IGCSE Maths · Algebra and Graphs · Core & Extended

Indices II: Applying the Laws to Algebraic Terms

This is the same set of index laws from the Number unit’s Indices and Surds note — multiplying adds powers, dividing subtracts them, a power outside a bracket multiplies in — now applied to terms with coefficients and letters together, and to equations where the unknown sits in the exponent. Fractional indices in algebraic expressions are Extended-only; everything else is shared.

Quick recap aᵐ × aⁿ = aᵐ⁺ⁿ    aᵐ ÷ aⁿ = aᵐ⁻ⁿ    (aᵐ)ⁿ = aᵐⁿ    a⁰ = 1    a⁻ⁿ = 1/aⁿ
If any of these feel unfamiliar, it’s worth reviewing the Indices and Surds note before continuing — everything below assumes you’re already comfortable with the laws themselves.

1. Multiplying and Dividing Algebraic Terms with Indices

Deal with the number part and each letter separately: multiply/divide the coefficients as ordinary numbers, and apply the index laws to each variable on its own.

Worked example — multiplying Simplify 6x⁷y⁴ × 5x⁻⁵y.
Coefficients: 6 × 5 = 30
x terms: x⁷ × x⁻⁵ = x⁷⁻⁵ = x²
y terms: y⁴ × y¹ = y⁴⁺¹ = y⁵
Answer: 30x²y⁵
Worked example — dividing Simplify 12a⁵ ÷ 3a⁻².
Coefficients: 12 ÷ 3 = 4
a terms: a⁵ ÷ a⁻² = a⁵⁻⁽⁻²⁾ = a⁷
Answer: 4a⁷
Try it — Multiplying & Dividing
  1. Simplify 4x³y² × 3x²y⁵.
  2. Simplify 20a⁸ ÷ 5a³.
  3. Simplify 6p⁴q⁻² × 2p⁻¹q⁵.
  4. Simplify 15x⁶y³ ÷ 3x⁻²y⁵.
Show answers
  1. 12x⁵y⁷
  2. 4a⁵
  3. 12p³q³
  4. 5x⁸y⁻² (can also be written as 5x⁸/y²)

2. Raising an Algebraic Term to a Power

When a whole term is raised to a power, every part inside — the coefficient and every letter — gets raised to that power.

(a^m x^n)^k = a^(mk) x^(nk)
Worked examples Simplify (5x³)².
5² × x^(3×2) = 25x⁶

Simplify (3x²y)³.
3³ × x^(2×3) × y³ = 27x⁶y³
Try it — Powers of Terms
  1. Simplify (4x²)³.
  2. Simplify (2a³b)⁴.
  3. Simplify (x⁴y²)³.
Show answers
  1. 64x⁶
  2. 16a¹²b⁴
  3. x¹²y⁶

3. Fractional Indices in Algebraic Expressions Extended only

The same fractional index rules from the Number unit apply here — a^(1/n) is the nth root of a — just combined with coefficients and the multiplying/dividing laws.

Worked example — multiplying with a fractional index Simplify 3x⁻⁴ × (2/3)x^(1/2).
Coefficients: 3 × 2/3 = 2
x terms: x⁻⁴ × x^(1/2) = x^(−4 + 1/2) = x^(−7/2)
Answer: 2x^(−7/2)
Worked example — dividing with a fractional index Simplify (2/5)x^(1/2) ÷ 2x⁻².
Coefficients: (2/5) ÷ 2 = 1/5
x terms: x^(1/2) ÷ x⁻² = x^(1/2 − (−2)) = x^(5/2)
Answer: (1/5)x^(5/2)
Worked example — power applied to a fraction Simplify [(2x⁵)/3]³.
Cube every part: numerator 2³x^(5×3) = 8x¹⁵; denominator 3³ = 27
Answer: 8x¹⁵/27
Try it — Fractional Indices Extended
  1. Simplify 4x^(1/2) × 2x⁻³.
  2. Simplify (3/4)x⁻¹ ÷ (1/2)x^(3/2).
  3. Simplify [(3x²)/2]².
Show answers
  1. 8x^(−5/2)
  2. (3/4)÷(1/2) = 3/2; x⁻¹÷x^(3/2)=x^(−5/2); answer: (3/2)x^(−5/2)
  3. 9x⁴/4

4. Solving Equations With a Variable Base to Convert

Same strategy as before: write both sides as powers of the same base, then set the exponents equal. The extra layer here is that the base you’re given may itself need converting first.

