Indices and Equations

This guide covers subtopics 2.4 Indices II and 2.5 Equations, from Topic 2, Algebra and graphs, for Cambridge O Level Mathematics (Syllabus D) 4024, 2025–2027 series.

Where this fits in 4024

Indices and equation-solving are the working tools the rest of Topic 2 depends on: exponential equations use index rules directly, and every later subtopic — inequalities, sequences, graphs — assumes fluent equation manipulation. This page follows on from Algebraic Manipulation, which covers 2.1–2.3.

Syllabus coverage

CAMBRIDGE O LEVEL MATHEMATICS (SYLLABUS D) 4024

• Understand and use indices — positive, zero, negative and fractional (2.4)
• Understand and use the rules of indices, including solving equations such as 32^x = 2 or 5^(x+1) = 25^x, and simplifying expressions with fractional and negative indices (2.4)
• Construct expressions, equations and formulas (2.5)
• Solve linear equations in one unknown (2.5)
• Solve fractional equations with numerical and linear algebraic denominators (2.5)
• Solve simultaneous linear equations in two unknowns, including constructing the simultaneous equations from a worded problem (2.5)
• Solve quadratic equations by factorisation, completing the square, and using the quadratic formula — solutions may be required in surd form (2.5)
• Change the subject of formulas, including where the subject appears twice or as a power or root (2.5)

4024 is not tiered — every candidate covers all of the above. Knowledge of logarithms is not required for 2.4. The quadratic formula is provided in the exam’s List of Formulas.

Indices

The rules of indices apply to positive, zero, negative and fractional powers alike:

a^m × a^n = a^(m+n)        a^m ÷ a^n = a^(m−n)        (a^m)^n = a^(mn)
a^0 = 1        a^(−n) = 1/a^n        a^(1/n) = ⁿ√a

Worked example. Simplify (x³ × x⁻¹) ÷ x^(1/2).

x³ × x⁻¹ = x²
x² ÷ x^(1/2) = x^(2 − 1/2) = x^(3/2)

Worked example. Solve 32^x = 2.

Rewrite both sides so they share the same base. Since 32 = 2⁵:

32^x = 2
(2⁵)^x = 2¹
2^(5x) = 2¹
so 5x = 1
x = 1/5

Worked example. Solve 5^(x+1) = 25^x.

Rewrite 25 as 5² first, so both sides share base 5:

5^(x+1) = (5²)^x = 5^(2x)
so x + 1 = 2x
x = 1

Matching the bases first, then equating the indices, is the standard technique for this style of question — logarithms are not required or expected.

Solving equations

Linear equations in one unknown are solved by collecting terms:

5 − 2x = 3(x + 7)
5 − 2x = 3x + 21
5 − 21 = 3x + 2x
−16 = 5x
x = −3.2

Fractional equations with numerical or linear algebraic denominators are solved by multiplying through to clear the fractions, then solving the resulting linear or quadratic equation.

Worked example. Solve x / (x + 2) = 3 / (x − 6).

x(x − 6) = 3(x + 2)
x² − 6x = 3x + 6
x² − 9x − 6 = 0

(This reduces to a quadratic — solve it with the quadratic formula, as below, remembering to reject any solution that makes an original denominator zero.)

Simultaneous linear equations in two unknowns are solved by elimination or substitution — construct the two equations from the problem first if they are not already given.

Quadratic equations can be solved three ways: factorisation (where the expression factorises over rational numbers), completing the square, or the quadratic formula:

for ax² + bx + c = 0:        x = (−b ± √(b² − 4ac)) / 2a

Worked example. Solve x² − 4x − 3 = 0, giving answers in surd form.

x = (4 ± √(16 + 12)) / 2 = (4 ± √28) / 2 = 2 ± √7

Since 28 doesn’t have an integer square root, the answer is left as a surd rather than approximated — exactly what “surd form” instructs.

Changing the subject of a formula

Rearranging a formula to make a different variable the subject follows the same rules as solving an equation, but two situations need extra care: when the subject appears twice, collect every term containing it on one side and factorise it out before dividing; when the subject appears as a power or root, undo the power/root as the last step, once the subject is isolated apart from that power or root.

Worked example. Make x the subject of y = √(x + 3) − 2.

y + 2 = √(x + 3)
(y + 2)² = x + 3
x = (y + 2)² − 3

This rearrangement is only valid when y + 2 ≥ 0, since a square root cannot itself be negative – squaring both sides is a step that must be checked for reversibility whenever it is used.

Common mistakes

• Forgetting a^0 = 1 for any nonzero a, or mishandling negative indices (a⁻ⁿ = 1/aⁿ, not −aⁿ).
• Cross-multiplying a fractional equation incorrectly, especially when one side has a sum of two fractions — multiply every term on both sides by the full common denominator, not just isolated terms.
• Forgetting to check for excluded values in fractional equations — any solution that makes an original denominator zero must be rejected.
• Approximating a surd answer when the question explicitly asks for surd form, or vice versa.
• Dividing before factorising when the subject of a formula appears twice — the subject must be isolated as a common factor first.

Quick revision checklist

• The index rules, including zero, negative and fractional indices
• Solving index equations by matching bases, then equating indices
• Solving linear, fractional and simultaneous equations
• The three quadratic-solving methods, and when to give an answer in surd form
• Changing the subject of a formula, including when it appears twice or as a power/root

Related resources

• Algebraic Manipulation — subtopics 2.1–2.3, the foundation this page builds on
• Inequalities — the next subtopic, using the same equation-solving skills
• Cambridge O Level Mathematics subject hub

Written against Cambridge O Level Mathematics (Syllabus D) 4024, 2025–2027 series. Always check the current syllabus for your examination year.