OpenRevise GCSE

Viewing as

Student Teacher

Teaching this? The teacher view adds 2 files, the mark schemes and the model answers.

Maths · Algebra

Algebraic indices

The index laws work just as well with letters as with numbers. Deal with the numbers first, then each letter in turn, and even a long expression simplifies in a few lines.

  • 6 key terms
  • All boards
Download the full pack · 3 files

From the Number Chapter

Answer each one, then check.

  1. 1

    Write \(2^5 \times 2^3\) as a single power of 2.

    Show answerHide answer

    \(2^8\)

  2. 2

    Write \((3^2)^4\) as a single power of 3.

    Show answerHide answer

    \(3^8\)

  3. 3

    What is \(5^0\)?

    Show answerHide answer

    \(1\)

  4. 4

    What is \(3^{-2}\)?

    Show answerHide answer

    \(\frac{1}{9}\)

  5. 5

    Simplify \(3a \times 4b\).

    Show answerHide answer

    \(12ab\)

Learning Objectives

  1. 1Use the index laws with letters.
  2. 2Simplify expressions with numbers and several letters.
  3. 3Raise a bracket to a power, such as \((2x^3)^4\).
  4. 4Use negative and fractional indices in algebra. (Higher)
  5. 5Solve equations with unknown powers, such as \(9^x = 27^{x-1}\). (Higher)

The Index Laws with Letters

Exactly the laws from the Number chapter - the base is now a letter.

  • Multiplying

    In words: Add the indices. Example: \(x^5 \times x^2 = x^7\)

  • Dividing

    In words: Subtract the indices. Example: \(y^8 \div y^2 = y^6\)

  • Power of a power

    In words: Multiply the indices. Example: \((z^3)^5 = z^{15}\)

  • Zero index

    In words: Anything (except 0) to the power 0 is 1. Example: \(x^0 = 1\)

  • A letter on its own

    In words: Has an index of 1. Example: \(x \times x^4 = x^1 \times x^4 = x^5\)

Multiplying Terms

Simplify \(3x^2y^4 \times 5x^3y\)

Show the solutionHide the solution
  1. 1 Multiply the numbers \(3 \times 5 = 15\)
  2. 2 Multiply the \(x\) terms: add the indices \(x^2 \times x^3 = x^5\)
  3. 3 Multiply the \(y\) terms: \(y\) is \(y^1\) \(y^4 \times y^1 = y^5\)
  4. 4 Put them together \(15x^5y^5\)

Answer\(15x^5y^5\)

Dividing Terms

Simplify \(24a^7b^3 \div 8a^2b^3\)

Show the solutionHide the solution
  1. 1 Divide the numbers \(24 \div 8 = 3\)
  2. 2 Divide the \(a\) terms: subtract the indices \(a^7 \div a^2 = a^5\)
  3. 3 Divide the \(b\) terms \(b^3 \div b^3 = b^0 = 1\)
  4. 4 Put them together \(3 \times a^5 \times 1\)

Answer\(3a^5\)

A Bracket to a Power

Simplify \((3x^2y^5)^3\)

Show the solutionHide the solution
  1. 1 Everything inside the bracket is raised to the power 3 \(3^3 \times (x^2)^3 \times (y^5)^3\)
  2. 2 The number \(3^3 = 27\)
  3. 3 Power of a power: multiply the indices \((x^2)^3 = x^6\) and \((y^5)^3 = y^{15}\)

Answer\(27x^6y^{15}\)

Mistakes to Avoid

Wrong

  • \((2x)^3 = 2x^3\)
  • \(3x^2 \times 2x = 6x^2\)
  • \(x^2 + x^3 = x^5\)
  • \((x^3)^2 = x^5\)
  • \(12x^6 \div 4x^2 = 3x^3\)

Right

  • \((2x)^3 = 8x^3\) - cube the 2 as well.
  • \(3x^2 \times 2x = 6x^3\) - \(x\) is \(x^1\).
  • \(x^2 + x^3\) cannot be simplified: they are not like terms.
  • \((x^3)^2 = x^6\) - multiply the indices.
  • \(12x^6 \div 4x^2 = 3x^4\) - subtract the indices.

Negative and Fractional Powers of Letters

The meanings are the same as for numbers.

