Powers, Roots & Indices | Cambridge O Level Maths Revision Notes 2025

Powers & Roots

What are powers/indices?

Powers of a number is when that number is multiplied by itself repeatedly
- 5¹ means 5
- 5² means 5 × 5
- 5³ means 5 × 5 × 5
- Therefore the powers of 5 are 5, 25, 125, etc
The big number on the bottom is sometimes called the base number
The small number that is raised is called the index or the exponent
Any non-zero number to the power of 0 is equal to 1
- 5⁰ = 1

What are roots?

Roots of a number are the opposite of powers
A square root of 25 is a number that when squared equals 25
- The two square roots are 5 and -5
- Every positive number has two square roots
  - They will have the same digits but one is positive and one is negative
- The notation $\sqrt{}$ refers to the positive square root of a number
  - $\sqrt{25} = 5$
  - You can show both roots at once using the plus or minus symbol ±
  - Square roots of 25 are $\pm \sqrt{25} = \pm 5$
- The square root of a negative number is not a real number
- The positive square root can be written as an index of $\frac{1}{2}$ so $25^{\frac{1}{2}} = 5$
A cube root of 125 is a number that when cubed equals 125
- A cube root of 125 is 5
- Every positive and negative number always has a cube root
- The notation $\sqrt[3]{}$ refers to the cube root of a number
  - $\sqrt[3]{125} = 5$
- The cube root can be written as an index of $\frac{1}{3}$ so $125^{\frac{1}{3}} = 5$
A nth root of a number ( $\sqrt[n]{}$ )is a number that when raised to the power n equals the original number
- If n is even then they work the same way as square roots
  - Every positive number will have a positive and negative nth root
  - The notation $\sqrt[n]{}$ refers to the positive nth root of a number
- If n is odd then they work the same way as cube roots
  - Every positive and negative number will have an nth root
- The nth root can be written as an index of $\frac{1}{n}$
If you know your powers of numbers then you can use them to find roots of numbers
- e.g. $2^{5} = 32$ means $\sqrt[5]{32} = 2$
  - You could write this using an index $32^{\frac{1}{5}} = 2$
You can also estimate roots by finding the closest powers
- e.g. $2^{3} = 8$ and $3^{3} = 27$ therefore $2 < \sqrt[3]{20} < 3$

What are reciprocals?

The reciprocal of a number is the number that you multiply it by to get 1
- The reciprocal of 2 is $\frac{1}{2}$
- The reciprocal of 0.25 or $\frac{1}{4}$ is 4
- The reciprocal of $\frac{3}{2}$ is $\frac{2}{3}$
The reciprocal of a number can be written as a power with an index of -1
- 5^-1 means the reciprocal of 5
This idea can be extended to other negative indices
- 5^-2 means the reciprocal of 5²

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Laws of Indices

What are the laws of indices?

There are lots of very important laws (or rules)
It is important that you know and can apply these
Understanding the explanations will help you remember them

Law	Description	Why
$a^{1} = a$	anything to the power 1 is itself	$6^{1} = 6$
$a^{m} \times a^{n} = a^{m + n}$	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	$7^{5} \div 7^{3} = \frac{7 \times 7 \times 7 \times 7 \times 7}{7 \times 7} = 7^{3}$
${(a^{m})}^{n} = a^{m n}$	to raise indices to a new power, multiply their powers	${(14^{3})}^{2} = (14 \times 14 \times 14) \times (14 \times 14 \times 14) = 14^{6}$
$a^{0} = 1$	anything to the power 0 is 1	$8^{0} = 8^{2 - 2} = 8^{2} \div 8^{2} = \frac{8^{2}}{8^{2}} = 1$
$a^{- n} = \frac{1}{a^{n}}$	a negative power is "1 over" the positive power	$11^{- 3} = 11^{0 - 3} = 11^{0} \div 11^{3} = \frac{11^{0}}{11^{3}} = \frac{1}{11^{3}}$
$a^{\frac{1}{n}} = \sqrt[n]{a}$	a power of an nth is an nth root	${(5^{\frac{1}{2}})}^{2} = 5^{\frac{1}{2} \times 2} = 5^{1} = 5 so 5^{\frac{1}{2}} = \sqrt{5}$
$a^{\frac{m}{n}} = {(\sqrt[n]{a})}^{m} = \sqrt[n]{a^{m}}$	a fractional power of m over n means either - do the the nth root first, then raise it to the power m or - raise it to the power m, then take the nth root (depending on what's easier)	$9^{\frac{3}{2}} = 9^{\frac{1}{2} \times 3} = {(9^{\frac{1}{2}})}^{3} = {(\sqrt{9})}^{3} or 9^{\frac{3}{2}} = 9^{3 \times \frac{1}{2}} = {(9^{3})}^{\frac{1}{2}} = \sqrt{9^{3}}$
${(\frac{a}{b})}^{n} = \frac{a^{n}}{b^{n}}$	a power outside a fraction applies to both the numerator and the denominator	${(\frac{5}{6})}^{3} = \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^{3}}{6^{3}}$
${(\frac{a}{b})}^{- n} = {(\frac{b}{a})}^{n} = \frac{b^{n}}{a^{n}}$	flipping the fraction inside changes a negative power into a positive power	${(\frac{5}{6})}^{- 2} = \frac{1}{{(\frac{5}{6})}^{2}} = 1 \div {(\frac{5}{6})}^{2} = 1 \div \frac{5^{2}}{6^{2}} = 1 \times \frac{6^{2}}{5^{2}} = \frac{6^{2}}{5^{2}} = {(\frac{6}{5})}^{2}$

