Lesson 4 ~25 min Unit 1 · Index Laws +85 XP

The Quotient Rule

When dividing powers with the same base, subtract the indices: $a^m \div a^n = a^{m-n}$.

Today's hook: $2^7 \div 2^3$ — you could expand everything and cancel sevens of $2$ against threes of $2$. Or you could just subtract indices: $2^{7-3} = 2^4$. Which is smarter?

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Think First

warm-up

Expand $\dfrac{2^5}{2^2}$ as repeated multiplication and cancel pairs of $2$. How many $2$s remain? What is the link between this and the indices $5$ and $2$?

Record in your workbook.

The Big Idea

+5 XP

The quotient rule says: when dividing powers with the same base, subtract the indices.

Why? Because $\dfrac{a^m}{a^n}$ has $m$ copies of $a$ on top and $n$ copies on the bottom. $n$ of them cancel, leaving $m - n$ copies on top: that's $a^{m-n}$.

$a^m \div a^n = a^{m-n}$ (same base, $a \ne 0$)

Same base only

$2^7 \div 3^4$ canNOT be combined — different bases.

Subtract

Top minus bottom: $\dfrac{a^7}{a^3} = a^{7-3} = a^4$.

Base stays

The base does not change. Only the indices change.

What You'll Master

objectives

Know

$a^m \div a^n = a^{m-n}$ (same base, $a \ne 0$)
Order matters: $m$ (top) minus $n$ (bottom)
If $m < n$, the result has a negative index

Understand

Why subtracting indices follows from cancelling factors
Why the rule requires the same base
How division relates to subtraction in indices

Can Do

Simplify $\dfrac{5^9}{5^4}$, $\dfrac{a^{10}}{a^3}$
Decide when the rule applies
Predict whether the result has a positive or negative index

Words You Need

vocabulary

Quotient rule$a^m \div a^n = a^{m-n}$. Divide same-base powers $\to$ subtract indices.

NumeratorThe top of a fraction. In $\dfrac{a^5}{a^2}$ the numerator is $a^5$.

DenominatorThe bottom of a fraction. Must not be zero.

CancelDivide both top and bottom by the same factor.

QuotientThe result of a division.

Negative indexWhen $m < n$, $a^{m-n}$ is negative — meaning a reciprocal (later).

Spot the Trap

heads-up

Wrong: "$\dfrac{2^7}{2^3} = 2^{7/3}$" — dividing the indices.

Right: Subtract the indices: $\dfrac{2^7}{2^3} = 2^{7-3} = 2^4 = 16$.

Wrong: "$\dfrac{a^3}{a^7} = a^{7-3} = a^4$" — reversing the order.

Right: Top index $-$ bottom index: $3 - 7 = -4$, so $a^{-4}$.

Why It Works — Cancellation

+5 XP

Let's verify by expanding. Take $\dfrac{a^5}{a^2}$:

$\dfrac{a^5}{a^2} = \dfrac{a \cdot a \cdot a \cdot a \cdot a}{a \cdot a}$. Cancel 2 from each: leaves $a \cdot a \cdot a = a^3$. We cancelled 2, leaving $5 - 2 = 3$ copies. The subtraction of indices is just bookkeeping for the cancellation.

$5 - 2 = 3$ — subtraction matches cancellation.

When the Bottom Is Bigger

+5 XP

What if $m < n$? Then $m - n$ is negative. We'll see in Lesson 8 that this gives a reciprocal: $a^{-k} = \dfrac{1}{a^k}$.

Example: $\dfrac{2^3}{2^5} = 2^{3-5} = 2^{-2}$. Equivalently, by cancellation: $\dfrac{2 \cdot 2 \cdot 2}{2 \cdot 2 \cdot 2 \cdot 2 \cdot 2} = \dfrac{1}{2 \cdot 2} = \dfrac{1}{4}$. So $2^{-2} = \dfrac{1}{4}$ — preview of negative indices.

$\dfrac{2^3}{2^5} = 2^{-2} = \dfrac{1}{4}$

Watch Me Solve It · 3 examples

Watch Me Solve It · Simplify a quotient

+15 XP per step

PROBLEM

Simplify $\dfrac{8^9}{8^4}$.

1
Same base?
Both bases are $8$.
Quotient rule applies.
2
Subtract indices
$9 - 4 = 5$
3
Result
$8^5$

Answer$8^5$

Watch Me Solve It · Algebraic quotient

+15 XP per step

PROBLEM

Simplify $\dfrac{x^{10}}{x^3}$.

1
Subtract indices
$10 - 3 = 7$
2
Apply rule
$x^{10-3} = x^7$
3
Verify by expanding
$10$ $x$s on top, $3$ cancel, $7$ remain. $\checkmark$

Answer$x^7$

Watch Me Solve It · Negative index preview

+15 XP per step

PROBLEM

Simplify $\dfrac{a^4}{a^6}$.