  • Negative powers

    \(x^{-1} = \dfrac{1}{x}\) and \(x^{-3} = \dfrac{1}{x^3}\). So \(\dfrac{1}{x^4}\) can be written as \(x^{-4}\).

  • Fractional powers

    \(x^{\frac{1}{2}} = \sqrt{x}\) and \(x^{\frac{1}{3}} = \sqrt[3]{x}\). So \(\dfrac{1}{\sqrt{x}} = x^{-\frac{1}{2}}\).

  • "Write in the form \(x^n\)"

    Turn roots and fractions into single powers: \(\dfrac{1}{x^2} = x^{-2}\), \(\sqrt{x^3} = x^{\frac{3}{2}}\).

  • Brackets

    Raise each part inside the bracket to the power: \((16x^8)^{\frac{1}{2}} = 16^{\frac{1}{2}} \times x^4 = 4x^4\).

A Negative Fractional Power

Simplify \((27x^6y^{-3})^{-\frac{1}{3}}\)

Show the solutionHide the solution
  1. 1 Raise each part to the power \(-\frac{1}{3}\) \(27^{-\frac{1}{3}} \times (x^6)^{-\frac{1}{3}} \times (y^{-3})^{-\frac{1}{3}}\)
  2. 2 The number: flip and cube root \(27^{-\frac{1}{3}} = \dfrac{1}{3}\)
  3. 3 The \(x\): multiply the indices \(6 \times \left(-\frac{1}{3}\right) = -2\), so \(x^{-2} = \dfrac{1}{x^2}\)
  4. 4 The \(y\): multiply the indices \(-3 \times \left(-\frac{1}{3}\right) = 1\), so \(y^1 = y\)
  5. 5 Put them together \(\dfrac{1}{3} \times \dfrac{1}{x^2} \times y\)

Answer\(\dfrac{y}{3x^2}\)

Solving an Equation with Powers

Solve \(2^{x+1} = 8^x\)

Show the solutionHide the solution
  1. 1 Write both sides with the same base \(8 = 2^3\), so \(8^x = 2^{3x}\)
  2. 2 So \(2^{x+1} = 2^{3x}\)
  3. 3 Same base, so the indices are equal \(x + 1 = 3x\)
  4. 4 Solve \(1 = 2x\)

Answer\(x = \dfrac{1}{2}\)

Index Pyramids

Simplify each. (a) \(a^4 \times a^6\) (b) \(6m^5 \times 2m\) (c) \(18p^8q^3 \div 6p^2q\) (d) \((4x^3)^2\) (e) \((2a^2b)^3 \times 3ab^4\) (f) \(\dfrac{x^7 \times x^2}{x^5}\). Higher: (g) \((9x^4)^{\frac{1}{2}}\) (h) \((8y^6)^{-\frac{1}{3}}\) (i) solve \(4^x = 2^{x+3}\).

1. Numbers first.

2. Then each letter in turn.

3. Check every bracket's power applies to everything inside.

A good answer shows: (a) \(a^{10}\) (b) \(12m^6\) (c) \(3p^6q^2\) (d) \(16x^6\) (e) \(8a^6b^3 \times 3ab^4 = 24a^7b^7\) (f) \(x^4\) (g) \(3x^2\) (h) \(\dfrac{1}{2y^2}\) (i) \(2^{2x} = 2^{x+3}\), so \(x = 3\).

Can I...?

  1. 1Multiply terms using the index laws.
  2. 2Divide terms using the index laws.
  3. 3Raise a power to a power.
  4. 4Raise a bracket to a power.
  5. 5Use \(x^0 = 1\).
  6. 6Use negative powers of letters. (Higher)
  7. 7Use fractional powers of letters. (Higher)
  8. 8Solve equations with unknown powers. (Higher)

Summary & Exam Focus

  • Multiply: add indices. Divide: subtract indices. Power of a power: multiply indices.
  • A power on a bracket applies to everything inside, including the number.
  • Only like terms can be added; \(x^2 + x^3\) does not simplify.
  • (Higher) \(x^{-n} = \dfrac{1}{x^n}\) and \(x^{\frac{1}{n}} = \sqrt[n]{x}\).

