Algebraic Roots & Indices (Cambridge IGCSE Maths): Revision notes

Section 1

What do the index laws say?

An index (or power) shows how many times a base is multiplied by itself. The key laws, which apply to algebraic terms as well as numbers, are:

Law	Rule
Multiplying	$a^m \times a^n = a^{m+n}$
Dividing	$a^m \div a^n = a^{m-n}$
Power of a power	$(a^m)^n = a^{mn}$

These apply to coefficients and letters together, e.g. $6x^7y^4 \times 5x^{-5}y = 30x^2y^5$ (multiply the numbers, add the indices of matching letters).

Key termsindexbase

Example

Simplify $(5x^3)^2$ : apply the power of a power to both the number and the letter: $5^2 \times x^{3\times2} = 25x^6$ .

Section 2

What do zero, negative and fractional indices mean?

Zero index: any non-zero base to the power 0 equals 1, so $a^0 = 1$
Negative index: means reciprocal, so $a^{-n} = \frac{1}{a^n}$
Fractional index: the denominator is a root and the numerator is a power, so $a^{\frac{1}{n}} = \sqrt[n]{a}$ and $a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^m$

For example, $16^{\frac{1}{4}} = \sqrt[4]{16} = 2$ .

Key termsnegative indexfractional index

Common mistake

A negative index does NOT make the value negative — $a^{-2} = \frac{1}{a^2}$ , which is positive if $a$ is positive.

Section 3

How do you simplify algebraic expressions using indices?

Apply the index laws term by term, dealing with numerical coefficients and each letter separately.

$12a^5 \div 3a^{-2} = (12 \div 3) \times a^{5-(-2)} = 4a^7$

Steps:

Divide or multiply the numerical coefficients
Add (multiplying) or subtract (dividing) the indices of each matching letter
Combine into a single simplified term

Key termscoefficient

Exam tip

Always show the coefficient and index working separately in your answer — e.g. ' $(12\div3)=4$ , $5-(-2)=7$ ' — as method marks are awarded for each correct step.

Section 4

How do you solve equations involving indices?

To solve an equation like $2^x = 32$ , write both sides with the same base, then equate the indices.

$2^x = 32 \Rightarrow 2^x = 2^5 \Rightarrow x = 5$

For equations like $5^{x+1} = 25^x$ : rewrite $25$ as $5^2$ , so $5^{x+1} = 5^{2x}$ , giving $x + 1 = 2x$ , so $x = 1$ .

Key termsexponential equation

Exam tip

The core strategy for these equations is always the same: express every term with the same base first, then simply equate the indices. Knowledge of logarithms is not required.

Must Know

$a^m \times a^n = a^{m+n}$ ; $a^m \div a^n = a^{m-n}$ ; $(a^m)^n = a^{mn}$
$a^0 = 1$ (for non-zero $a$ )
$a^{-n} = \frac{1}{a^n}$ — a negative index means reciprocal, not a negative value
$a^{\frac{1}{n}} = \sqrt[n]{a}$ and $a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^m$
To solve exponential equations, write both sides with the same base then equate the indices

That's the notes covered.

Carry on to the next subtopic.