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Indices - Edexcel GCSE Higher Maths

Typically 3-5 marks per paper, usually early in the paper · Spec 1MA1

Indices are the shorthand for repeated multiplication. On Edexcel's 1MA1 Higher tier they sit in two places in the specification: N6 and N7 in Number (positive integer powers, the associated roots, and calculating with integer and fractional indices) and A4 in Algebra (simplifying expressions using the laws of indices). Foundation candidates meet the three index laws, the zero index and negative integer powers. Positive and negative fractional powers such as 8238^{\frac{2}{3}} and 16−3416^{-\frac{3}{4}} are Higher tier only.

Edexcel tends to put these near the front of a paper, in short parts worth 1-2 marks each: write down the value of 303^0, then simplify x9x4\frac{x^9}{x^4}, then work out 27−1327^{-\frac{1}{3}}. Because Paper 1 is non-calculator, the powers of 2, 3, 4 and 5 and the cube roots of 8, 27, 64 and 125 have to be in your head rather than in the machine. Papers 2 and 3 allow a calculator, which shortens the arithmetic but changes nothing about the algebraic simplifications, which are still done by hand.

The lost marks are predictable. Multiplying the indices instead of adding them turns 34×323^4 \times 3^2 into 383^8; a negative index gets read as a negative answer, so 2−32^{-3} comes back as −8-8; the number in front of a bracket is left behind, so (3x2)3(3x^2)^3 is written as 3x63x^6. There is also time lost: on Paper 1 a candidate who works out 3123^{12} longhand instead of subtracting indices usually runs out of room and out of minutes. Practise saying which law applies before you write a single symbol.

Worked example

(a) Work out the value of 16−3416^{-\frac{3}{4}} (b) Simplify fully (2c3)4×3c2(2c^3)^4 \times 3c^2
[4 marks]
  1. (a) The denominator of the index is the root, so deal with that first: 1614=216^{\frac{1}{4}} = 2, because 2×2×2×2=162 \times 2 \times 2 \times 2 = 16.
    M1: taking the fourth root before the power, which keeps the numbers small
  2. The numerator is the power and the minus sign means reciprocal: 1634=23=816^{\frac{3}{4}} = 2^3 = 8, so 16−34=1816^{-\frac{3}{4}} = \frac{1}{8}.
    A1: 18\frac{1}{8}; a negative index gives a fraction, never a negative answer
  3. (b) The power outside the bracket applies to every factor inside it: (2c3)4=24×c12=16c12(2c^3)^4 = 2^4 \times c^{12} = 16c^{12}.
    M1: raising the 2 as well as the letter, and multiplying the indices 3×43 \times 4
  4. Now multiply by 3c23c^2: the numbers give 16×3=4816 \times 3 = 48 and the letters give c12×c2=c14c^{12} \times c^2 = c^{14}, so the answer is 48c1448c^{14}.
    A1: 48c1448c^{14}, coefficient and index both correct

Practice questions

These are original questions written in Edexcel 1MA1 Higher style. Indices can appear on any of the three papers, but the negative and fractional powers are most at home on Paper 1, so work through these without a calculator first and give exact answers (you can type a fraction such as 1/8). Several questions ask for the index rather than the value, which is exactly how Edexcel keeps the arithmetic short while still testing the law.

No account needed: the questions below are always free to check, plus 8 extra questions today. Your score so far updates as you go.

  1. Question 1 Warm-up [1 mark]
    56×535^6 \times 5^3 can be written in the form 5n5^n. Write down the value of nn.
  2. Question 2 Exam pace [1 mark]
    Work out the value of 271327^{\frac{1}{3}}.
  3. Question 3 Exam pace [2 marks]
    Work out the value of 8238^{\frac{2}{3}}.
  4. Question 4 Exam pace [2 marks]
    2102n=23\frac{2^{10}}{2^n} = 2^3 Find the value of nn.
  5. Question 5 Stretch [3 marks]
    Work out the value of (254)−12\left(\frac{25}{4}\right)^{-\frac{1}{2}}.
  6. Question 6 Stretch [3 marks]
    xx is a positive number. Simplify fully (8x9)23(8x^9)^{\frac{2}{3}}.

Common mistakes examiners see

  • Multiplying the indices when two powers are multiplied, so 34×323^4 \times 3^2 is written as 383^8 instead of 363^6.

    Sort the three laws by the operation between the terms: multiply the terms, add the indices; divide the terms, subtract the indices; multiply the indices only when a power sits outside a bracket. Say which of the three you are using before you write the answer.

  • Reading a negative index as a negative answer, giving 2−3=−82^{-3} = -8 or −6-6.

    Write the reciprocal line out in full: 2−3=123=182^{-3} = \frac{1}{2^3} = \frac{1}{8}. That extra line is often the method mark, and it makes a negative answer impossible when the base is positive.

  • Leaving the coefficient outside the bracket untouched, so (3x2)3(3x^2)^3 becomes 3x63x^6 rather than 27x627x^6.

    Split the bracket into its factors before you raise it: (3x2)3=33×(x2)3(3x^2)^3 = 3^3 \times (x^2)^3. Cube the 3 first, then deal with the letter.

  • Applying the zero index to the whole term, so 5x05x^0 is given as 1 instead of 5, or writing x0=0x^0 = 0.

    The index attaches only to what it sits on. In 5x05x^0 only the xx is raised to the power 0, so the term is 5×1=55 \times 1 = 5. Any non-zero base to the power 0 is 1, not 0.

  • Treating a fractional index as a division, so 251225^{\frac{1}{2}} is answered as 12.5, or doing the power before the root and getting stuck computing 82=648^2 = 64 before rooting.

    Read the denominator as the root and the numerator as the power, then take the root first: 8238^{\frac{2}{3}} is (83)2=22=4(\sqrt[3]{8})^2 = 2^2 = 4. Rooting first keeps every number on a non-calculator paper small enough to handle.

Frequently asked questions

Are fractional indices on the Foundation paper?
No. Foundation tier covers the index laws, the zero index and negative integer powers. Positive and negative fractional indices, under spec reference N7, are Higher tier only. A Higher question can mix the two, for example asking for 16−3416^{-\frac{3}{4}}, so you need the negative-index rule and the fractional-index rule in the same line of working.
How many marks are indices worth on Edexcel Higher?
Individual parts are usually 1-2 marks and a whole question 3-4, so budget roughly 3-5 marks a paper. The real value is larger than that, because the same rules are used in standard form, surds, algebraic fractions and expanding brackets, none of which are labelled as indices questions.
Do I get the laws of indices in the exam?
No. The Edexcel 1MA1 formulae sheet does not list them, so the three laws plus the zero, negative and fractional index rules have to be memorised. They are short enough to write out from memory at the top of your working before you start.
Why does anything to the power of zero equal 1?
Divide 535^3 by 535^3. Any non-zero number divided by itself is 1, and the division law says subtract the indices, which gives 505^0. Both answers describe the same calculation, so 50=15^0 = 1. The same argument works for every non-zero base.

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