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Maths · Number

Calculating with powers (indices)

Powers are a shorthand for repeated multiplication. Three index laws let you multiply, divide and raise powers without ever writing out the long multiplication - as long as the bases are the same.

  • 7 key terms
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Last Lesson and Before

Answer each one, then check.

  1. 1

    What is \(5^2\)?

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    \(25\)

  2. 2

    What is \(\sqrt{81}\)?

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    \(9\)

  3. 3

    Last lesson: write 72 as a product of prime factors in index form.

    Show answerHide answer

    \(2^3 \times 3^2\)

  4. 4

    Last lesson: find the HCF of 12 and 18.

    Show answerHide answer

    \(6\)

  5. 5

    Last lesson: find the LCM of 4 and 10.

    Show answerHide answer

    \(20\)

Learning Objectives

  1. 1Know the square numbers to \(15^2\) and the cubes of 1, 2, 3, 4, 5 and 10.
  2. 2Find square roots and cube roots, including negative roots.
  3. 3Use index notation for powers.
  4. 4Use the index laws to multiply, divide and raise powers.
  5. 5Use the order of operations with powers and roots, with and without a calculator.

Powers and Roots

A power (or index) tells you how many times a number is multiplied by itself.

  • Index notation

    \(5^3 = 5 \times 5 \times 5 = 125\). The 5 is the base; the 3 is the index or power.

  • Squares and square roots

    \(7^2 = 49\), so \(\sqrt{49} = 7\). But the equation \(x^2 = 49\) has two answers: \(x = 7\) or \(x = -7\), since \((-7)^2 = 49\) too.

  • Cubes and cube roots

    \(4^3 = 64\), so \(\sqrt[3]{64} = 4\). A cube root of a negative number is negative: \(\sqrt[3]{-8} = -2\).

  • Roots undo powers

    Squaring and square-rooting are inverse operations, like \(\times\) and \(\div\).

Numbers to Know by Heart

  • Square numbers

    1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225 (up to \(15^2\)).

  • Cube numbers

    1, 8, 27, 64, 125 (\(1^3\) to \(5^3\)) and \(10^3 = 1000\).

  • Powers of 2

    2, 4, 8, 16, 32, 64, 128, 256, 512, 1024 (\(2^1\) to \(2^{10}\)).

  • Powers of 10

    10, 100, 1000, 10 000 - the index is the number of zeros.

The Three Index Laws

They only work when the base is the same.

  • Multiplying

    In words: Add the indices. Example: \(3^4 \times 3^2 = 3^6\)

  • Dividing

    In words: Subtract the indices. Example: \(5^7 \div 5^3 = 5^4\)

  • Power of a power

    In words: Multiply the indices. Example: \((2^3)^4 = 2^{12}\)

  • Different bases

    In words: The laws do not apply: work each power out. Example: \(2^3 \times 5^2 = 8 \times 25 = 200\)

Why the Laws Work

Write the powers out in full and count.

  • Multiplying

    \(3^4 \times 3^2 = (3 \times 3 \times 3 \times 3) \times (3 \times 3)\): six 3s multiplied, so \(3^6\).

  • Dividing

    \(5^7 \div 5^3\): three of the seven 5s cancel, leaving four: \(5^4\).

  • Power of a power

    \((2^3)^4 = 2^3 \times 2^3 \times 2^3 \times 2^3\): four lots of three 2s, so \(2^{12}\).

  • In letters

    \(a^m \times a^n = a^{m+n}\), \(a^m \div a^n = a^{m-n}\) and \((a^m)^n = a^{mn}\).

Using the Index Laws

Work out the value of \(\dfrac{7^5 \times 7^3}{7^6}\)

Show the solutionHide the solution
  1. 1 Multiplying: add the indices \(7^5 \times 7^3 = 7^8\)
  2. 2 Dividing: subtract the indices \(7^8 \div 7^6 = 7^2\)
  3. 3 Work out the value \(7^2 = 49\)

Answer\(49\)

Changing the Base

Write \(4^3 \times 8^2\) as a single power of 2.