Worked example — basic case Solve 2ˣ = 32.
32 = 2⁵, so x = 5
Worked example — base needs converting Extended Solve 32ˣ = 2.
Write 32 as a power of 2: 32 = 2⁵
So (2⁵)ˣ = 2¹ → 2^(5x) = 2¹ → 5x = 1 → x = 1/5
Worked example — unknown on both sides Extended Solve 5^(x+1) = 25ˣ.
Write 25 as a power of 5: 25 = 5²
So 5^(x+1) = (5²)ˣ = 5^(2x)
x + 1 = 2x → x = 1
Worked example — a third common base Extended Solve 9ˣ = 27^(x−1).
Write both as powers of 3: 9 = 3², 27 = 3³
3^(2x) = 3^(3(x−1)) = 3^(3x−3)
2x = 3x − 3 → x = 3
Try it — Solving Index Equations
  1. Solve 3ˣ = 243.
  2. Solve 16ˣ = 4.
  3. Solve 2^(x+3) = 8ˣ.
  4. Solve 4^(x) = 8^(x−1). Ext.
Show answers
  1. x = 5
  2. 16 = 2⁴, 4 = 2². So 2^(4x) = 2² → 4x = 2 → x = 1/2
  3. 2^(x+3) = (2³)ˣ = 2^(3x) → x + 3 = 3x → x = 3/2
  4. 4 = 2², 8 = 2³. 2^(2x) = 2^(3(x−1)) = 2^(3x−3) → 2x = 3x − 3 → x = 3
Common mistakes to watch for
  • Applying a power to only the coefficient, or only the variable, instead of both — (5x³)² is 25x⁶, not 5x⁶ and not 25x³.
  • Adding fractional and negative indices incorrectly — always write them with a common denominator first, e.g. −4 + 1/2 = −8/2 + 1/2 = −7/2.
  • Forgetting to convert the base on both sides of an equation before comparing exponents — every term must share the same base before you can equate the powers.
  • Losing track of brackets in the exponent when a side has the form a^(k(x−1)) — expand 3(x−1) to 3x−3 before setting exponents equal.
  • Mixing up multiplying and raising to a power — x³ × x² = x⁵ (add), but (x³)² = x⁶ (multiply).

5. Quick-Fire Challenge Round

True or False?
  1. (2x²)³ = 6x⁶.
  2. x⁷ ÷ x⁻² = x⁵.
  3. To solve 8ˣ = 4, both sides can be written as powers of 2.
  4. x⁻³ × x⁻² = x⁻⁵.
Show answers
  1. False — the 2 must also be cubed: 2³ × x⁶ = 8x⁶, not 6x⁶.
  2. False — dividing means subtracting the powers: 7 − (−2) = 9, so it’s x⁹.
  3. True — 8 = 2³ and 4 = 2², both are powers of 2.
  4. True — multiplying adds the powers: −3 + (−2) = −5.
Spot the Error
  1. “6x⁴ × 3x² = 18x⁸”
  2. “(2x³)⁴ = 2x¹²”
  3. “Solve 27ˣ = 3: since 27 = 3³, x = 3.”
Show answers
  1. Multiplying means adding the powers, not multiplying them: x⁴ × x² = x⁴⁺² = x⁶, so the answer is 18x⁶, not 18x⁸.
  2. The coefficient must also be raised to the power: 2⁴ = 16, so the answer is 16x¹², not 2x¹².
  3. 27ˣ = (3³)ˣ = 3^(3x), and this equals 3¹, so 3x = 1, giving x = 1/3, not x = 3.

6. Mixed Practice — Bringing It All Together

Core & Extended
  1. Simplify 8x⁵y³ × 2x⁻²y⁴.
  2. Simplify 18a⁷ ÷ 6a⁻¹.
  3. Simplify (3x²y)³.
  4. Solve 5ˣ = 125.
Show answers
  1. 16x³y⁷
  2. 3a⁸
  3. 27x⁶y³
  4. x = 3
Extended-level mixed set Extended only
  1. Simplify 5x^(1/2) × 2x⁻².
  2. Simplify [(4x³)/5]².
  3. Solve 4^(x−1) = 16ˣ.
Show answers
  1. 10x^(−3/2)
  2. 16x⁶/25
  3. 4 = 2², 16 = 2⁴. 2^(2(x−1)) = 2^(4x) → 2x−2 = 4x → x = −1

Practice at Home

A large set again — take it steadily.

Part A — Multiplying and dividing algebraic terms
  1. Simplify 5x²y⁴ × 4x³y⁻¹.
  2. Simplify 24a⁶b³ ÷ 6a²b⁻²
  3. Simplify 3p⁻²q⁵ × 4p⁴q⁻³.
Part B — Powers of algebraic terms
  1. Simplify (2x⁴)³.
  2. Simplify (5a²b³)².
  3. Simplify (x³y⁵)⁴.
Part C — Fractional indices Extended
  1. Simplify 6x^(1/3) × 2x⁻¹.
  2. Simplify (3/8)x⁻² ÷ (3/4)x^(1/2).
  3. Simplify [(2x⁴)/3]³.
Part D — Solving index equations
  1. Solve 2ˣ = 128.
  2. Solve 9ˣ = 3.
  3. Solve 3^(x+2) = 9ˣ.
  4. Solve 25^(x) = 5^(x+3). Ext.
  5. Solve 8^(x) = 16^(x−2). Ext.
Part E — Applied / challenge
  1. Simplify (2x²)³ × (3x)² fully — combine both parts into a single term.
  2. Solve 4^(2x) = 8^(x+1) for x.
  3. Show that 9^(1/2) × 27^(1/3) is a whole number, by simplifying each part first.

Key Vocabulary Recap

Coefficient — the number multiplying a variable term, e.g. 5 in 5x³
Common base — writing all terms as powers of the same number, to compare exponents directly
Fractional index — a power that represents a root, e.g. x^(1/2) = √x

Syllabus Reference

This note covers Cambridge IGCSE Mathematics (0580) section C2.4 (Core subject content) and section E2.4 (Extended subject content) — Indices II. Core covers positive, zero, and negative indices applied to algebraic terms; Extended adds fractional indices in algebraic expressions and equations requiring a base conversion with fractional exponent results, both marked “Extended only” above.

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