Indices and Equations: Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Indices and Equations revision notes

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Section A

1. Simplify: (a) x⁵ × x³ (b) y⁷ ÷ y² (c) (a³)⁴ (d) 5x⁰ [4]

2. Evaluate: (a) 3⁻² (b) 27^(2/3) (c) (16/81)^(1/2) [3]

Section B

3. Solve:

(a) 5(2x − 3) = 3x + 13 [3] (b) (x + 2)/3 − (x − 1)/4 = 2 [4]

4. Solve the quadratic equations:

(a) x² − 7x + 12 = 0 by factorising [3] (b) 2x² + 5x − 4 = 0 using the formula, to 2 d.p. [4]

5. Solve the simultaneous equations 3x + 2y = 16 and 5x − y = 18, checking your answer in the equation you did not substitute into. [5]

6. Make r the subject of V = ⅓πr²h, given that r must be positive. [3]

7. Solve 2^(3x) = 32, showing your method clearly. [3]

8. The area of a rectangle is 40 cm². Its length is 3 cm more than its width. Form an equation and find the dimensions, explaining why one algebraic solution must be rejected. [5]

9. Solve x/(x + 2) = 3/(x − 6), giving your answers as exact surds where necessary, and state which values of x had to be excluded from the domain before solving. [6]

10. Solve x² − 4x − 3 = 0, giving your answers in surd form. [3]

11. Solve 5^(x+1) = 25^x. [3]

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Answers

1. (a) x⁸ [1]. (b) y⁵ [1]. (c) a¹² [1]. (d) 5 [1].

2. (a) 1/9 [1]. (b) (∛27)² = 3² = 9 [1]. (c) 4/9 [1].

3. (a) 10x − 15 = 3x + 13 [1]; 7x = 28 [1]; x = 4 [1]. (b) Multiply through by 12: 4(x + 2) − 3(x − 1) = 24 [1]; 4x + 8 − 3x + 3 = 24 [1]; x + 11 = 24 [1]; x = 13 [1].

4. (a) (x − 3)(x − 4) = 0 [1] [1]; x = 3 or 4 [1]. (b) a = 2, b = 5, c = −4; discriminant = 25 + 32 = 57 [1]; x = (−5 ± √57) ÷ 4 [1] [1]; x = 0.64 or −3.14 [1].

5. From the second equation, y = 5x − 18 [1]. Substituting into the first equation: 3x + 2(5x − 18) = 16 [1]; 13x = 52, so x = 4 [1]; y = 20 − 18 = 2 [1]. Check in the equation not substituted into (the second equation): 5(4) − 2 = 20 − 2 = 18 ✓, confirming the solution [1].

6. 3V = πr²h [1]; r² = 3V ÷ (πh) [1]; r = √(3V ÷ πh) [1].

7. 32 = 2⁵ [1]; so 3x = 5 [1]; x = 5/3 [1].

8. Let the width be x, so the length is x + 3 [1]. x(x + 3) = 40 [1]; x² + 3x − 40 = 0 [1]; (x + 8)(x − 5) = 0, so x = 5 (rejecting x = −8 as a length cannot be negative) [1]. Width 5 cm, length 8 cm [1].

9. The excluded values are x = −2 and x = 6, since these make one of the original denominators zero [1]. Cross-multiplying: x(x − 6) = 3(x + 2) [1]; x² − 6x = 3x + 6 [1]; x² − 9x − 6 = 0 [1]. Using the formula, x = (9 ± √(81 + 24)) ÷ 2 = (9 ± √105) ÷ 2 [1]. Neither root equals −2 or 6, so both solutions are valid [1].

10. a = 1, b = −4, c = −3; x = (4 ± √(16 + 12)) ÷ 2 [1] = (4 ± √28) ÷ 2 [1]; since 28 has no integer square root, leave the answer as a surd rather than approximating: x = 2 ± √7 [1].

11. Rewrite 25 as 5² so both sides share the same base: 5^(x+1) = (5²)^x = 5^(2x) [1]. Once the bases match, the indices must be equal: x + 1 = 2x [1]; x = 1 [1].

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Where marks are usually lost

• Writing x⁵ × x³ as x¹⁵, adding the powers only when multiplying, never when the two terms are added or subtracted.
• Multiplying only part of the expression when clearing fractions.
• Forgetting the ± in the quadratic formula.
• Not rejecting the negative root in a context question.
• Approximating a surd answer when a question explicitly asks for surd form, or vice versa — check the exact wording before rounding anything.
• Forgetting to check a fractional equation’s solutions against the values that were originally excluded (those that make a denominator zero) — a valid-looking root can still need rejecting.
• Cross-multiplying only one side of a fractional equation, rather than every term on both sides.
• Confusing a fractional-index question, e.g. 27^(2/3), with a negative-index question, e.g. 3⁻² — the first is a root-then-power calculation, the second a reciprocal.
• Trying to equate indices before the bases actually match — 5^(x+1) = 25^x only becomes solvable once 25 is rewritten as 5², not before.

For condensed recall notes on this topic, see the Indices and Equations revision notes; for the full explanation with worked examples, see the Indices and Equations study guide.