Mathematics • Year 9 • Unit 1 • Lesson 9
Index Laws — Mixed Challenge
Pull together every index rule so far (L1-L9). Flip fractions, combine negative and positive indices, handle negative bases, and always finish with positive indices. Then spot a mistake and design your own challenge.
1. Mixed problems — choose the right rule
Each question uses a different combination of rules from Lessons 1-9. Decide which rule applies before you start writing. Show your working and always finish with a positive index. 3 marks each
1.1 Evaluate $\left(\dfrac{2}{5}\right)^{-3}$.
1.2 Simplify $\dfrac{a^{-2} \times a^5}{a^{-4}}$, leaving your answer with a positive index ($a \ne 0$).
1.3 Evaluate $(-3)^{-2}$ and $-3^{-2}$. State both values and explain why they differ.
1.4 Simplify $b^{-3} \times b^{-2}$ and leave your answer with a positive index ($b \ne 0$).
1.5 Find $n$ if $\left(\dfrac{1}{4}\right)^{-n} = 64$.
1.6 Simplify $\dfrac{(x^{-1})^3 \times x^7}{x^2}$, leaving your answer with a positive index ($x \ne 0$).
2. Find the mistake
Another student has tried to evaluate $\left(\dfrac{2}{3}\right)^{-2}$. 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 — evaluate $\left(\dfrac{2}{3}\right)^{-2}$:
Line 1: $\left(\dfrac{2}{3}\right)^{-2} = \dfrac{1}{(2/3)^2}$
Line 2: $(2/3)^2 = \dfrac{2^2}{3^2} = \dfrac{4}{9}$
Line 3: $\left(\dfrac{2}{3}\right)^{-2} = \dfrac{1}{4/9} = \dfrac{4}{9}$
Line 4: Final answer $= \dfrac{4}{9}$.
(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? Dividing by a fraction is the same as multiplying by its reciprocal. So $\dfrac{1}{4/9}$ should equal $1 \times \dfrac{9}{4}$, not just $\dfrac{4}{9}$.3. Open-ended challenge — design your own evaluation
This question has more than one valid answer — there are many different expressions that work. 4 marks
3.1 Build two different expressions that each evaluate to $\mathbf{27}$, with these rules: each expression must contain at least one negative index, and they must not be the same as each other. Trivial answers (like just writing $27$ in disguise) are not allowed — your expression must include at least one application of a Lesson 9 rule (flip fraction, combine + and - indices, or convert a negative index back to positive).
For each expression you build:
(i) Write it down clearly.
(ii) Show step-by-step working that confirms it equals $27$.
(iii) Name the index law(s) you used.
Hint to start: $27 = 3^3$. So $\left(\dfrac{1}{3}\right)^{-3} = 27$ works, and so does $3^{-1} \times 3^4$, and so does $\dfrac{3^5}{3^2}$ written using a negative power somewhere ...
How did this worksheet feel?
What I'll revisit before next class:
1.1 — $\left(\dfrac{2}{5}\right)^{-3}$
Flip: $\left(\dfrac{2}{5}\right)^{-3} = \left(\dfrac{5}{2}\right)^3 = \dfrac{5^3}{2^3} = \mathbf{\dfrac{125}{8}}$.
1.2 — $\dfrac{a^{-2} \times a^5}{a^{-4}}$
Numerator: $a^{-2+5} = a^3$.
Whole expression: $\dfrac{a^3}{a^{-4}} = a^{3 - (-4)} = a^{3+4} = \mathbf{a^7}$.
1.3 — $(-3)^{-2}$ vs $-3^{-2}$
$(-3)^{-2} = \dfrac{1}{(-3)^2} = \dfrac{1}{9} = \mathbf{\dfrac{1}{9}}$ — the WHOLE bracket is the base; squaring removes the minus.
$-3^{-2} = -(3^{-2}) = -\dfrac{1}{9} = \mathbf{-\dfrac{1}{9}}$ — no brackets, so the index only applies to $3$; the minus is outside.
The brackets decide what the index applies to.
1.4 — $b^{-3} \times b^{-2}$
Product rule: $b^{-3 + (-2)} = b^{-5}$.
Convert to positive index: $b^{-5} = \mathbf{\dfrac{1}{b^5}}$.
1.5 — Solve $\left(\dfrac{1}{4}\right)^{-n} = 64$
Flip: $\left(\dfrac{1}{4}\right)^{-n} = 4^n$. So $4^n = 64$. Try $n = 3$: $4^3 = 64$ ✓. So $\mathbf{n = 3}$.
1.6 — $\dfrac{(x^{-1})^3 \times x^7}{x^2}$
Step 1 — power-of-a-power: $(x^{-1})^3 = x^{-1 \times 3} = x^{-3}$.
Step 2 — product on numerator: $x^{-3} \times x^7 = x^{-3+7} = x^4$.
Step 3 — quotient: $\dfrac{x^4}{x^2} = x^{4-2} = \mathbf{x^2}$.
2 — Find the mistake
(a) The mistake is on Line 3.
(b) The student computed $\dfrac{1}{4/9}$ as just $\dfrac{4}{9}$, but dividing $1$ by a fraction means multiplying by its reciprocal: $\dfrac{1}{4/9} = 1 \times \dfrac{9}{4} = \dfrac{9}{4}$. They didn't flip the fraction in the reciprocal step.
(c) Corrected working:
$\left(\dfrac{2}{3}\right)^{-2} = \dfrac{1}{(2/3)^2} = \dfrac{1}{4/9} = 1 \times \dfrac{9}{4} = \mathbf{\dfrac{9}{4}}$.
Or much more efficient: flip first, then square — $\left(\dfrac{2}{3}\right)^{-2} = \left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4}$.
3 — Open-ended challenge (sample solutions)
Expression 1 (flip rule): $\left(\dfrac{1}{3}\right)^{-3}$.
Working: Flip the fraction: $\left(\dfrac{1}{3}\right)^{-3} = 3^3 = 27$ ✓.
Rule used: flip rule for fractions with negative index.
Expression 2 (combine negative + positive): $3^{-1} \times 3^4$.
Working: Product rule: $3^{-1+4} = 3^3 = 27$ ✓.
Rule used: product rule (works even with one negative index).
Other valid approaches: $\dfrac{3^5}{3^2}$ rewritten as $3^5 \times 3^{-2} = 3^3$, $\dfrac{3^{-1}}{3^{-4}} = 3^{-1-(-4)} = 3^3$, $\left(\dfrac{1}{9}\right)^{-1} \times 3 = 9 \times 3 = 27$, and so on. Award full marks for any two valid distinct expressions that use a negative index meaningfully.
Marking: 2 marks per valid expression (one for the expression and the named rule, one for clear working that lands at $27$). Up to 4 in total.