Mathematics • Year 9 • Unit 1 • Lesson 5
Index Laws — Mixed Challenge
Pull together every index law from Unit 1 so far: index notation (L1), evaluating powers (L2), product rule (L3), quotient rule (L4) and power-of-a-power (L5). You'll have to choose the right tool for each problem, spot a mistake in someone else's working, and tackle an open-ended challenge.
1. Mixed problems — choose the right rule
Each question uses a different combination of the index laws from Lessons 1-5. Decide which rule applies before you start writing. Show your working. 3 marks each
1.1 Simplify $\dfrac{(x^4)^3}{x^5}$. Leave your answer in index form.
1.2 Simplify $(2a^3)^4 \times a^2$ fully.
1.3 Simplify $\left(\dfrac{3m^2}{n}\right)^3$, stating any restriction on $n$.
1.4 Show that $(-2)^4$ and $-2^4$ give different answers, and explain why in one sentence. (Recall Lesson 2 — brackets matter.)
1.5 Find $n$ if $(5^3)^n = 5^{15}$, and check your answer by also evaluating $(5^3)^n$ directly.
1.6 Simplify $\dfrac{(2x^2)^3 \times (3x)^2}{6x^4}$ fully.
2. Find the mistake
Another student has tried to simplify $(2a^3)^4$. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — simplify $(2a^3)^4$:
Line 1: $(2a^3)^4 = 2^4 \times (a^3)^4$
Line 2: $2^4 = 16$
Line 3: $(a^3)^4 = a^{3+4} = a^7$
Line 4: So $(2a^3)^4 = 16 a^7$.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? Compare Line 3's working to the lesson's rule for $(a^m)^n$ — are they adding or multiplying the indices? Which one should it be?3. Open-ended challenge — the same answer, two ways
This question has more than one valid answer — there are several different expressions that work. 4 marks
3.1 Find two different expressions of the form $(a^m)^n$ — where $a$, $m$ and $n$ are positive whole numbers and $a > 1$ — that both simplify to $2^{12}$.
For each expression you find:
(i) Write it down.
(ii) Show the working that confirms it equals $2^{12}$.
(iii) Briefly say which index law(s) you used.
Bonus: Your two expressions must not be the same as each other and must not just be $(2^{12})^1$ or $(2^1)^{12}$.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — $\dfrac{(x^4)^3}{x^5}$
Numerator: $(x^4)^3 = x^{4 \times 3} = x^{12}$ (power-of-a-power).
Whole expression: $\dfrac{x^{12}}{x^5} = x^{12-5} = \mathbf{x^7}$ (quotient rule).
1.2 — $(2a^3)^4 \times a^2$
$(2a^3)^4 = 2^4 \times (a^3)^4 = 16 \times a^{12} = 16 a^{12}$ (power-of-a-product, then power-of-a-power).
Multiply by $a^2$: $16 a^{12} \times a^2 = 16 a^{12+2} = \mathbf{16 a^{14}}$ (product rule).
1.3 — $\left(\dfrac{3m^2}{n}\right)^3$
Cube top and bottom: $\dfrac{(3m^2)^3}{n^3} = \dfrac{3^3 \times (m^2)^3}{n^3} = \dfrac{27 m^6}{n^3} = \mathbf{\dfrac{27 m^6}{n^3}}$, with $n \ne 0$ (otherwise the original is undefined).
1.4 — $(-2)^4$ vs $-2^4$
$(-2)^4 = (-2) \times (-2) \times (-2) \times (-2) = \mathbf{16}$ (four negatives multiplied gives a positive).
$-2^4 = -(2 \times 2 \times 2 \times 2) = \mathbf{-16}$ (the power only applies to the $2$ — the minus sign is "out the front").
Why different: brackets show what the index applies to. In $(-2)^4$ the index applies to the whole "$-2$". In $-2^4$ the index applies only to $2$.
1.5 — Solve $(5^3)^n = 5^{15}$
By power-of-a-power, $(5^3)^n = 5^{3n}$. Match indices: $3n = 15$, so $\mathbf{n = 5}$.
Check: $(5^3)^5 = 5^{3 \times 5} = 5^{15}$. ✓
1.6 — $\dfrac{(2x^2)^3 \times (3x)^2}{6x^4}$
Step 1 — power-of-a-product on each bracket:
$(2x^2)^3 = 2^3 \times (x^2)^3 = 8 x^6$.
$(3x)^2 = 3^2 x^2 = 9 x^2$.
Step 2 — multiply the numerator:
$8 x^6 \times 9 x^2 = 72 x^{6+2} = 72 x^8$.
Step 3 — divide by $6x^4$:
$\dfrac{72 x^8}{6 x^4} = \dfrac{72}{6} \times x^{8-4} = \mathbf{12 x^4}$.
2 — Find the mistake
(a) The mistake is on Line 3.
(b) The student has added the indices ($3 + 4 = 7$) instead of multiplying them. The rule for $(a^m)^n$ is power-of-a-power, which multiplies: $(a^3)^4 = a^{3 \times 4} = a^{12}$. Adding indices is the product rule, which applies to $a^m \times a^n$ — a different situation.
(c) Corrected working:
$(2a^3)^4 = 2^4 \times (a^3)^4$
$= 16 \times a^{3 \times 4}$
$= 16 \times a^{12}$
$= \mathbf{16 a^{12}}$.
This is precisely the trap flagged in the lesson's "Common Pitfalls" card: confusing product (add) with power-of-a-power (multiply).
3 — Open-ended challenge (sample solution)
We need $(a^m)^n = 2^{12}$, so we need $a$ to be a power of $2$ and the resulting index $mn$ (after applying any rewriting) to equal $12$. The simplest approach is to keep $a = 2$ and pick $m, n$ with $m \times n = 12$.
Expression 1: $(2^2)^6$.
Working: $(2^2)^6 = 2^{2 \times 6} = 2^{12}$ ✓.
Rule used: power-of-a-power.
Expression 2: $(2^3)^4$.
Working: $(2^3)^4 = 2^{3 \times 4} = 2^{12}$ ✓.
Rule used: power-of-a-power.
Other valid approaches: $(2^4)^3$, $(2^6)^2$ — any pair $(m, n)$ with $m \times n = 12$ and $m, n \ge 2$. Students who notice that $4 = 2^2$ can also write $(4^3)^2 = ((2^2)^3)^2 = 2^{12}$, which uses power-of-a-power twice.
Marking: 2 marks per valid expression (one for the expression itself, one for clear working showing it equals $2^{12}$). Up to 4 in total. Award full marks for any two distinct, valid expressions other than the trivial $(2^{12})^1$ or $(2^1)^{12}$.