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.

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) = y ⁶.
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.