Mathematics • Year 9 • Unit 1 • Lesson 20
Unit 1 Boss Challenge — Every Index Law
The full-unit synthesis. Six mixed problems pulling from every lesson in Unit 1 (product, quotient, power-of-a-power, zero, negative, fractional, scientific notation), one "find the mistake", and an open-ended boss expression that requires all the laws at once.
1. Mixed problems — pull from every lesson
Each question uses a different combination of the index laws from Lessons 1-19. Decide which rule(s) apply before you start writing. Show your working and give answers with positive indices (and standard form for scientific notation). 3 marks each
1.1 Simplify $a^5 \cdot a^{-2}$, expressing the answer with a positive index.
1.2 Expand $(2 x^3 y)^4$.
1.3 Simplify $\dfrac{12 m^2 n^{-1}}{4 m^{-3} n^2}$, expressing the answer with positive indices only.
1.4 Calculate $(6 \times 10^5) \times (4 \times 10^{-2})$ in scientific notation, then state the order of magnitude.
1.5 Evaluate $16^{3/4}$ using "root first, then power".
1.6 Simplify $\dfrac{(2 x^2)^3 \cdot x^{-1}}{4 x^{1/2}}$, then evaluate your answer when $x = 4$.
2. Find the mistake
Another student has tried to simplify $\dfrac{(2 a^3)^2}{a^{-1}}$. 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 $\dfrac{(2 a^3)^2}{a^{-1}}$:
Line 1: Bracket: $(2 a^3)^2 = 2^2 \cdot (a^3)^2 = 4 a^6$. ✓
Line 2: Negative index: $a^{-1} = -a$.
Line 3: Substitute: $\dfrac{4 a^6}{-a} = -4 a^5$.
Line 4: So $\dfrac{(2 a^3)^2}{a^{-1}} = -4 a^5$.
(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? The negative-index rule is $a^{-n} = \dfrac{1}{a^n}$ — it flips the base, not the sign. The "$-$" in the index is not a minus on the value.3. Open-ended challenge — Unit Boss
This is the Unit 1 boss expression — every single law gets a turn. 4 marks
3.1 Simplify $\dfrac{\left(2 a^{1/2} b^{-1}\right)^4 \cdot a^0 b^3}{8 a^{-1} b^{-2}}$, leaving your answer with positive indices.
For each step in your working, write the name of the index law you used in a small note alongside (e.g. "power-of-a-product", "zero index", "product rule", "quotient rule", "negative index").
Finally, state whether your final answer is a polynomial in $a$ and $b$ — and briefly justify (a polynomial has only non-negative integer indices).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — $a^5 \cdot a^{-2}$
Product rule: $a^{5 + (-2)} = \mathbf{a^3}$. (Positive index already.)
1.2 — $(2 x^3 y)^4$
Power-of-a-product: $2^4 \cdot (x^3)^4 \cdot y^4 = 16 \cdot x^{12} \cdot y^4 = \mathbf{16 x^{12} y^4}$.
1.3 — $\dfrac{12 m^2 n^{-1}}{4 m^{-3} n^2}$
Coefficients: $\dfrac{12}{4} = 3$. On $m$: $m^{2 - (-3)} = m^5$. On $n$: $n^{-1 - 2} = n^{-3}$. So result is $3 m^5 n^{-3}$. Flip the negative index to the denominator: $\mathbf{\dfrac{3 m^5}{n^3}}$.
1.4 — $(6 \times 10^5) \times (4 \times 10^{-2})$
$6 \times 4 = 24$; $10^{5 + (-2)} = 10^3$; re-normalise $24 \times 10^3 = \mathbf{2.4 \times 10^4}$. Order of magnitude $\approx \mathbf{10^4}$.
1.5 — $16^{3/4}$
Root first: $\sqrt[4]{16} = 2$ (since $2^4 = 16$). Then $2^3 = \mathbf{8}$.
1.6 — Simplify, then evaluate at $x = 4$
Simplify: $(2 x^2)^3 = 8 x^6$ (power-of-a-product). Product rule on numerator: $8 x^6 \cdot x^{-1} = 8 x^5$. Quotient: $\dfrac{8 x^5}{4 x^{1/2}} = 2 x^{5 - 1/2} = 2 x^{9/2}$.
Evaluate at $x = 4$: $2 \cdot 4^{9/2} = 2 \cdot (\sqrt{4})^9 = 2 \cdot 2^9 = 2 \cdot 512 = \mathbf{1024}$.
2 — Find the mistake
(a) The mistake is on Line 2.
(b) The student treated the negative sign in the index as a negative sign on the base, but the negative-index rule is $a^{-n} = \dfrac{1}{a^n}$ — it flips the base to the denominator, not changes the sign. So $a^{-1} = \dfrac{1}{a}$, not $-a$.
(c) Corrected working:
$\dfrac{(2 a^3)^2}{a^{-1}} = \dfrac{4 a^6}{a^{-1}}$ (bracket done)
$= 4 a^{6 - (-1)}$ (quotient rule)
$= 4 a^{6 + 1}$
$= \mathbf{4 a^7}$.
Alternative: $\dfrac{4 a^6}{a^{-1}} = 4 a^6 \cdot a^1 = 4 a^7$ (negative on the bottom flips up to a positive on the top). Same answer.
3 — Unit Boss (sample solution)
Step 1 — Power-of-a-product on the bracket:
$\left(2 a^{1/2} b^{-1}\right)^4 = 2^4 \cdot a^{1/2 \times 4} \cdot b^{-1 \times 4} = 16 \cdot a^2 \cdot b^{-4} = 16 a^2 b^{-4}$.
Laws used: power-of-a-product, power-of-a-power (twice).
Step 2 — Zero index:
$a^0 = 1$. So the numerator becomes $16 a^2 b^{-4} \cdot 1 \cdot b^3 = 16 a^2 b^{-4} \cdot b^3$.
Law used: zero index.
Step 3 — Product rule on the numerator (for $b$):
$b^{-4} \cdot b^3 = b^{-4 + 3} = b^{-1}$. So numerator $= 16 a^2 b^{-1}$.
Law used: product rule.
Step 4 — Quotient rule:
$\dfrac{16 a^2 b^{-1}}{8 a^{-1} b^{-2}} = \dfrac{16}{8} \cdot a^{2 - (-1)} \cdot b^{-1 - (-2)} = 2 \cdot a^3 \cdot b^1 = 2 a^3 b$.
Laws used: quotient rule, plus careful handling of negative indices.
Final answer: $\mathbf{2 a^3 b}$.
Is it a polynomial? Yes — the final expression has only non-negative integer indices ($a^3$ and $b^1$), so it is a polynomial in $a$ and $b$. All the fractional and negative indices that appeared in intermediate steps cancelled out.
Marking: 1 mark for the correct simplified answer $2 a^3 b$; 1 for naming the laws used at each step; 1 for handling the double negative correctly in the $b$ index; 1 for the polynomial justification.