Mathematics • Year 9 • Unit 1 • Lesson 1
Index Notation — Mixed Challenge
Sharpen everything from Lesson 1: index notation, base and index, evaluating small powers, and perfect squares and cubes. Spot a mistake in another student's working, then take on an open-ended puzzle.
1. Mixed problems — pull it all together
Each question uses an idea from Lesson 1. Decide which idea applies before you start writing. Show your working. 3 marks each
1.1 Write $\;8 \times 8 \times 8 \times 8 \times 8 \times 8\;$ in index notation, then evaluate $8^2$ as a check on the base.
1.2 Evaluate $\;2^{10}\;$ by doubling step by step. Write each intermediate value.
1.3 Write $\;2 \times 2 \times 2 \times 7 \times 7 \times 7 \times 7\;$ in index notation. (Group the same bases.)
1.4 Without a calculator, decide which is bigger: $\;5^3\;$ or $\;3^5\;$. Show both values.
1.5 $144$ is a perfect square because $144 = 12^2$. Is $216$ a perfect square, a perfect cube, both, or neither? Justify your answer with a calculation.
1.6 A water bottle factory packs $10^3$ bottles per pallet, and $10^2$ pallets per delivery truck. How many bottles fit in one truck? Give your answer in index notation as a single power of $10$, then as a normal number.
2. Find the mistake
Another student has tried to evaluate $\;3^4\;$. 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 $3^4$:
Line 1: $3^4$ means four copies of $3$ multiplied.
Line 2: So $3^4 = 3 \times 3 \times 3 \times 3$.
Line 3: $3 \times 3 = 9$ and $3 \times 3 = 9$, so $3^4 = 9 + 9 = 18$.
Line 4: So $3^4 = 18$.
(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 student split the multiplication into pairs (good idea!) but then made one wrong operation between the pairs. Should you add or multiply the two $9$s?3. Open-ended challenge — three ways to make $64$
This question has more than one valid answer — there are several different powers that equal $64$. 4 marks
3.1 Find three different expressions of the form $\;a^n\;$ — where $a$ and $n$ are positive whole numbers and $a > 1$ — that all equal $64$.
For each expression you find:
(i) Write it down.
(ii) Show the multiplication that confirms it equals $64$.
(iii) State which is the base and which is the index.
Bonus: Your three expressions must not include the trivial $64^1$.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Six $8$s
Six copies of base $8$: $\mathbf{8^6}$. Base $= 8$, index $= 6$. Check: $8^2 = 64$ ✓.
1.2 — Evaluate $2^{10}$ by doubling
$2^1 = 2$; $2^2 = 4$; $2^3 = 8$; $2^4 = 16$; $2^5 = 32$; $2^6 = 64$; $2^7 = 128$; $2^8 = 256$; $2^9 = 512$; $\mathbf{2^{10} = 1024}$.
1.3 — Grouped bases
Three $2$s give $2^3$; four $7$s give $7^4$. So $2 \times 2 \times 2 \times 7 \times 7 \times 7 \times 7 = \mathbf{2^3 \times 7^4}$.
1.4 — Compare $5^3$ and $3^5$
$5^3 = 5 \times 5 \times 5 = 125$.
$3^5 = 3 \times 3 \times 3 \times 3 \times 3 = 9 \times 9 \times 3 = 81 \times 3 = 243$.
So $\mathbf{3^5 > 5^3}$ (i.e. $243 > 125$). The bigger index $5$ beats the bigger base $5$ here — a typical "growth of indices" surprise.
1.5 — Is $216$ square, cube, both, or neither?
Try cubes: $5^3 = 125$, $6^3 = 216$ ✓. So $216 = 6^3$ — it IS a perfect cube.
Try squares: $14^2 = 196$, $15^2 = 225$. Neither equals $216$, so $216$ is NOT a perfect square.
$216$ is a perfect cube but not a perfect square.
1.6 — Water bottle factory
$10^3$ bottles per pallet $\times\ 10^2$ pallets per truck $= 10^3 \times 10^2$. That's three $10$s times two $10$s — five $10$s in total: $\mathbf{10^5} = 100{,}000$ bottles per truck.
This previews the product rule from Lesson 3: same base, add indices.
2 — Find the mistake
(a) The mistake is on Line 3.
(b) The student split the multiplication into the pairs $(3 \times 3)$ and $(3 \times 3)$ correctly, but then added the two $9$s instead of multiplying them. The original expression $3^4 = 3 \times 3 \times 3 \times 3$ is all multiplication — so the two pairs of $9$s must be multiplied, not added.
(c) Corrected working:
$3^4 = 3 \times 3 \times 3 \times 3$
$= (3 \times 3) \times (3 \times 3)$
$= 9 \times 9$
$= \mathbf{81}$.
Quick check: doubling-style growth — $3^2 = 9$, $3^3 = 27$, $3^4 = 81$ ✓.
3 — Open-ended (sample solution)
We need $a^n = 64$ with $a > 1$ and $a, n$ positive whole numbers.
Expression 1: $2^6$.
Working: $2 \times 2 \times 2 \times 2 \times 2 \times 2 = 4 \times 4 \times 4 = 16 \times 4 = 64$ ✓.
Base $= 2$, index $= 6$.
Expression 2: $4^3$.
Working: $4 \times 4 \times 4 = 16 \times 4 = 64$ ✓.
Base $= 4$, index $= 3$.
Expression 3: $8^2$.
Working: $8 \times 8 = 64$ ✓.
Base $= 8$, index $= 2$.
Marking: 1 mark per correct expression (with working) — up to 3 marks. 1 extra mark for clearly labelling base and index in each. Other valid answers exist via more creative bases (e.g. $64^1$ excluded by the bonus rule).