A4cLaws of indices in algebra

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

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Revision notes

Using the laws of indices to simplify algebraic expressions such as \(3a^4 \times 5a^2\) or \((2x^3)^4\). Both tiers use the multiplying, dividing and power laws with zero and negative indices; Higher also uses fractional indices.

Grade by grade

What you need to be able to do, from the first marks up to the top grade.

4
Multiply powers of the same letterAdd the indices: \(a^4 \times a^3 = a^7\).
4
Divide powers of the same letterSubtract the indices: \(y^9 \div y^4 = y^5\).
5
Raise a power to a powerMultiply the indices: \((x^3)^5 = x^{15}\).
5
Simplify terms with numbers and lettersDeal with the numbers and each letter separately, e.g. \(4a^2b \times 3ab^5 = 12a^3b^6\).
5
Raise a whole term to a powerThe number is raised to the power too: \((3x^4)^2 = 9x^8\).
5
Use zero and negative indices\(x^0 = 1\) and \(x^{-n} = \frac{1}{x^n}\), e.g. \(x^{-2} = \frac{1}{x^2}\).

Notes

The three main laws

Multiplying: add the indices. \(a^m \times a^n = a^{m + n}\), e.g. \(x^5 \times x^3 = x^8\).
Dividing: subtract the indices. \(a^m \div a^n = a^{m - n}\), e.g. \(x^7 \div x^2 = x^5\).
Power of a power: multiply the indices. \((a^m)^n = a^{mn}\), e.g. \((x^4)^3 = x^{12}\).
These laws only work for the same base: \(x^2 \times y^3\) cannot be simplified.
A letter on its own has index 1: \(x = x^1\).

Numbers and letters together

Multiply or divide the numbers as normal, then use the laws on each letter: \(6a^5b^2 \div 2a^3b = 3a^2b\).
A power outside a bracket applies to everything inside: \((2x^3y)^4 = 2^4x^{12}y^4 = 16x^{12}y^4\).

Zero and negative indices

\(a^0 = 1\) for any non-zero a, so \(7x^0 = 7 \times 1 = 7\).
\(a^{-n} = \frac{1}{a^n}\), e.g. \(x^{-3} = \frac{1}{x^3}\).
\(5x^{-2} = \frac{5}{x^2}\): only the x has the negative index, not the 5.

Cheatsheet

\(a^m \times a^n = a^{m + n}\)
\(a^m \div a^n = a^{m - n}\)
\((a^m)^n = a^{mn}\)
\((ab)^n = a^nb^n\)
\(a^0 = 1\) and \(a^1 = a\)
\(a^{-n} = \frac{1}{a^n}\)

How to answer each type of question

Simplify using one law

1 mark each4

Decide which law applies: ×, ÷ or a power of a power.
Add, subtract or multiply the indices.

Example. Simplify (a) \(p^6 \times p^3\) (b) \(q^{10} \div q^2\) (c) \((r^4)^5\)

Show the model answer

(a) \(p^9\) B1
(b) \(q^8\) B1
(c) \(r^{20}\) B1

Simplify fully (numbers and letters)

2 marks5

Divide (or multiply) the numbers.
Use the index laws on each letter in turn.
Write one number and each letter once.

Example. Simplify fully \(\frac{12x^7y^3}{4x^2y}\)

Show the model answer

\(3x^5y^2\) B2 (B1 for two of 3, \(x^5\), \(y^2\) correct in a product)

Shortcuts and memory tricks

Multiplying terms means adding indices; dividing means subtracting; a power of a power means multiplying.
A negative index means 'one over': \(x^{-1} = \frac{1}{x}\).
Higher: in a fractional index, the denominator is the root (roots are at the bottom, like a tree's).
Check with a number: put \(x = 2\) into the question and your answer.

Where marks are lost

Multiplying indices instead of adding: \(x^3 \times x^4\) is \(x^7\), not \(x^{12}\).
Adding the numbers in front instead of multiplying: \(3x^2 \times 4x^3 = 12x^5\), not \(7x^5\).
Not applying the power to the number: \((2x^3)^4 = 16x^{12}\), not \(2x^{12}\).
Thinking a negative index makes the answer negative: \(x^{-2} = \frac{1}{x^2}\), not \(-x^2\).
Writing \(x^0 = 0\): it equals 1.

Exam technique

'Simplify fully' means one number and each letter written once with a single index.
If the question asks for a particular form (e.g. 'without negative indices'), rewrite your answer to match it.
In 'find the value of n' questions, write everything as a power of the same base before combining.

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 \(m^{4} \times m^{4}\)[1]

(b) Simplify \((x^{4})^{4}\)[1]

Show the answer and mark scheme

(a) Answer: \(m^{8}\)

B1 for \(m^{8}\)

(b) Answer: \(x^{16}\)

B1 for \(x^{16}\)

B2 for \(30p^{8}q^{6}\) (B1 for two of \(30\), \(p^{8}\), \(q^{6}\) correct in a product)

Worked solution: Multiply the numbers and add the indices: \(6 \times 5 = 30\), \(p^{4 + 4} = p^{8}\), \(q^{3 + 3} = q^{6}\)

Question 2Medium5 marks

(a) Simplify \((3a^{2}b^{4})^{2}\)[2]

(b) Simplify \(\frac{9a^{6}b}{3a^{3}b^{4}}\)[2]

Show the answer and mark scheme

(a) Answer: \(9a^{4}b^{8}\)

B2 for \(9a^{4}b^{8}\) (B1 for two of \(9\), \(a^{4}\), \(b^{8}\) correct in a product)

Worked solution: \((3a^{2}b^{4})^{2} = 3^{2} \times a^{2 \times 2} \times b^{4 \times 2} = 9a^{4}b^{8}\)

(b) Answer: \(\frac{3a^{3}}{b^{3}}\)

B2 for \(\frac{3a^{3}}{b^{3}}\) oe (B1 for two of \(3\), \(a^{3}\), \(b^{-3}\) correct)

Worked solution: Divide the numbers and subtract the indices: \(9 \div 3 = 3\), \(a^{6 - 3} = a^{3}\), \(b^{1 - 4} = b^{-3}\), so the answer is \(\frac{3a^{3}}{b^{3}}\)

B1 for \(y^{2}\)

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