Mathematics • Year 9 • Unit 1 • Lesson 5
The Power of a Power Rule
Build fluency with $(a^m)^n = a^{mn}$, $(ab)^n = a^n b^n$ and $\left(\dfrac{a}{b}\right)^n = \dfrac{a^n}{b^n}$, one step at a time, from a fully worked example through guided practice to independent problems.
1. I do, fully worked example
Read every line. Each step has a short reason on the right so you can see why, not just what.
Problem. Simplify $(2x^3)^4$. 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: $(ab)^n = a^n b^n$, distribute the outer index to every factor inside.
Step 2, Give each factor the outer power of 4.
$(2x^3)^4 = 2^4 \times (x^3)^4$
Reason: there are TWO factors inside, the 2 and the $x^3$. Both get raised to the 4.
Step 3, Evaluate the numerical factor.
$2^4 = 2 \times 2 \times 2 \times 2 = 16$
Reason: numbers we can fully evaluate, we should. Pronumerals we leave in index form.
Step 4, Apply the power-of-a-power rule to $(x^3)^4$.
$(x^3)^4 = x^{3 \times 4} = x^{12}$
Reason: $(a^m)^n = a^{mn}$, MULTIPLY the inner and outer indices.
Step 5, Put it together.
$(2x^3)^4 = 16 x^{12}$
Reason: a numerical coefficient out the front, then each pronumeral with its simplified index.
Answer: $\mathbf{16x^{12}}$.
2. We do, fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank line. 4 marks
Problem. Simplify $(3y^2)^3$.
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:
$(3y^2)^3 = \_\_\_\_^3 \times (\_\_\_\_\)^3$
Step 3, Evaluate the number:
$3^3 = \_\_\_\_\_$
Step 4, Apply $(a^m)^n = a^{mn}$:
$(y^2)^3 = y^{\,\_\_\,\times\,\_\_\,} = y^{\_\_\_}$
Step 5, Put it together:
$(3y^2)^3 = \_\_\_\_\_\_\_\_$
3. You do, independent practice
Show your working in the space under each problem. The first four are foundation (single rule). The middle two are standard (combine 2 ideas). The last two are extension (combine 3+ ideas or include a quotient).
Foundation, single rule
3.1 Simplify $(a^6)^2$. 1 mark
3.2 Simplify $(5^2)^4$. Leave your answer as a power of 5. 1 mark
3.3 Expand $(4m)^3$. 1 mark
3.4 Simplify $\left(\dfrac{k}{3}\right)^2$. 1 mark
Standard, combine two ideas
3.5 Simplify $(2p^4)^3$. 2 marks
3.6 Simplify $\left(\dfrac{2}{x}\right)^3$, stating any restriction on $x$. 2 marks
Extension, push your thinking
3.7 Simplify $(3a^2 b)^4$. 3 marks
3.8 Find the value of $n$ in $(x^n)^5 = x^{20}$. Explain how you used the power-of-a-power rule. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2, We do (faded $(3y^2)^3$)
Step 1: power of a product.
Step 2: $(3y^2)^3 = \mathbf{3}^3 \times (\mathbf{y^2})^3$.
Step 3: $3^3 = \mathbf{27}$.
Step 4: $(y^2)^3 = y^{2 \times 3} = \mathbf{y^6}$.
Step 5: $(3y^2)^3 = \mathbf{27 y^6}$.
3.1, $(a^6)^2$
$(a^m)^n = a^{mn}$, so $(a^6)^2 = a^{6 \times 2} = \mathbf{a^{12}}$.
3.2, $(5^2)^4$
$5^{2 \times 4} = \mathbf{5^8}$. (As a number that's $390{,}625$, but leaving it as $5^8$ is the asked form.)
3.3, $(4m)^3$
Every factor gets the 3: $(4m)^3 = 4^3 m^3 = \mathbf{64 m^3}$.
3.4, $\left(\dfrac{k}{3}\right)^2$
Square top and bottom: $\dfrac{k^2}{3^2} = \mathbf{\dfrac{k^2}{9}}$.
3.5, $(2p^4)^3$
$2^3 \times (p^4)^3 = 8 \times p^{12} = \mathbf{8 p^{12}}$. (Common slip: writing $2 p^{12}$, the 2 must also be cubed.)
3.6, $\left(\dfrac{2}{x}\right)^3$
$\dfrac{2^3}{x^3} = \mathbf{\dfrac{8}{x^3}}$, with $x \ne 0$ (otherwise the original is undefined).
3.7, $(3a^2 b)^4$
Three factors inside the brackets: 3, $a^2$ and $b$. Each gets the 4.
$(3a^2 b)^4 = 3^4 \times (a^2)^4 \times b^4 = 81 \times a^8 \times b^4 = \mathbf{81 a^8 b^4}$.
3.8, Solve $(x^n)^5 = x^{20}$
By the power-of-a-power rule $(x^n)^5 = x^{5n}$. Match the indices: $5n = 20$, so $\mathbf{n = 4}$.
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.