IGCSE Maths 0580 Indices Past Paper Questions and Worked Solutions

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IGCSE Maths (0580) · Number · 1.7

Every indices question from 21 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 1.7. 25 questions covering Core and Extended tiers, ordered from Core to Extended. Each question is labelled with its question paper code (for example 0580/13/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Number past paper questions.

0580 Indices past paper questions: Core (Papers 1 and 3)

12 marksCore

May/June 2026 · Paper 1 · 0580/13/M/J/26 · Q18(b)

Write 5⁻¹ as a decimal.

Show solution
  1. A power of −1 means the reciprocal: 5⁻¹ = 1⁄5.
  2. 1 ÷ 5 = 0.2
Answer 0.2
21 markCore

May/June 2026 · Paper 3 · 0580/31/M/J/26 · Q2(a)

Write 7 × 7 × 7 × 7 × 7 × 7 as a power of 7.

Show solution
  1. Count how many 7s are multiplied together: there are 6.
  2. 7 × 7 × 7 × 7 × 7 × 7 = 7⁶
Answer 7⁶
32 marksCore

May/June 2026 · Paper 3 · 0580/32/M/J/26 · Q13

(a) 8ʸ = 1. Find the value of y. (b) x⁴ × x⁵ = xʷ. Find the value of w.

Show solution
Part (a)
  1. Any non-zero number to the power 0 is 1, so 8⁰ = 1.
  2. y = 0
Answer y = 0
Part (b)
  1. When multiplying powers of the same letter, add the indices: x⁴ × x⁵ = x4 + 5 = x⁹.
  2. w = 9
Answer w = 9
42 marksCore

Feb/Mar 2026 · Paper 1 · 0580/12/F/M/26 · Q12

Write 9 × 27 as a power of 3.

Show solution
  1. Write each number as a power of 3: 9 = 3 × 3 = 3² and 27 = 3 × 3 × 3 = 3³.
  2. When multiplying powers of the same number, add the indices: 3² × 3³ = 32 + 3.
  3. 9 × 27 = 3⁵
Answer 3⁵
53 marksCore

Feb/Mar 2026 · Paper 3 · 0580/32/F/M/26 · Q21

(a) Write 1⁄k × k as a power of k. (b) Simplify 3w⁵ × 4w³.

Show solution
Part (a)
  1. k × k = k², so 1⁄k × k = 1⁄k².
  2. 1 over a power is a negative power: k⁻²
Answer k⁻²
Part (b)
  1. Multiply the numbers: 3 × 4 = 12.
  2. Multiply the powers of w by adding the indices: w⁵ × w³ = w⁸.
  3. 12w⁸
Answer 12w⁸
62 marksCore

Oct/Nov 2025 · Paper 1 · 0580/11/O/N/25 · Q23

(a) a⁸ × aˣ = a¹⁵. Find the value of x. (b) (b⁷)ʸ = b²¹. Find the value of y.

Show solution
Part (a)
  1. When multiplying powers of the same letter, add the indices: a8 + x = a15.
  2. 8 + x = 15, so x = 7
Answer x = 7
Part (b)
  1. For a power of a power, multiply the indices: (b⁷)ʸ = b7y.
  2. 7y = 21, so y = 3
Answer y = 3
72 marksCore

Oct/Nov 2025 · Paper 1 · 0580/12/O/N/25 · Q21

Simplify 18x⁸⁄9x².

Show solution
  1. Divide the numbers: 18 ÷ 9 = 2.
  2. Divide the powers of x by subtracting the indices: x⁸ ÷ x² = x8 − 2 = x⁶.
  3. 2x⁶
Answer 2x⁶
82 marksCore

Oct/Nov 2025 · Paper 1 · 0580/13/O/N/25 · Q9

Find the value of (a) 2⁵ (b) 6⁰.

Show solution
Part (a)
  1. 2⁵ = 2 × 2 × 2 × 2 × 2 = 4 × 4 × 2 = 32
Answer 32
Part (b)
  1. Any number (except 0) to the power 0 equals 1.
  2. 6⁰ = 1
Answer 1
91 markCore

May/June 2025 · Paper 1 · 0580/11/M/J/25 · Q14

3p × 34 = 310. Find the value of p.

