numbers, and multiply	${(f g)}^{2} = {(f \times g)}^{2} = f^{2} \times g^{2} = f^{2} g^{2}$
${(\frac{a}{b})}^{m} = \frac{a^{m}}{b^{m}}$	To raise a fraction to a power, apply the power to both the numerator and denominator	${(\frac{h}{i})}^{2} = \frac{h^{2}}{i^{2}}$
$a^{- 1} = \frac{1}{a}$ $a^{- n} = \frac{1}{a^{n}}$	A negative power is the reciprocal	$j^{- 1} = \frac{1}{j}$ $k^{- 3} = \frac{1}{k^{3}}$
${(\frac{a}{b})}^{- n} = {(\frac{b}{a})}^{n} = \frac{b^{n}}{a^{n}}$	A fraction to a negative power, is the reciprocal of the fraction, to the positive power	${(\frac{l}{m})}^{- 3} = {(\frac{m}{l})}^{3} = \frac{m^{3}}{l^{3}}$
$a^{\frac{1}{n}} = \sqrt[n]{a}$	The fractional power $\frac{1}{n}$ is the n^th root ( $\sqrt[n]{}$ )	$n^{\frac{1}{2}} = \sqrt{n}$ $p^{\frac{1}{3}} = \sqrt[3]{p}$
$a^{- \frac{1}{n}} = {(a^{\frac{1}{n}})}^{- 1} = {(\sqrt[n]{a})}^{- 1} = \frac{1}{\sqrt[n]{a}}$	A negative, fractional power is one over a root	$q^{- \frac{1}{2}} = \frac{1}{\sqrt{q}}$ $r^{- \frac{1}{3}} = \frac{1}{\sqrt[3]{r}}$
$\begin{array}{rcl} a^{\frac{m}{n}} & = & a^{\frac{1}{n} \times m} \\ = & {(a^{\frac{1}{n}})}^{m} = {(\sqrt[n]{a})}^{m} \\ or & = & {(a^{m})}^{\frac{1}{n}} = \sqrt[n]{a^{m}} \end{array}$	The fractional power $\frac{m}{n}$ is the n^th root all to the power m, ${(\sqrt[n]{})}^{m}$ , or the n^th root of the power m, $\sqrt[n]{{()}^{m}}$ (both are the same)	$s^{\frac{2}{3}} = {(s^{\frac{1}{3}})}^{2} = {(\sqrt[3]{s})}^{2}$ $s^{\frac{2}{3}} = {(s^{2})}^{\frac{1}{3}} = \sqrt[3]{s^{2}}$

These can be used to simplify expressions
- Work out the number and algebra parts separately
  - $(3 x^{7}) \times (6 x^{4}) = (3 \times 6) \times (x^{7} \times x^{4}) = 18 x^{7 + 4} = 18 x^{11}$
  - $\frac{6 x^{7}}{3 x^{4}} = \frac{6}{3} \times \frac{x^{7}}{x^{4}} = 2 x^{7 - 4} = 2 x^{3}$
  - ${(3 x^{7})}^{2} = {(3)}^{2} \times {(x^{7})}^{2} = 9 x^{14}$

How do I find an unknown inside a power?

A term may have a power involving an unknown
- E.g. $7^{4 x}$
If both sides of an equation have the same base number, then the powers must be equal
- E.g. If $4^{3 x} = 4^{9}$ then $3 x = 9$
- And $x = 3$
You may have to do some simplifying first to reach this point
- E.g. $3^{2 x} \times 3^{4} = 3^{18}$ simplifies to $3^{2 x + 4} = 3^{18}$
- Therefore $2 x + 4 = 18$
- And $x = 7$

Worked Example

(a) Simplify ${(u^{5})}^{5}$ .

(b) Expand and simplify fully $x (x^{\frac{3}{2}} + x^{- 1})$ .

Answer:

Part (a)

Use ${(a^{m})}^{n} = a^{m n}$

$\begin{array}{rcl} {(u^{5})}^{5} & = & u^{5 \times 5} \end{array}$

$u^{25}$

Part (b)

Expand the brackets

$x (x^{\frac{3}{2}} + x^{- 1}) = x \times x^{\frac{3}{2}} + x \times x^{- 1}$

Use $a^{m} \times a^{n} = a^{m + n}$ to simplify the expressions

Remember that $x = x^{1}$

$= x^{1} \times x^{\frac{3}{2}} + x^{1} \times x^{- 1} = x^{1 + \frac{3}{2}} + x^{1 + (- 1)} = x^{\frac{5}{2}} + x^{0}$

Use $a^{0} = 1$

$= x^{\frac{5}{2}} + 1$

$x^{\frac{5}{2}} + 1$

Worked Example

(a) Simplify $\frac{m^{6} \times {(m^{4})}^{3}}{m^{7}}$ .

(b) Simplify ${(n^{- 3} \times n^{6})}^{- 4}$ . Give the answer with a positive power.

Answer:

Part (a)

Use ${(a^{m})}^{n} = a^{m n}$ on the second term in the denominator

$\begin{array}{rcl} \frac{m^{6} \times {(m^{4})}^{3}}{m^{7}} & = & \frac{m^{6} \times m^{4 \times 3}}{m^{7}} \\ = & \frac{m^{6} \times m^{12}}{m^{7}} \end{array}$

Use $a^{m} \times a^{n} = a^{m + n}$ to simplify the numerator

$= \frac{m^{6 + 12}}{m^{7}} = \frac{m^{18}}{m^{7}}$

Use $a^{m} \div a^{n} = \frac{a^{m}}{a^{n}} = a^{m - n}$ to finish simplifying

$= m^{18 - 7}$

$m^{11}$

Part (b)

Use $a^{m} \times a^{n} = a^{m + n}$ to simplify the terms in the brackets

$\begin{array}{rcl} {(n^{- 3} \times n^{6})}^{- 4} & = & {(n^{- 3 + 6})}^{- 4} \\ = & {(n^{3})}^{- 4} \end{array}$

Use ${(a^{m})}^{n} = a^{m n}$ to get rid of the brackets

$= n^{3 \times (- 4)} = n^{- 12}$

Use $a^{- n} = \frac{1}{a^{n}}$ to rewrite with a positive power

$= \frac{1}{n^{12}}$

$\frac{1}{n^{12}}$

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Laws of Indices (SQA National 5 Maths): Revision Note

Simplifying expressions using the laws of indices

What are the laws of indices?

How do I find an unknown inside a power?

Worked Example

Worked Example

Unlock more, it's free!

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

Author: Roger B

Reviewer: Dan Finlay