Mathematics • Year 9 • Unit 1 • Lesson 14
Mixed Index Laws — Mixed Challenge
Pull together every index law from Unit 1 so far. Choose the right rule, work the order — power $\to$ product $\to$ quotient $\to$ zero — spot a Year 9 mistake, then design a problem whose answer is exactly $1$.
1. Mixed problems — order matters
Each question uses at least three rules. Decide on the ORDER before you start. Show working. 3 marks each
1.1 Simplify $\dfrac{2 x^3 \cdot 3 x^2}{6 x^4}$.
1.2 Simplify $\dfrac{(2 x^2)^3 \cdot 5 x}{10 x^7}$.
1.3 Simplify $\dfrac{(3 a^2 b)^2 \cdot 4 a b^3}{6 a^3 b^4}$.
1.4 Simplify $\dfrac{(2 m^2 n)^3 \cdot 3 m n^2}{12 m^5 n^4}$.
1.5 Simplify $\dfrac{(2 x^2 y)^3 \cdot 6 x y^2}{(3 x^3 y)^2}$. Then evaluate it when $x = 1$, $y = 2$.
1.6 Simplify $\dfrac{8 p^5 q^2 \cdot p}{(2 p^3 q)^2}$ and explain in one sentence why the answer is a plain number with no $p$ or $q$.
2. Find the mistake
A student has tried to simplify $\dfrac{(2 x^2)^3 \cdot 5 x}{10 x^7}$. Exactly one line contains a mistake. Spot it, explain it, then redo the working correctly. 3 marks
Student's working — simplify $\dfrac{(2 x^2)^3 \cdot 5 x}{10 x^7}$:
Line 1: $(2 x^2)^3 = 8 x^6$
Line 2: Numerator $= 8 x^6 \cdot 5 x = 40 x^7$
Line 3: $\dfrac{40 x^7}{10 x^7} = 4 x^0 = 0$
Line 4: So $\dfrac{(2 x^2)^3 \cdot 5 x}{10 x^7} = 0$.
(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? Revisit lesson § "Spot the Trap" — what does $x^0$ actually equal?3. Open-ended challenge — design for the answer 1
This question has many valid answers. 4 marks
3.1 Design two different expressions, each of the form $\dfrac{\text{numerator}}{\text{denominator}}$ where at least one part contains a bracketed power (e.g. $(\ldots)^2$ or $(\ldots)^3$), that simplify to exactly $\mathbf{1}$.
For each expression:
(i) Write the expression.
(ii) Show the four-step working (power $\to$ product $\to$ quotient $\to$ zero).
(iii) Confirm the final answer is $1$.
Bonus: At least one of your expressions must involve two different variables (e.g. both $x$ and $y$).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — $\dfrac{2 x^3 \cdot 3 x^2}{6 x^4}$
Product (top): $6 x^5$. Quotient: $\dfrac{6 x^5}{6 x^4} = 1 \cdot x^1 = \mathbf{x}$.
1.2 — $\dfrac{(2 x^2)^3 \cdot 5 x}{10 x^7}$
Power: $(2x^2)^3 = 8 x^6$. Product: $8 x^6 \cdot 5 x = 40 x^7$. Quotient: $\dfrac{40 x^7}{10 x^7} = 4 x^0 = \mathbf{4}$.
1.3 — $\dfrac{(3 a^2 b)^2 \cdot 4 a b^3}{6 a^3 b^4}$
Power: $(3 a^2 b)^2 = 9 a^4 b^2$. Product: $9 a^4 b^2 \cdot 4 a b^3 = 36 a^5 b^5$. Quotient: $\dfrac{36 a^5 b^5}{6 a^3 b^4} = 6 \cdot a^2 \cdot b = \mathbf{6 a^2 b}$.
1.4 — $\dfrac{(2 m^2 n)^3 \cdot 3 m n^2}{12 m^5 n^4}$
Power: $(2 m^2 n)^3 = 8 m^6 n^3$. Product: $8 m^6 n^3 \cdot 3 m n^2 = 24 m^7 n^5$. Quotient: $\dfrac{24 m^7 n^5}{12 m^5 n^4} = 2 \cdot m^2 \cdot n = \mathbf{2 m^2 n}$.
1.5 — $\dfrac{(2 x^2 y)^3 \cdot 6 x y^2}{(3 x^3 y)^2}$
Powers: $(2 x^2 y)^3 = 8 x^6 y^3$; $(3 x^3 y)^2 = 9 x^6 y^2$.
Product (top): $8 x^6 y^3 \cdot 6 x y^2 = 48 x^7 y^5$.
Quotient: $\dfrac{48 x^7 y^5}{9 x^6 y^2} = \dfrac{48}{9} \cdot x \cdot y^3 = \mathbf{\dfrac{16}{3} x y^3}$.
At $x = 1, y = 2$: $\dfrac{16}{3} \times 1 \times 8 = \dfrac{128}{3} \approx \mathbf{42.7}$.
1.6 — $\dfrac{8 p^5 q^2 \cdot p}{(2 p^3 q)^2}$
Power (bottom): $(2 p^3 q)^2 = 4 p^6 q^2$. Product (top): $8 p^5 q^2 \cdot p = 8 p^6 q^2$. Quotient: $\dfrac{8 p^6 q^2}{4 p^6 q^2} = 2 \cdot p^0 \cdot q^0 = \mathbf{2}$.
The $p^6$ on top and bottom cancel (giving $p^0 = 1$) and the $q^2$ on top and bottom cancel (giving $q^0 = 1$), so only the coefficient ratio $\tfrac{8}{4} = 2$ remains.
2 — Find the mistake
(a) The mistake is on Line 3.
(b) The student wrote $x^0 = 0$, but the zero index rule says $x^0 = 1$ for any non-zero $x$, NOT zero. The coefficient stays as $4$, and $4 \times 1 = 4$.
(c) Corrected working:
$(2 x^2)^3 = 8 x^6$
Numerator $= 8 x^6 \cdot 5 x = 40 x^7$
$\dfrac{40 x^7}{10 x^7} = 4 x^0 = 4 \times 1 = \mathbf{4}$.
3 — Open-ended challenge (sample solution)
For the answer to be $1$ we need the coefficient ratio to be $1$ AND every variable's index to cancel completely.
Expression 1 (single variable): $\dfrac{(2 x^2)^2 \cdot x}{4 x^5}$.
Power: $(2 x^2)^2 = 4 x^4$. Product: $4 x^4 \cdot x = 4 x^5$. Quotient: $\dfrac{4 x^5}{4 x^5} = 1 \cdot x^0 = \mathbf{1}$ ✓.
Expression 2 (two variables — bonus): $\dfrac{(3 x y)^2 \cdot 2 x y^3}{18 x^3 y^5}$.
Power: $(3 x y)^2 = 9 x^2 y^2$. Product: $9 x^2 y^2 \cdot 2 x y^3 = 18 x^3 y^5$. Quotient: $\dfrac{18 x^3 y^5}{18 x^3 y^5} = 1 \cdot x^0 \cdot y^0 = \mathbf{1}$ ✓.
Other valid expressions: Any quotient where the simplified numerator equals the denominator. Examples: $\dfrac{(2x)^3}{8 x^3}$, $\dfrac{(5 a^2)^2}{25 a^4}$, $\dfrac{(2 m n)^2 \cdot m}{4 m^3 n^2}$.
Marking: 2 marks per valid expression (1 for the expression, 1 for the four-step working). Up to 4 in total. Bonus only awarded if at least one expression has two different variables.