1
Subtract indices
$4 - 6 = -2$
2
Write the result
$a^{-2}$
This is shorthand for $\dfrac{1}{a^2}$.
3
Equivalent form
$\dfrac{a^4}{a^6} = \dfrac{1}{a^2}$
Cancel 4 $a$s; 2 remain on the bottom.

Answer$a^{-2}$ or $\dfrac{1}{a^2}$

Common Pitfalls

heads-up

Dividing the indices

$\dfrac{2^6}{2^2} \ne 2^3$. Subtract, not divide: $6 - 2 = 4$ gives $2^4$.

Fix: Multiplication adds; division subtracts. Memorise this pairing.

Wrong subtraction order

$\dfrac{a^3}{a^7} \ne a^4$. Top minus bottom: $3 - 7 = -4$.

Fix: Always top index $-$ bottom index, in that order.

Changing the base

$\dfrac{6^4}{2^4} \ne 3^4$. Different bases — the rule fails.

Fix: Same base required. Evaluate numerically if bases differ.

Copy Into Your Books

Quotient rule

$\dfrac{a^m}{a^n} = a^{m-n}$
Same base, SUBTRACT indices
Top minus bottom

Why?

Cancel $n$ factors from $m$
Leaves $m - n$ factors
Base unchanged

Negative index

If $m < n$, result negative
$a^{-k} = \dfrac{1}{a^k}$
Full treatment: Lesson 8

Restriction

$a \ne 0$ (cannot divide by 0)
Bases must match

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Quotient drill

4 problems

Use the quotient rule to simplify each.

1 Simplify $\dfrac{5^8}{5^3}$.
Same base, subtract: $8 - 3$.$5^5$
2 Simplify $\dfrac{a^{12}}{a^5}$.
$12 - 5 = 7$.$a^7$
3 Simplify $\dfrac{7^4}{7^4}$.
$4 - 4 = 0$. $7^0 = 1$ (lesson 7!).$7^0 = 1$
4 Simplify $\dfrac{2^3}{2^7}$.
$3 - 7 = -4$. Negative index.$2^{-4} = \dfrac{1}{16}$

Complete in your workbook.

Quick Check · 5 questions

Simplify $\dfrac{6^9}{6^4}$.

+10 XP

Simplify $\dfrac{x^{15}}{x^7}$.

+10 XP

Which expression equals $\dfrac{a^5}{a^9}$?

+10 XP

Evaluate $\dfrac{4^7}{4^4}$.

+10 XP

Which CANNOT be simplified using the quotient rule?

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

ApplyMedium3 MARKS

Q6. Simplify each: (a) $\dfrac{m^9}{m^2}$, (b) $\dfrac{4^7}{4^3}$, (c) $\dfrac{y^5}{y^5}$.

Answer in your workbook.

UnderstandEasy2 MARKS

Q7. By expanding and cancelling, prove that $\dfrac{a^6}{a^2} = a^4$.

Answer in your workbook.

ReasonHard4 MARKS

Q8. Show that $\dfrac{2^3 \times 2^5}{2^4} = 2^4$. Use BOTH the product and quotient rules in your working.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. D — $6^5$.

2. A — $x^8$.

3. B — $a^{-4}$.

4. C — $64$.

5. D — different bases cannot combine.

Show Your Working Model Answers

Q6 (3 marks): (a) $m^7$ [1]; (b) $4^4 = 256$ [1]; (c) $y^0 = 1$ [1].

Q7 (2 marks): $\dfrac{a^6}{a^2} = \dfrac{a \cdot a \cdot a \cdot a \cdot a \cdot a}{a \cdot a}$ [1] $= a \cdot a \cdot a \cdot a = a^4$ [1].

Q8 (4 marks): Numerator: $2^3 \times 2^5 = 2^{3+5} = 2^8$ [2]. Quotient: $\dfrac{2^8}{2^4} = 2^{8-4} = 2^4$ [2].

Stretch Challenge · +25 XP, +10 coins

The Hidden Index II

If $\dfrac{5^n}{5^3} = 5^{8}$, find the value of $n$.

Reveal solution

By the quotient rule, $n - 3 = 8$, so $n = 11$.

Quick Review

Quotient rule

$\dfrac{a^m}{a^n} = a^{m-n}$

Subtract

Top minus bottom

Same base

Always required

Cancel proof

$\dfrac{a^5}{a^2}$: 2 cancel, 3 remain

$m < n$

Gives negative index (lesson 8)

$m = n$

Gives $a^0 = 1$ (lesson 7)

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