How do I apply more than one of the laws of indices?

Powers can include negatives and fractions
- These can be dealt with in any order
- However the following order is easiest as it avoids large numbers
If there is a negative sign in the power then deal with that first
- Take the reciprocal of the base number
- ${(\frac{8}{27})}^{- \frac{2}{3}} = {(\frac{27}{8})}^{\frac{2}{3}}$
Next deal with the denominator of the fraction of the power
- Take the root of the base number
- ${(\frac{27}{8})}^{\frac{2}{3}} = {(\frac{\sqrt[3]{27}}{\sqrt[3]{8}})}^{2} = {(\frac{3}{2})}^{2}$
Finally deal with the numerator of the fraction of the power
- Take the power of the base number
- ${(\frac{3}{2})}^{2} = \frac{3^{2}}{2^{2}} = \frac{9}{4}$

How do I deal with different bases?

Sometimes expressions involve different base values
You can use index laws to change the base of a term to simplify an expression involving terms with different bases
- For example $9^{4} = {(3^{2})}^{4} = 3^{2 \times 4} = 3^{8}$
- Using the above can then help with problems like $9^{4} \div 3^{7} = 3^{8} \div 3^{7} = 3^{8 - 7} = 3^{1} = 3$

Examiner Tip

Index laws only work with terms that have the same base, so something like 2³ × 5² cannot be simplified using index laws

Worked example

(a)

\begin{matrix} ^{} \end{matrix}

Without using a calculator, write

\frac{7^{3} \times 7^{2}}{7^{8}}

in the form

\frac{1}{7^{k}}

where k is a positive whole number.

Use $a^{m} \times a^{n} = a^{m + n}$ on the numerator.

$7^{3} \times 7^{2} = 7^{3 + 2} = 7^{5}$

Use $a^{m} \div a^{n} = a^{m - n}$

$\frac{7^{5}}{7^{8}} = 7^{5 - 8} = 7^{- 3}$

Use $a^{- n} = \frac{1}{a^{n}}$ .

$7^{- 3} = \frac{1}{7^{3}}$

The value of k is 3.

$\frac{1}{7^{3}}$

(b)

\begin{matrix}  \end{matrix}

Without using a calculator, simplify

{(\frac{4}{25})}^{- \frac{3}{2}}

Flip the fraction to change the negative outside power into a positive outside power, ${(\frac{a}{b})}^{- n} = {(\frac{b}{a})}^{n}$ .

${(\frac{4}{25})}^{- \frac{3}{2}} = {(\frac{25}{4})}^{\frac{3}{2}}$

Use that a power outside a fraction applies to both the numerator and denominator, ${(\frac{a}{b})}^{n} = \frac{a^{n}}{b^{n}}$ .

${(\frac{25}{4})}^{\frac{3}{2}} = \frac{25^{\frac{3}{2}}}{4^{\frac{3}{2}}}$

Use that a fractional power of m over n is the nth root all to the power m, $a^{\frac{m}{n}} = {(\sqrt[n]{a})}^{m}$ .

$25^{\frac{3}{2}} = {(\sqrt{25})}^{3} = 5^{3} = 125$ and $4^{\frac{3}{2}} = {(\sqrt{4})}^{3} = 2^{3} = 8$

$\frac{125}{8}$

Powers, Roots & Indices (Cambridge O Level Maths)

Revision Note