Show the solutionHide the solution
  1. 1 Write each base as a power of 2 \(4 = 2^2\) and \(8 = 2^3\)
  2. 2 Power of a power: multiply \(4^3 = (2^2)^3 = 2^6\) and \(8^2 = (2^3)^2 = 2^6\)
  3. 3 Multiplying: add \(2^6 \times 2^6 = 2^{12}\)
  4. 4 Check \(4^3 \times 8^2 = 64 \times 64 = 4096 = 2^{12}\)

Answer\(2^{12}\)

Order of Operations

Powers and roots come before multiplying, dividing, adding and subtracting.

  • BIDMAS

    Brackets, Indices (powers and roots), Division and Multiplication, Addition and Subtraction.

  • Negative numbers

    \(-3^2 = -9\) (square 3, then make it negative), but \((-3)^2 = 9\).

  • A root sign is a bracket

    \(\sqrt{9 + 16} = \sqrt{25} = 5\). It is NOT \(\sqrt{9} + \sqrt{16} = 7\).

  • On a calculator

    Use brackets for the top and bottom of a fraction, and write down the full display before rounding.

A Non-Calculator Calculation

Work out \(2^3 \times 5^2 - \sqrt{144}\)

Show the solutionHide the solution
  1. 1 Powers and roots first \(2^3 = 8\), \(5^2 = 25\), \(\sqrt{144} = 12\)
  2. 2 Then multiply \(8 \times 25 = 200\)
  3. 3 Then subtract \(200 - 12 = 188\)

Answer\(188\)

Mistakes to Avoid

Wrong

  • \(3^4 \times 3^2 = 9^6\)
  • \(2^3 = 6\)
  • \(5^2 + 5^3 = 5^5\)
  • \((-4)^2 = -16\)

Right

  • \(3^4 \times 3^2 = 3^6\) - the base stays the same.
  • \(2^3 = 2 \times 2 \times 2 = 8\)
  • \(5^2 + 5^3 = 25 + 125 = 150\) - no law for adding.
  • \((-4)^2 = 16\) - negative times negative is positive.

Case study

Rice on a Chessboard

An old legend tells of a clever inventor who showed a king the game of chess. As a reward he asked for one grain of rice on the first square of the board, two on the second, four on the third, and so on, doubling each time. The king laughed at such a small request - until his treasurers did the maths. The 64th square alone needs \(2^{63}\) grains, and the whole board needs \(2^{64} - 1\): about 18 quintillion grains, far more rice than has ever been grown. The story is a legend, but the numbers are real, and they show how fast powers grow.

\(2^{63}\) Grains on the last square alone
\(1.8 \times 10^{19}\) Grains on the whole board

Power Pyramids

Write each as a single power, then as an ordinary number. (a) \(2^5 \times 2^3\) (b) \(10^8 \div 10^5\) (c) \((3^2)^3\) (d) \(5^9 \div (5^2 \times 5^4)\) (e) \(9^2 \times 3^3\) as a power of 3 (f) find \(n\) if \(2^n = 4^5 \div 2^3\)

1. Use the index laws.

2. Change a base where you need to.

3. Check one answer the long way.

A good answer shows: (a) \(2^8 = 256\) (b) \(10^3 = 1000\) (c) \(3^6 = 729\) (d) \(5^3 = 125\) (e) \(3^4 \times 3^3 = 3^7 = 2187\) (f) \(4^5 = 2^{10}\), so \(2^{10} \div 2^3 = 2^7\) and \(n = 7\)

Can I...?

  1. 1Recall square numbers to \(15^2\).
  2. 2Recall the cubes of 1, 2, 3, 4, 5 and 10.
  3. 3Find square roots and cube roots.
  4. 4Give both square roots of a number.
  5. 5Multiply powers of the same base.
  6. 6Divide powers of the same base.
  7. 7Raise a power to a power.
  8. 8Use BIDMAS with powers and roots.

Summary & Exam Focus

  • \(5^3\) means \(5 \times 5 \times 5\); a root undoes a power.
  • Same base: multiply, add indices; divide, subtract indices; power of a power, multiply indices.
  • Different bases: work each power out separately.
  • Powers and roots come before \(\times\), \(\div\), \(+\) and \(-\); a root sign acts as a bracket.

