Mathematics • Year 9 • Unit 1 • Lesson 8
Index Laws — Mixed Challenge
Pull together index notation (L1), evaluating powers (L2), product rule (L3), quotient rule (L4), power-of-a-power (L5), the index laws drill (L6), zero index (L7) and the new negative-index rule (L8). Choose the right tool, spot a mistake, 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 1-8. Decide which rule applies before you start writing. Show your working. 3 marks each
1.1 Evaluate $5^{-2}$.
1.2 Simplify $\dfrac{a^3}{a^7}$ for $a \ne 0$, leaving your answer with a positive index.
1.3 Rewrite $(2x)^{-3}$ with a positive index ($x \ne 0$). (Hint: $(2x)^{-3} = \dfrac{1}{(2x)^3}$, then expand the bracket.)
1.4 Find the value of $n$ in $\dfrac{1}{8} = 2^n$.
1.5 Complete the table for powers of $3$, then state the rule that connects each value to the next: $3^3, 3^2, 3^1, 3^0, 3^{-1}, 3^{-2}$.
1.6 Simplify $\dfrac{a^2 \times a^3}{a^7}$ for $a \ne 0$, leaving your answer with a positive index.
2. Find the mistake
Another student has tried to evaluate $\dfrac{a^4}{a^9}$ (with $a \ne 0$) using the index laws. 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 $\dfrac{a^4}{a^9}$:
Line 1: $\dfrac{a^4}{a^9} = a^{4-9}$
Line 2: $= a^{-5}$
Line 3: $= -a^5$
Line 4: So $\dfrac{a^4}{a^9} = -a^5$.
(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? Compare Line 3 with the negative-index rule from § "The Big Idea" — does $a^{-5}$ equal $-a^5$ or $\dfrac{1}{a^5}$?3. Open-ended challenge — build an expression that equals $\dfrac{1}{16}$
This question has more than one valid answer — there are several different expressions that work. 4 marks
3.1 Build two different expressions that each simplify to $\dfrac{1}{16}$, with the rule that at least one of your expressions must use a negative index, and the other must use a quotient that produces a negative index naturally. The expressions must not be the same and must not just be the bare expression "$\dfrac{1}{16}$".
For each expression you find:
(i) Write it down.
(ii) Show step-by-step working that confirms it equals $\dfrac{1}{16}$.
(iii) Name the index law(s) you used.
Hint to start: $16 = 2^4 = 4^2$. So $\dfrac{1}{16}$ could come from $2^{-4}$, or from $\dfrac{2^3}{2^7}$, or from $\dfrac{4^1}{4^3}$, or ...
How did this worksheet feel?
What I'll revisit before next class:
1.1 — $5^{-2}$
$5^{-2} = \dfrac{1}{5^2} = \mathbf{\dfrac{1}{25}}$.
1.2 — $\dfrac{a^3}{a^7}$
Quotient rule: $a^{3-7} = a^{-4}$.
Negative-index rule: $a^{-4} = \mathbf{\dfrac{1}{a^4}}$ (with $a \ne 0$).
1.3 — $(2x)^{-3}$
$(2x)^{-3} = \dfrac{1}{(2x)^3}$. Expand: $(2x)^3 = 2^3 x^3 = 8x^3$. So the answer is $\mathbf{\dfrac{1}{8x^3}}$ (with $x \ne 0$).
1.4 — Solve $\dfrac{1}{8} = 2^n$
$\dfrac{1}{8} = \dfrac{1}{2^3}$, which by the negative-index rule equals $2^{-3}$. So $\mathbf{n = -3}$.
1.5 — Table of powers of $3$
$3^3 = 27, \; 3^2 = 9, \; 3^1 = 3, \; 3^0 = 1, \; 3^{-1} = \dfrac{1}{3}, \; 3^{-2} = \dfrac{1}{9}$.
Rule: as the index drops by $1$, the value is divided by $3$ (the base). Once past $3^0$, the values become fractions.
1.6 — $\dfrac{a^2 \times a^3}{a^7}$
Step 1 — product rule on numerator: $a^2 \times a^3 = a^{2+3} = a^5$.
Step 2 — quotient rule: $\dfrac{a^5}{a^7} = a^{5-7} = a^{-2}$.
Step 3 — negative-index rule: $a^{-2} = \mathbf{\dfrac{1}{a^2}}$.
2 — Find the mistake
(a) The mistake is on Line 3.
(b) The student has converted $a^{-5}$ to $-a^5$ — that is, they have moved the minus sign from the INDEX to the VALUE. The negative-index rule says $a^{-n} = \dfrac{1}{a^n}$, NOT $-a^n$. A negative index produces a fraction, not a negative number.
(c) Corrected working:
$\dfrac{a^4}{a^9} = a^{4-9}$
$= a^{-5}$
$= \dfrac{1}{a^5}$
$= \mathbf{\dfrac{1}{a^5}}$.
This is exactly the trap flagged in the lesson's "Spot the Trap" card.
3 — Open-ended challenge (sample solutions)
Expression 1 (negative index): $2^{-4}$.
Working: $2^{-4} = \dfrac{1}{2^4} = \dfrac{1}{16}$ ✓.
Rule used: negative-index rule.
Expression 2 (quotient that creates a negative index): $\dfrac{2^3}{2^7}$.
Working: Quotient rule gives $2^{3-7} = 2^{-4}$. Negative-index rule: $2^{-4} = \dfrac{1}{2^4} = \dfrac{1}{16}$ ✓.
Rules used: quotient rule, then negative-index rule.
Other valid approaches: $4^{-2}$, $\dfrac{4^1}{4^3}$, $\dfrac{2^5}{2^9}$, $(2^{-2})^2$ from L5, and so on. Award full marks for any two valid expressions that satisfy the brief (one must be a direct negative index; the other must come from a quotient that produces a negative index).
Marking: 2 marks per valid expression (one for the expression and the named rule, one for clear step-by-step working that lands at $\dfrac{1}{16}$). Up to 4 in total.