Mathematics • Year 9 • Unit 1 • Lesson 6
Combining Index Laws
Build fluency with combining the index laws in one expression: brackets first (power-of-a-power), then multiply (product rule), then divide (quotient rule). 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.
Problem. Simplify $\;(2^3)^2 \times 2^4 \div 2^5$.
Step 1 — Brackets first (power-of-a-power).
$(2^3)^2 = 2^{3 \times 2} = 2^6$
Reason: power of a power MULTIPLIES indices: $(a^m)^n = a^{mn}$.
Step 2 — Multiply (product rule).
$2^6 \times 2^4 = 2^{6+4} = 2^{10}$
Reason: same base $2$ multiplied — ADD indices.
Step 3 — Divide (quotient rule).
$2^{10} \div 2^5 = 2^{10-5} = 2^5$
Reason: same base $2$ divided — SUBTRACT indices.
Step 4 — Evaluate & check.
$2^5 = 32$.
Direct check: $(2^3)^2 = 64$; $64 \times 16 = 1024$; $1024 \div 32 = 32$ ✓
Answer: $\mathbf{2^5 = 32}$.
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 $\;(a^4)^2 \times a^3 \div a^6$.
Step 1 — Brackets first. Power-of-a-power MULTIPLIES indices:
$(a^4)^2 = a^{4 \times 2} = a^{\_\_}$
Step 2 — Multiply. Product rule ADDS indices:
$a^{\_\_} \times a^3 = a^{\_\_ + 3} = a^{\_\_}$
Step 3 — Divide. Quotient rule SUBTRACTS (top minus bottom):
$a^{\_\_} \div a^6 = a^{\_\_ - 6} = a^{\_\_}$
Answer: $\mathbf{a^{\_\_}}$.
3. You do — independent practice
Show your working in the space under each problem. The first four are foundation (two rules). The middle two are standard (three rules). The last two are extension (push your thinking).
Foundation — two rules combined
3.1 Simplify $\;2^4 \times 2^3 \div 2^2$. 1 mark
3.2 Simplify $\;(x^3)^2 \times x^4$. 1 mark
3.3 Simplify $\;\dfrac{(a^2)^3}{a^4}$. 1 mark
3.4 Evaluate $\;\dfrac{3^5 \times 3^2}{3^4}$ as a single power of $3$, then as a number. 1 mark
Standard — three rules combined
3.5 Simplify $\;(y^2)^3 \times y^5 \div y^4$. 2 marks
3.6 Simplify $\;\dfrac{(2^3)^2 \times 2^4}{2^5}$. Evaluate the final result as a number. 2 marks
Extension — push your thinking
3.7 Simplify $\;\dfrac{(2 a^3)^2 \times a^4}{4 a^5}$ fully, including the coefficients. 3 marks
3.8 Find $n$ if $\;(x^3)^2 \times x^n = x^{10}$. Explain which two index laws you used. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded $(a^4)^2 \times a^3 \div a^6$)
Step 1: $(a^4)^2 = a^{\mathbf{8}}$ (multiply: $4 \times 2 = 8$).
Step 2: $a^{\mathbf{8}} \times a^3 = a^{\mathbf{8 + 3}} = a^{\mathbf{11}}$ (add).
Step 3: $a^{\mathbf{11}} \div a^6 = a^{\mathbf{11 - 6}} = a^{\mathbf{5}}$ (subtract).
Answer: $\mathbf{a^5}$.
3.1 — $2^4 \times 2^3 \div 2^2$
$2^{4+3-2} = \mathbf{2^5 = 32}$.
3.2 — $(x^3)^2 \times x^4$
Bracket first: $(x^3)^2 = x^6$.
Product rule: $x^6 \times x^4 = \mathbf{x^{10}}$.
3.3 — $(a^2)^3 / a^4$
Bracket: $(a^2)^3 = a^6$.
Quotient: $\dfrac{a^6}{a^4} = \mathbf{a^2}$.
3.4 — $\dfrac{3^5 \times 3^2}{3^4}$
Product rule on top: $3^5 \times 3^2 = 3^7$.
Quotient: $\dfrac{3^7}{3^4} = 3^3 = \mathbf{27}$.
3.5 — $(y^2)^3 \times y^5 \div y^4$
Bracket: $(y^2)^3 = y^6$.
Product: $y^6 \times y^5 = y^{11}$.
Quotient: $y^{11} \div y^4 = \mathbf{y^7}$.
3.6 — $\dfrac{(2^3)^2 \times 2^4}{2^5}$
Bracket: $(2^3)^2 = 2^6$.
Top product: $2^6 \times 2^4 = 2^{10}$.
Quotient: $\dfrac{2^{10}}{2^5} = 2^5 = \mathbf{32}$.
3.7 — $\dfrac{(2 a^3)^2 \times a^4}{4 a^5}$
Step 1 — resolve the bracket: $(2 a^3)^2 = 2^2 \times (a^3)^2 = 4 a^6$.
Step 2 — multiply on top: $4 a^6 \times a^4 = 4 a^{10}$.
Step 3 — divide: $\dfrac{4 a^{10}}{4 a^5} = \dfrac{4}{4} \times a^{10 - 5} = 1 \times a^5 = \mathbf{a^5}$.
3.8 — Solve $(x^3)^2 \times x^n = x^{10}$
Bracket (power-of-a-power): $(x^3)^2 = x^6$.
Product rule: $x^6 \times x^n = x^{6+n} = x^{10}$, so $6 + n = 10$, giving $\mathbf{n = 4}$.
Laws used: power-of-a-power then product rule.