Exam focus

Show that \(9^4 \times 27^2 = 3^{14}\) (3 marks) (3 marks)

In a "show that" question the answer is given, so the marks are all for the working. Write every step, including \(9 = 3^2\) and \(27 = 3^3\).

Key terms

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

Power (index)
The small number that says how many times the base is multiplied by itself.
Base
The number being multiplied, e.g. the 5 in \(5^3\).
Square number
A number made by multiplying a whole number by itself, e.g. \(49 = 7^2\).
Cube number
A number made by multiplying a whole number by itself three times, e.g. \(64 = 4^3\).
Square root
The number that squares to give a number: \(\sqrt{49} = 7\).
Cube root
The number that cubes to give a number: \(\sqrt[3]{64} = 4\).
Index laws
The rules for multiplying, dividing and raising powers of the same base.

Practice questions

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

  1. Question 1 Non-calculator 1 mark

    Write down the value of \(3^4\).

    Show answerHide answer

    Model answer

    \(81\)

    Mark scheme

    • \(81\) — B1
  2. Question 2 Non-calculator 2 marks

    Write \(\dfrac{5^6 \times 5^3}{5^4}\) as a single power of 5.

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    Model answer

    \(5^6 \times 5^3 = 5^9\), and \(5^9 \div 5^4 = 5^5\)

    Mark scheme

    • \(5^9\) seen, or a correct method for dividing — M1
    • \(5^5\) — A1
  3. Question 3 Non-calculator 2 marks

    Work out \(2^3 \times 5^2 - \sqrt{144}\)

    Show answerHide answer

    Model answer

    \(8 \times 25 - 12 = 200 - 12 = 188\)

    Mark scheme

    • Two of 8, 25 and 12 seen, or 200 seen — M1
    • \(188\) — A1
  4. Question 4 Non-calculator 2 marks

    \(2^n = 4^5 \div 2^3\). Find the value of \(n\).

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    Model answer

    \(4^5 = (2^2)^5 = 2^{10}\), so \(2^{10} \div 2^3 = 2^7\) and \(n = 7\).

    Mark scheme

    • \(4^5\) written as \(2^{10}\), or \(1024 \div 8 = 128\) — M1
    • \(n = 7\) — A1
  5. Question 5 Non-calculator · Show that 3 marks

    Show that \(9^4 \times 27^2 = 3^{14}\)

    Show answerHide answer

    Model answer

    \(9 = 3^2\), so \(9^4 = (3^2)^4 = 3^8\). \(27 = 3^3\), so \(27^2 = (3^3)^2 = 3^6\). \(3^8 \times 3^6 = 3^{14}\).

    Mark scheme

    • \(9 = 3^2\) or \(27 = 3^3\) used — M1
    • \(3^8\) and \(3^6\) — M1
    • Fully correct working leading to \(3^{14}\) — A1
  6. Question 6 Calculator 2 marks

    (a) Work out \(\dfrac{2.7^3 - \sqrt{18.5}}{1.4^2}\). Write down all the figures on your calculator display. (b) Give your answer to part (a) correct to 3 significant figures.

    Show answerHide answer

    Model answer

    (a) \(7.847876207\) (b) \(7.85\)

    Mark scheme

    • (a) \(7.8478\ldots\) (at least 5 figures) — B1
    • (b) \(7.85\), follow through from (a) — B1

Quick check

  1. What is the value of \(4^3\)?

    1. A12
    2. B16
    3. C64
    4. D81
    Show answerHide answer

    C: 64

    \(4^3 = 4 \times 4 \times 4 = 64\).

  2. Which is \(2^5 \times 2^3\) as a single power?

    1. A\(2^8\)
    2. B\(2^{15}\)
    3. C\(4^8\)
    4. D\(4^{15}\)
    Show answerHide answer

    A: \(2^8\)

    Multiplying powers of the same base: add the indices, \(5 + 3 = 8\).

  3. What is the value of \(-5^2\)?

    1. A\(25\)
    2. B\(-25\)
    3. C\(-10\)
    4. D\(10\)
    Show answerHide answer

    B: \(-25\)

    The power comes first: \(5^2 = 25\), then the minus sign makes it \(-25\). \((-5)^2\) would be 25.

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