Show solution
  1. When you multiply powers of the same number, add the indices: 3p × 34 = 3p + 4.
  2. So p + 4 = 10.
  3. p = 10 − 4 = 6
Answer 6
102 marksCore

May/June 2025 · Paper 1 · 0580/12/M/J/25 · Q7

(a) Find the value of the reciprocal of 1⁄3.

(b) Write 2−3 as a fraction.

Show solution
Part (a)
  1. The reciprocal of a fraction is the fraction turned upside down: 1⁄3 → 3⁄1.
  2. 3⁄1 = 3
Answer 3
Part (b)
  1. A negative power means “1 over”: 2−3 = 1⁄2³.
  2. 2³ = 2 × 2 × 2 = 8, so 2−3 = 1⁄8
Answer 1⁄8
113 marksCore

May/June 2025 · Paper 1 · 0580/13/M/J/25 · Q17

Simplify. (a) y⁴ × y⁶ (b) p⁵⁄p⁸ (c) (w⁴)³

Show solution
Part (a)
  1. When multiplying powers of the same letter, add the indices: 4 + 6 = 10.
  2. y⁴ × y⁶ = y10
Answer y10
Part (b)
  1. When dividing, subtract the indices: 5 − 8 = −3.
  2. p⁵⁄p⁸ = p−3 (or 1⁄p³)
Answer p−3
Part (c)
  1. For a power of a power, multiply the indices: 4 × 3 = 12.
  2. (w⁴)³ = w12
Answer w12

0580 Indices past paper questions: Extended (Papers 2 and 4)

121 markExtended

May/June 2026 · Paper 2 · 0580/21/M/J/26 · Q1

Write down the value of 143⁰.

Show solution
  1. Any number (except 0) raised to the power 0 equals 1.
  2. 143⁰ = 1
Answer 1
134 marksExtended

May/June 2026 · Paper 2 · 0580/21/M/J/26 · Q20

Find the value of 27−2/3 × (6⁴)0.5.

Show solution
  1. Work out 27−2/3: cube root first, 271/3 = 3. Then square: 3² = 9. The minus sign gives the reciprocal: 1⁄9.
  2. Work out (6⁴)0.5: multiply the powers, 4 × 0.5 = 2, so it is 6² = 36.
  3. Multiply: 1⁄9 × 36 = 36 ÷ 9 = 4
Answer 4
143 marksExtended

May/June 2026 · Paper 2 · 0580/22/M/J/26 · Q24

72x + 1 = 492 − x. Find the value of x.

Show solution
  1. Write 49 as a power of 7: 49 = 7², so 492 − x = 72(2 − x) = 74 − 2x.
  2. The bases are the same, so the powers are equal: 2x + 1 = 4 − 2x.
  3. Add 2x to both sides: 4x + 1 = 4. Subtract 1: 4x = 3.
  4. x = 3⁄4
Answer x = 3⁄4
156 marksExtended

May/June 2026 · Paper 2 · 0580/23/M/J/26 · Q22

(a) Simplify (i) (10b⁴)⁻³ (ii) (8p¹⁵)2/3

(b) 16ⁿ = 64. Find the value of n.

Show solution
Part (a)(i)
  1. Raise each part to the power 3: 10³ = 1000 and (b⁴)³ = b¹². So (10b⁴)³ = 1000b¹².
  2. The negative power means the reciprocal.
  3. (10b⁴)⁻³ = 1⁄1000b¹²
Answer 1⁄1000b¹²
Part (a)(ii)
  1. 82/3: cube root of 8 is 2, squared is 4.
  2. (p¹⁵)2/3: multiply the powers, 15 × 2⁄3 = 10, so p¹⁰.
  3. 4p¹⁰
Answer 4p¹⁰
Part (b)
  1. Write both numbers as powers of 4: 16 = 4² and 64 = 4³.
  2. (4²)ⁿ = 4³, so 4²ⁿ = 4³ and 2n = 3.
  3. n = 3⁄2
Answer n = 3⁄2
166 marksExtended

Feb/Mar 2026 · Paper 2 · 0580/22/F/M/26 · Q20

(a) Simplify (27x⁹)−2/3. (b) Solve 7ˣ = 49(1 − 2x).

