Laws of indices | Mathematics

Unit 1 · Number & Algebra

5 Laws of Indices

Writing x × x × x × x × x × x × x is exhausting. Writing x⁷ is not. This lesson is about the shortcut — and the three rules that let you work with it without ever writing the long version again.

🎯 By the end of this you can…

✓Multiply powers by adding the indices.
✓Divide powers by subtracting the indices.
✓Raise a power to a power by multiplying the indices.
✓Handle the numbers in front separately from the letters.

💡 The idea

What an index actually means

$x^{5}$ just means five x’s multiplied together: $x \times x \times x \times x \times x$ . The little number counts how many.

Every rule in this lesson comes from that one sentence. If you ever forget a rule, write the x’s out and count them. You will never be wrong.

📋 Rule 1 · Multiplying — ADD the indices

1
Take $x^{3} \times x^{4}$ . Write it out longhand.
2
$(x \times x \times x) \times (x \times x \times x \times x)$
3
Now just count them. Three x’s, then four x’s — that is seven x’s.
4
So $x^{3} \times x^{4} = x^{7}$ . And 3 + 4 = 7. You add the indices.

✏️ Worked example

Simplify $5 e^{2} f \times 4 e^{4} g$

1
Numbers first: 5 × 4 = 20. Numbers multiply normally — never add them.
2
Then each letter separately. For e: $e^{2} \times e^{4} = e^{6}$ (2 + 4).
3
For f: there is only one f, so it stays $f$ . Same for g.
4
Put it together.

✅ Answer: $20 e^{6} f g$

⚠️ The classic trap

Do not add the numbers in front. In $5 e^{2} \times 4 e^{4}$ the answer is $20 e^{6}$ — the 5 and 4 multiply, only the indices add.

Two different jobs happening at once. Do the numbers, then do the letters, and never let the rules cross over.

🎮 Rule 1 · Multiplying

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Write

t \times t \times t \times t \times t \times t

in power form

Write

k \times k \times k \times k \times k \times k \times k

in power form

n^{4} \times n^{3}

n^{5} \times n

n^{2} \times n^{3}

n^{6} \times n^{2} \times n^{4}

x^{5} \times x^{6}

x^{7} \times x \times x^{3}

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📋 Rule 2 · Dividing — SUBTRACT the indices

1
Take $\frac{x ^{5}}{x ^{2}}$ . Write both out longhand.
2
$\frac{x \times x \times x \times x \times x}{x \times x}$
3
Cancel one x on top with one x on the bottom. Do it twice.
4
Three x’s survive, so the answer is $x^{3}$ . And 5 − 2 = 3. You subtract.

✏️ Worked example

Simplify $\frac{14 x ^{4} y ^{7}}{6 x ^{5} y ^{4}}$

1
Numbers first: $\frac{14}{6}$ cancels to $\frac{7}{3}$ .
2
Now x: 4 − 5 = −1. A negative index means it belongs on the bottom.
3
Now y: 7 − 4 = 3, so $y^{3}$ stays on top.
4
Put it together with only positive powers.

✅ Answer: $\frac{7 y ^{3}}{3 x}$

💡 The idea

What a negative index means

A negative index is not a negative number. It means “this belongs on the bottom of a fraction”.

x^{- 3} = \frac{1}{x ^{3}}

Why? Because $\frac{x ^{2}}{x ^{5}}$ leaves three x’s on the bottom, and the rule says 2 − 5 = −3. Both descriptions must agree, so $x^{- 3}$ and $\frac{1}{x ^{3}}$ are the same thing.

And $x^{0} = 1$ — always. Because $\frac{x ^{3}}{x ^{3}}$ is obviously 1, and the rule says 3 − 3 = 0.

🎮 Rule 2 · Dividing

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x^{12} \div x^{6}

\frac{x ^{3} \times x ^{4}}{x ^{5}}

\frac{5 ^{4}}{5}

— give the number

\frac{2 ^{4}}{2 ^{2}}

— give the number

\frac{2 n ^{2}}{n}

\frac{5 n ^{8}}{20 n ^{8}}

\frac{7 n ^{3}}{2 n ^{5}}

\frac{12 y x ^{4}}{10 y x ^{8}}

\frac{18 x ^{8} y ^{8}}{10 x ^{3}}

\frac{8 x ^{3}}{10 x ^{5}}

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📋 Rule 3 · Power of a power — MULTIPLY the indices

1
Take $(x^{3})^{2}$ . The outside 2 means “two of these, multiplied”.
2
$x^{3} \times x^{3}$
3
By Rule 1 that is $x^{6}$ . And 3 × 2 = 6. You multiply.
4
Careful: the outside power hits everything inside the bracket, including the number in front. $(3 a^{2})^{3}$ means $3^{3} \times (a^{2})^{3}$ .

✏️ Worked example

Simplify $(3 a^{2})^{3}$

1
The 3 outside applies to both the number and the letter.
2
Number: $3^{3} = 27$ — not 3 × 3 = 9, and definitely not just 3.
3
Letter: $(a^{2})^{3} = a^{6}$ because 2 × 3 = 6.

✅ Answer: $27 a^{6}$

⚠️ The classic trap

The number in front gets the power too. $(3 a^{2})^{3}$ is $27 a^{6}$ , not $3 a^{6}$ .

Test it on something small: $(2 a)^{3} = 2 a \times 2 a \times 2 a = 8 a^{3}$ . Three 2s multiplied is 8. The bracket really does hit everything.

🎮 Rule 3 · Power of a power

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(3 a^{2})^{3}

(2 n^{4})^{4}

(3 x^{4})^{4}

(6 b^{2})^{2}

(7 y^{4})^{2}

(3 a b^{4})^{4}

(2 x^{4} y^{4})^{3}

(5 m n^{3})^{3}

(x^{2} y^{2})^{2}

(6 y x^{4})^{2}

(u^{4} v^{3})^{2}

(2 x^{4} y^{4})^{4}

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🤔 What is $(4 m^{3})^{2}$ ?

💡 The idea

Putting all three together

Every question from here mixes the rules. Work in this order and nothing goes wrong:

1 · Brackets first (Rule 3) → 2 · Numbers → 3 · Each letter on its own → 4 · Tidy any negative powers to the bottom

🎮 Mixed — including negative and zero powers

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4 m^{4} n^{3} \cdot 4 m n^{4}

2 x^{2} y^{3} \cdot 4 x^{3} y^{2}

3 y^{- 4} \cdot 4 y^{4}

4 m n^{4} \cdot 3 m^{0}

2 y^{4} \cdot 2 x^{0}

3 u v^{- 2} \cdot 3 u

a^{- 4} \cdot 3 ab

2 u^{0} v^{- 3} \cdot 4 u^{- 3} v^{3}

x^{0} y^{- 4} \cdot 4 x^{- 4} y^{2}

2 m^{2} n^{- 1} \cdot 2 n

u^{- 1} \cdot u^{3} v^{- 2}

3 a b^{4} \cdot 4 a^{4} b^{4}

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🤔 What is $7^{0}$ ?

🏅

You finished this lesson!

Three rules, all from counting letters: × → add, ÷ → subtract, bracket → multiply. Plus two facts worth memorising: $x^{0} = 1$ and $x^{- n} = \frac{1}{x ^{n}}$ .

And if you ever blank in an exam — write the letters out and count them. It always works.

Keep reading — this lesson marks itself done automatically.