Exam focus

Simplify \((2a^3b)^5\) (2 marks) (2 marks)

Work through the numbers first, then each letter in alphabetical order. The two marks are usually one for the number and one for the letters.

Key terms

The vocabulary this lesson expects you to use. Each one is linked from the first place it appears above.

Index (power)
The small number showing how many times the base is multiplied by itself.
Coefficient
The number in front of a letter, e.g. 5 in \(5x^3\).
Like terms
Terms with exactly the same letters and powers, e.g. \(3x^2\) and \(7x^2\).
Simplify
Write an expression in its shortest form.
Negative index (Higher)
\(x^{-n} = \dfrac{1}{x^n}\).
Fractional index (Higher)
\(x^{\frac{1}{n}} = \sqrt[n]{x}\).

Practice questions

Have a go at each one before you open its answer.

  1. Question 1 Non-calculator 1 mark

    Simplify \(m^5 \times m^3\)

    Show answerHide answer

    Model answer

    \(m^8\)

    Mark scheme

    • \(m^8\) — B1
  2. Question 2 Non-calculator 1 mark

    Simplify \((p^4)^3\)

    Show answerHide answer

    Model answer

    \(p^{12}\)

    Mark scheme

    • \(p^{12}\) — B1
  3. Question 3 Non-calculator 2 marks

    Simplify \(4x^2y^3 \times 3x^5y\)

    Show answerHide answer

    Model answer

    \(12x^7y^4\)

    Mark scheme

    • \(12x^7y^4\) — B2
    • Two of \(12\), \(x^7\) and \(y^4\) in a single term — B1
  4. Question 4 Non-calculator 2 marks

    Simplify \((2a^3b)^5\)

    Show answerHide answer

    Model answer

    \(32a^{15}b^5\)

    Mark scheme

    • \(32a^{15}b^5\) — B2
    • Two of \(32\), \(a^{15}\) and \(b^5\) in a single term — B1
  5. Question 5 Non-calculator · Higher 2 marks

    Simplify \((64x^6)^{\frac{2}{3}}\)

    Show answerHide answer

    Model answer

    \(64^{\frac{2}{3}} = \left(\sqrt[3]{64}\right)^2 = 4^2 = 16\) and \((x^6)^{\frac{2}{3}} = x^4\), so \(16x^4\).

    Mark scheme

    • \(16x^4\) — B2
    • \(16\) or \(x^4\) in a single term — B1
  6. Question 6 Non-calculator · Higher 3 marks

    Solve \(9^x = 27^{x-1}\)

    Show answerHide answer

    Model answer

    \(9 = 3^2\) and \(27 = 3^3\), so \(3^{2x} = 3^{3x-3}\). Then \(2x = 3x - 3\), so \(x = 3\).

    Mark scheme

    • Both sides written as powers of 3 — M1
    • \(2x = 3x - 3\) — M1
    • \(x = 3\) — A1

Quick check

  1. Simplify \(x^3 \times x^4\)

    1. A\(x^7\)
    2. B\(x^{12}\)
    3. C\(2x^7\)
    4. D\(x^1\)
    Show answerHide answer

    A: \(x^7\)

    Multiplying: add the indices, \(3 + 4 = 7\).

  2. Simplify \((3x)^2\)

    1. A\(3x^2\)
    2. B\(6x^2\)
    3. C\(9x^2\)
    4. D\(9x\)
    Show answerHide answer

    C: \(9x^2\)

    Everything in the bracket is squared: \(3^2 \times x^2 = 9x^2\).

  3. (Higher) Write \(\dfrac{1}{\sqrt{x}}\) as a power of \(x\).

    1. A\(x^2\)
    2. B\(x^{\frac{1}{2}}\)
    3. C\(x^{-2}\)
    4. D\(x^{-\frac{1}{2}}\)
    Show answerHide answer

    D: \(x^{-\frac{1}{2}}\)

    \(\sqrt{x} = x^{\frac{1}{2}}\), and one over it is \(x^{-\frac{1}{2}}\).

Downloads

Free to keep, print and annotate.

Download all 3 as a zip
Something here looks wrong?

Tell us what and we will go and look. It goes to whoever writes these pages, nobody else, and we do not ask who you are — so there is nothing to sign and nothing comes back to you.