Mathematics • Year 10 • Unit 1 • Lesson 9

Power of a Power, Product and Quotient

Build fluency with the three "bracket" index laws from Lesson 9: (aᵐ)ⁿ = aᵐⁿ, (ab)ⁿ = aⁿbⁿ, and (a/b)ⁿ = aⁿ/bⁿ. From a worked example through guided practice to independent problems.

Build · I Do / We Do / You Do

1. I do — fully worked example

Read every line. Each step has a short reason.

Problem. Simplify (2x³)⁴. Leave your answer in index form.

Step 1 — Spot the rule.

Brackets with a power outside → power of a product. Every factor inside gets the outer power.

Reason: (ab)ⁿ = aⁿbⁿ — distribute the outer index to every factor inside.

Step 2 — Give each factor the outer power of 4.

(2x³)⁴ = 2⁴ × (x³)⁴

Reason: there are TWO factors inside — the 2 and the x³. Both get raised to the 4.

Step 3 — Evaluate the numerical factor.

2⁴ = 2 × 2 × 2 × 2 = 16

Reason: numbers we can fully evaluate, we should.

Step 4 — Apply power-of-a-power to (x³)⁴.

(x³)⁴ = x³ˣ⁴ = x¹²

Reason: (aᵐ)ⁿ = aᵐⁿ — MULTIPLY the inner and outer indices.

Step 5 — Put it together.

(2x³)⁴ = 16x¹²

Reason: a numerical coefficient out the front, then each pronumeral with its simplified index.

Answer: 16x¹².

Stuck? Revisit lesson § "Power of a Product and Quotient" — forgetting that the 2 also gets the outer power is the most common slip.

2. We do — fill in the missing steps

Same structure as Section 1, but with the working faded. Fill in each blank. 4 marks

Problem. Simplify (3y²)³.

Step 1 — Spot the rule: brackets with a power outside → power of a __________________. Every factor inside gets the outer power.

Step 2 — Give each factor the outer power of 3:

(3y²)³ = ____³ × (____)³

Step 3 — Evaluate the number:

3³ = ______

Step 4 — Apply (aᵐ)ⁿ = aᵐⁿ:

(y²)³ = y^(__ × __) = y____

Step 5 — Put it together:

(3y²)³ = __________

Stuck? Revisit lesson § "Worked Example 3 — Power of a Product" for the same pattern.

3. You do — independent practice

Show your working. The first four are foundation (single rule). The middle two are standard (combine 2 rules). The last two are extension (chain multiple rules).

Foundation — single rule

3.1 Simplify (a⁶)².    1 mark

3.2 Simplify (5²)⁴. Leave as a power of 5.    1 mark

3.3 Expand (4m)³.    1 mark

3.4 Simplify (k/3)².    1 mark

Standard — combine two rules

3.5 Simplify (2p⁴)³.    2 marks

3.6 Simplify (2/x)³, stating any restriction on x.    2 marks

Extension — push your thinking

3.7 Simplify (3a²b)⁴.    3 marks

3.8 Find the value of n in (xⁿ)⁵ = x²⁰. Explain how you used (aᵐ)ⁿ = aᵐⁿ.    2 marks

Stuck on 3.7? Treat the bracket as having THREE factors: 3, a² and b. Each gets the outer index of 4.

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

Section 2 — We do (3y²)³

Step 1: power of a product.
Step 2: (3y²)³ = 3³ × (y²)³.
Step 3: 3³ = 27.
Step 4: (y²)³ = y^(2 × 3) = y⁶.
Step 5: (3y²)³ = 27y⁶.

3.1 — (a⁶)²

(aᵐ)ⁿ = aᵐⁿ, so (a⁶)² = a⁶ˣ² = a¹².

3.2 — (5²)⁴

5²ˣ⁴ = 5⁸. (As a number that's 390,625, but leaving it as 5⁸ is the asked form.)

3.3 — (4m)³

Every factor gets the 3: (4m)³ = 4³m³ = 64m³.

3.4 — (k/3)²

Square top and bottom: k²/3² = k²/9.

3.5 — (2p⁴)³

2³ × (p⁴)³ = 8 × p¹² = 8p¹². (Common slip: writing 2p¹² — the 2 must also be cubed.)

3.6 — (2/x)³

2³/x³ = 8/x³, with x ≠ 0 (otherwise the original is undefined).

3.7 — (3a²b)⁴

Three factors inside the brackets: 3, a² and b. Each gets the 4.
(3a²b)⁴ = 3⁴ × (a²)⁴ × b⁴ = 81 × a⁸ × b⁴ = 81a⁸b⁴.

3.8 — Solve (xⁿ)⁵ = x²⁰

By the power-of-a-power rule (xⁿ)⁵ = x⁵ⁿ. Match the indices: 5n = 20, so n = 4.
The rule says we multiply the inner and outer indices, so the resulting index is 5n. Set that equal to the given index of 20 and solve.

HSCScience • Mathematics • Year 10 • Unit 1 • Lesson 9 • Worksheet 1 (Build)