Show solution
Part (a)
  1. Deal with the number and the letter separately.
  2. 272/3: cube root of 27 is 3, squared is 9. (x⁹)2/3: multiply the powers, 9 × 2⁄3 = 6, so x⁶.
  3. So (27x⁹)2/3 = 9x⁶, and the negative power gives the reciprocal.
  4. 1⁄9x⁶
Answer 1⁄9x⁶
Part (b)
  1. Write 49 as a power of 7: 49 = 7², so 491 − 2x = 72(1 − 2x) = 72 − 4x.
  2. The bases are the same, so the powers are equal: x = 2 − 4x.
  3. Add 4x: 5x = 2, so x = 2⁄5
Answer x = 2⁄5
173 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/21/O/N/25 · Q6

Find the value of (a) 5⁻⁵ × 5⁵ (b) 1252/3

Show solution
Part (a)
  1. When multiplying powers of the same number, add the indices: −5 + 5 = 0.
  2. 5⁻⁵ × 5⁵ = 5⁰ = 1
Answer 1
Part (b)
  1. The bottom of the fraction (3) means cube root, and the top (2) means square.
  2. ∛125 = 5 (because 5 × 5 × 5 = 125).
  3. 5² = 25
Answer 25
183 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/21/O/N/25 · Q19

Solve (1⁄3)ˣ = 9x + 4.

Show solution
  1. Write both sides as powers of 3.
  2. 1⁄3 = 3⁻¹, so (1⁄3)ˣ = 3⁻ˣ.
  3. 9 = 3², so 9x + 4 = 32(x + 4) = 32x + 8.
  4. The bases are the same, so the powers must be equal: −x = 2x + 8.
  5. Subtract 2x from both sides: −3x = 8.
  6. x = −8⁄3
Answer x = −8⁄3
194 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/22/O/N/25 · Q14

Find the value of (a) 3² ÷ 3⁻² (b) 16−3/2

Show solution
Part (a)
  1. When dividing powers of the same number, subtract the indices: 2 − (−2) = 4.
  2. 3² ÷ 3⁻² = 3⁴ = 81
Answer 81
Part (b)
  1. The 2 on the bottom means square root, the 3 on top means cube, and the minus sign means reciprocal.
  2. √16 = 4, then 4³ = 64.
  3. 16−3/2 = 1⁄64
Answer 1⁄64
202 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/22/O/N/25 · Q24

27ⁿˣ = (9ˣ)². Find the value of n.

Show solution
  1. Write both sides as powers of 3. 27 = 3³ and 9 = 3².
  2. 27ⁿˣ = (3³)ⁿˣ = 3³ⁿˣ. (9ˣ)² = 9²ˣ = (3²)²ˣ = 3⁴ˣ.
  3. The powers must be equal: 3nx = 4x.
  4. Divide by x: 3n = 4, so n = 4⁄3
Answer n = 4⁄3
211 markExtended

Oct/Nov 2025 · Paper 2 · 0580/23/O/N/25 · Q9

n¹⁵ ÷ nˣ = n⁵. Find the value of x.

Show solution
  1. When dividing powers of the same letter, subtract the indices: n¹⁵ ÷ nˣ = n15 − x.
  2. 15 − x = 5, so x = 10
Answer x = 10
225 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/23/O/N/25 · Q12

(a) Write down the value of 73⁰. (b) Find the value of 7⁄3⁻². (c) Write 27 × 81² in the form 3ⁿ.

