Mathematics • Year 9 • Unit 1 • Lesson 7
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) and the new zero-index rule (L7). 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 1-7. Decide which rule applies before you start writing. Show your working. 3 marks each
1.1 Evaluate $7^0 + 11^0 + 2^3$.
1.2 Simplify $\dfrac{x^7}{x^7}$ for $x \ne 0$ — give your answer using the zero-index rule.
1.3 Simplify $(2x^3)^0 + 5 - 3$ for $x \ne 0$. Leave your answer as a number.
1.4 Evaluate $4 \times 5^0 + (3y)^0$ for $y \ne 0$.
1.5 Show that $(-9)^0$ and $-9^0$ give different answers. Explain why in one sentence. (Recall Lesson 2 — brackets matter.)
1.6 Simplify $\dfrac{(a^4)^2 \times a^0}{a^6}$ for $a \ne 0$. Show every step.
2. Find the mistake
Another student has tried to simplify $5x^0 + (4x)^0$ for $x \ne 0$. 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 $5x^0 + (4x)^0$:
Line 1: $5x^0 = 5 \times x^0 = 5 \times 1 = 5$
Line 2: $(4x)^0 = 4 \times x^0 = 4 \times 1 = 4$
Line 3: So $5x^0 + (4x)^0 = 5 + 4 = 9$.
(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? Look at where the brackets are in $(4x)^0$. Is the $4$ inside the bracket or outside?3. Open-ended challenge — build an expression that equals $1$
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 $1$ and that each use at least one zero index. The expressions must not be identical to each other, and must not just be the bare expression "$a^0$".
For each expression you build:
(i) Write it down clearly.
(ii) Show step-by-step working that confirms it equals $1$.
(iii) Name the index law(s) you used.
Examples of valid styles: a product like $a^n \times a^{-n}$ that lands on $a^0$, a quotient like $\dfrac{a^m}{a^m}$, or a power-of-a-power like $(a^0)^k$.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — $7^0 + 11^0 + 2^3$
$7^0 = 1$ and $11^0 = 1$ (zero-index rule). $2^3 = 8$. Sum: $1 + 1 + 8 = \mathbf{10}$.
1.2 — $\dfrac{x^7}{x^7}$
Quotient rule: $\dfrac{x^7}{x^7} = x^{7-7} = x^0$. Zero-index rule: $x^0 = \mathbf{1}$ (since $x \ne 0$).
1.3 — $(2x^3)^0 + 5 - 3$
The brackets enclose the WHOLE base $2x^3$. With $x \ne 0$, that base is non-zero, so $(2x^3)^0 = 1$.
Then $1 + 5 - 3 = \mathbf{3}$.
1.4 — $4 \times 5^0 + (3y)^0$
$5^0 = 1$, so $4 \times 5^0 = 4 \times 1 = 4$. $(3y)^0 = 1$ (with $y \ne 0$, the whole bracket is non-zero). Sum: $4 + 1 = \mathbf{5}$.
1.5 — $(-9)^0$ vs $-9^0$
$(-9)^0 = \mathbf{1}$ — the brackets show the WHOLE base is $-9$, and $-9$ is non-zero.
$-9^0 = -(9^0) = -1 \times 1 = \mathbf{-1}$ — there are no brackets, so the index $0$ only applies to $9$; the minus sign sits outside.
The brackets decide what the index applies to.
1.6 — $\dfrac{(a^4)^2 \times a^0}{a^6}$
Step 1 — power-of-a-power: $(a^4)^2 = a^{4 \times 2} = a^8$.
Step 2 — zero-index rule: $a^0 = 1$. Numerator becomes $a^8 \times 1 = a^8$.
Step 3 — quotient rule: $\dfrac{a^8}{a^6} = a^{8-6} = \mathbf{a^2}$.
2 — Find the mistake
(a) The mistake is on Line 2.
(b) The student treated $(4x)^0$ as $4 \times x^0$, but the brackets mean the WHOLE base is $4x$. So $(4x)^0 = 1$ directly, not $4 \times 1$. Confusing $(4x)^0$ with $4x^0$ is the classic "brackets matter" trap.
(c) Corrected working:
$5x^0 = 5 \times x^0 = 5 \times 1 = 5$ (the $5$ is OUTSIDE the index, only $x$ has the $0$).
$(4x)^0 = 1$ (the WHOLE bracket has index $0$).
$5x^0 + (4x)^0 = 5 + 1 = \mathbf{6}$.
This is exactly the trap flagged in the lesson's "Spot the Trap" card and "Common Pitfalls" — missing brackets change what the index applies to.
3 — Open-ended challenge (sample solutions)
Expression 1: $\dfrac{3^5}{3^5}$.
Working: Quotient rule gives $3^{5-5} = 3^0$. Zero-index rule gives $3^0 = 1$ ✓.
Rules used: quotient rule, zero-index rule.
Expression 2: $(7^0)^4$.
Working: Zero-index rule gives $7^0 = 1$. Then $1^4 = 1$ ✓. (Alternatively: power-of-a-power gives $7^{0 \times 4} = 7^0 = 1$.)
Rules used: zero-index rule (and optionally power-of-a-power).
Other valid approaches: any $\dfrac{a^n}{a^n}$ (with $a \ne 0$), any $(a^n)^0$ (with $a \ne 0$), $a^k \times a^{-k}$ (preview of Lesson 8), $\dfrac{x^0 + x^0}{2}$ which equals $1$, and so on. Award full marks for any two distinct, valid expressions other than just "$a^0$".
Marking: 2 marks per valid expression (one for the expression, one for clear step-by-step working that shows it equals $1$ AND names the rule used). Up to 4 in total.