IBMathsDPAnalysis & Approaches (AA)HLRevision Notes1. Number & AlgebraPartial Fractions, Indices & Standard FormLaws of Indices

Laws of Indices (DP IB Analysis & Approaches (AA): HL): Revision Note

Written by: Amber

Reviewed by: Mark Curtis

Updated on 26 November 2025

Laws of indices

What are the laws of indices?

Index laws are rules for doing operations with powers
- They work on 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	$6^{- 1} = \frac{1}{6}$ $11^{- 3} = \frac{1}{11^{3}}$
${(\frac{a}{b})}^{- n} = {(\frac{b}{a})}^{n} = \frac{b^{n}}{a^{n}}$	A fraction to a negative power, is the reciprocal of the fraction, to the positive power	${(\frac{2}{5})}^{- 3} = {(\frac{5}{2})}^{3} = \frac{5^{3}}{2^{3}} = \frac{125}{8}$
$a^{\frac{1}{n}} = \sqrt[n]{a}$	The fractional power $\frac{1}{n}$ is the n^th root ( $\sqrt[n]{}$ )	$25^{\frac{1}{2}} = \sqrt[2]{25} = 5$ $27^{\frac{1}{3}} = \sqrt[3]{27} = 3$
$a^{- \frac{1}{n}} = {(a^{\frac{1}{n}})}^{- 1} = {(\sqrt[n]{a})}^{- 1} = \frac{1}{\sqrt[n]{a}}$	A negative, fractional power is one over a root	$64^{- \frac{1}{2}} = \frac{1}{\sqrt[2]{64}} = \frac{1}{8}$ $125^{- \frac{1}{3}} = \frac{1}{\sqrt[3]{125}} = \frac{1}{5}$
$a^{\frac{m}{n}} = a^{\frac{1}{n} \times m} = {(a^{\frac{1}{n}})}^{m} = {(a^{m})}^{\frac{1}{n}}$	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)	$8^{\frac{2}{3}} = {(8^{\frac{1}{3}})}^{2} = {(\sqrt[3]{8})}^{2} = 2^{2} = 4$ $8^{\frac{2}{3}} = {(8^{2})}^{\frac{1}{3}} = \sqrt[3]{64} = 4$

Examiner Tips and Tricks

The index laws are not in the formula booklet so you must remember them!

How do I change the base?

Index laws only work with terms that have the same base
- $2^{3} \times 5^{2}$ cannot be simplified using index laws

You can sometimes rewrite a base as a power of another base
- $2^{5} \times 4^{3} = 2^{5} \times {(2^{2})}^{3}$
  - The $4$ changes to $(2^{2})$
  - This is called changing the base
- It can now be simplified using index laws
- $2^{5} \times {(2^{2})}^{3} = 2^{5} \times 2^{6} = 2^{11}$

Worked Example

Simplify the following expressions:

i) $\frac{(3 x^{2}) (2 x^{3} y^{2})}{(6 x^{2} y)}$

Answer:

ii) $(4 x^{2} y^{- 4})^{3} (2 x^{3} y^{- 1})^{- 2}$

Answer:

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