Show solution
Part (a)
  1. Any number (except 0) to the power 0 equals 1.
  2. 73⁰ = 1
Answer 1
Part (b)
  1. A negative power means the reciprocal: 3⁻² = 1⁄3² = 1⁄9.
  2. Dividing by 1⁄9 is the same as multiplying by 9: 7 ÷ 1⁄9 = 7 × 9 = 63
Answer 63
Part (c)
  1. Write each number as a power of 3: 27 = 3³ and 81 = 3⁴, so 81² = (3⁴)² = 3⁸.
  2. Multiply by adding the powers: 3³ × 3⁸ = 33 + 8 = 3¹¹
Answer 3¹¹
233 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/43/O/N/25 · Q6

Simplify. (a) y⁵⁄y² (b) 3x³ × 5x⁵

Show solution
Part (a)
  1. When dividing powers of the same letter, subtract the indices: y5 − 2.
  2. y³
Answer y³
Part (b)
  1. Multiply the numbers: 3 × 5 = 15.
  2. Multiply the powers of x by adding the indices: x³ × x⁵ = x3 + 5 = x⁸.
  3. 15x⁸
Answer 15x⁸
242 marksExtended

May/June 2025 · Paper 2 · 0580/21/M/J/25 · Q19

Find the value of 27−2⁄3.

Show solution
  1. Break the power into three parts: the 3 on the bottom means cube root, the 2 means square, and the minus sign means reciprocal.
  2. Cube root of 27 = 3.
  3. Square it: 3² = 9.
  4. The negative power means 1 over: 27−2/3 = 1⁄9
Answer 1⁄9
254 marksExtended

May/June 2025 · Paper 2 · 0580/22/M/J/25 · Q17

Find the value of (a) 1252⁄3 (b) 4−5⁄2.

Show solution
Part (a)
  1. The 3 on the bottom of the fraction means cube root, and the 2 on top means square.
  2. ∛125 = 5 (because 5 × 5 × 5 = 125).
  3. 5² = 25
Answer 25
Part (b)
  1. The 2 on the bottom means square root, the 5 on top means power 5, and the minus sign means reciprocal.
  2. √4 = 2, then 2⁵ = 32.
  3. 4−5/2 = 1⁄32
Answer 1⁄32

Question papers used on this page

All indices questions above, listed by 0580 question paper code.

QP codeFile namePaperQuestions
0580/13/M/J/260580_s26_qp_13May/June 2026 · Paper 1 (Core)Q18(b)
0580/21/M/J/260580_s26_qp_21May/June 2026 · Paper 2 (Extended)Q1, Q20
0580/22/M/J/260580_s26_qp_22May/June 2026 · Paper 2 (Extended)Q24
0580/23/M/J/260580_s26_qp_23May/June 2026 · Paper 2 (Extended)Q22
0580/31/M/J/260580_s26_qp_31May/June 2026 · Paper 3 (Core)Q2(a)
0580/32/M/J/260580_s26_qp_32May/June 2026 · Paper 3 (Core)Q13
0580/12/F/M/260580_m26_qp_12Feb/Mar 2026 · Paper 1 (Core)Q12
0580/22/F/M/260580_m26_qp_22Feb/Mar 2026 · Paper 2 (Extended)Q20
0580/32/F/M/260580_m26_qp_32Feb/Mar 2026 · Paper 3 (Core)Q21
0580/11/O/N/250580_w25_qp_11Oct/Nov 2025 · Paper 1 (Core)Q23
0580/12/O/N/250580_w25_qp_12Oct/Nov 2025 · Paper 1 (Core)Q21
0580/13/O/N/250580_w25_qp_13Oct/Nov 2025 · Paper 1 (Core)Q9
0580/21/O/N/250580_w25_qp_21Oct/Nov 2025 · Paper 2 (Extended)Q6, Q19
0580/22/O/N/250580_w25_qp_22Oct/Nov 2025 · Paper 2 (Extended)Q14, Q24
0580/23/O/N/250580_w25_qp_23Oct/Nov 2025 · Paper 2 (Extended)Q9, Q12
0580/43/O/N/250580_w25_qp_43Oct/Nov 2025 · Paper 4 (Extended)Q6
0580/11/M/J/250580_s25_qp_11May/June 2025 · Paper 1 (Core)Q14
0580/12/M/J/250580_s25_qp_12May/June 2025 · Paper 1 (Core)Q7
0580/13/M/J/250580_s25_qp_13May/June 2025 · Paper 1 (Core)Q17
0580/21/M/J/250580_s25_qp_21May/June 2025 · Paper 2 (Extended)Q19
0580/22/M/J/250580_s25_qp_22May/June 2025 · Paper 2 (Extended)Q17

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