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A4dLaws of indices in algebra

Edexcel GCSE Maths (1MA1), Higher tier · Algebra

Practise Laws of indices in algebra. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.

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Sample questions

Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.

Question 1Easy4 marks
(a) Simplify \(p^{6} \times p^{7}\)[1]
(b) Simplify \((t^{2})^{5}\)[1]
(c) Simplify \(5p^{3}q^{4} \times 4pq^{3}\)[2]
Show the answer and mark scheme
(a) Answer: \(p^{13}\)
  • B1 for \(p^{13}\)
(b) Answer: \(t^{10}\)
  • B1 for \(t^{10}\)
(c) Answer: \(20p^{4}q^{7}\)
  • B2 for \(20p^{4}q^{7}\) (B1 for two of \(20\), \(p^{4}\), \(q^{7}\) correct in a product)

Worked solution: Multiply the numbers and add the indices: \(5 \times 4 = 20\), \(p^{3 + 1} = p^{4}\), \(q^{4 + 3} = q^{7}\)

Question 2Medium5 marks
(a) Simplify \((3s^{3}t^{3})^{4}\)[2]
(b) Simplify \(\frac{18s^{8}t}{3s^{7}t^{5}}\)[2]
(c) Simplify \(\frac{t^{5} \times t^{3}}{t^{7}}\)[1]
Show the answer and mark scheme
(a) Answer: \(81s^{12}t^{12}\)
  • B2 for \(81s^{12}t^{12}\) (B1 for two of \(81\), \(s^{12}\), \(t^{12}\) correct in a product)

Worked solution: \((3s^{3}t^{3})^{4} = 3^{4} \times s^{3 \times 4} \times t^{3 \times 4} = 81s^{12}t^{12}\)

(b) Answer: \(\frac{6s}{t^{4}}\)
  • B2 for \(\frac{6s}{t^{4}}\) oe (B1 for two of \(6\), \(s\), \(t^{-4}\) correct)

Worked solution: Divide the numbers and subtract the indices: \(18 \div 3 = 6\), \(s^{8 - 7} = s\), \(t^{1 - 5} = t^{-4}\), so the answer is \(\frac{6s}{t^{4}}\)

(c) Answer: \(t\)
  • B1 for \(t\)
Question 3Hard5 marks
(a) Show that \(\frac{4^x \times 8^2}{2^{x + 1}}\) can be written as \(2^{x + 5}\).[3]
(b) Hence solve \(\frac{4^x \times 8^2}{2^{x + 1}} = \frac{1}{8}\).[2]
Show the answer and mark scheme
(a) Answer: \(\frac{2^{2x} \times 2^6}{2^{x + 1}} = 2^{2x + 6 - x - 1} = 2^{x + 5}\)
  • M1 for \(4^x = 2^{2x}\) or \(8^2 = 2^6\)
  • M1 for \(2^{2x + 6 - (x + 1)}\) oe
  • C1 for \(2^{x + 5}\) from fully correct working

Worked solution: \(4^x = (2^2)^x = 2^{2x}\) and \(8^2 = (2^3)^2 = 2^6\).
\(\frac{2^{2x} \times 2^6}{2^{x + 1}} = 2^{2x + 6 - x - 1} = 2^{x + 5}\)

(b) Answer: \(x = -8\)
  • M1 for \(2^{x + 5} = 2^{-3}\) or \(x + 5 = -3\)
  • A1 for −8

Worked solution: \(\frac{1}{8} = 2^{-3}\), so \(x + 5 = -3\) and \(x = -8\).

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