The shape of the lesson
By the end of the lesson, learners can
- The three laws: In the lab, choose a base \(a\) and the powers \(m\) and \(n\), and watch the factors line up: multiplying joins the two strings of factors, dividing cancels them in pairs.
- Zero and negative indices: Dividing \(5^{3}\) by itself gives 1, but the law gives \(5^{3-3} = 5^{0}\), so \(5^{0} = 1\).
- Fractional indices: Extended.
- Multiply: add indices. Divide: subtract. Power of a power: multiply.
- Zero: \(a^{0} = 1\). Negative: \(a^{-n} = \dfrac{1}{a^{n}}\); a fraction flips.
- Fractional (Extended): bottom = root, top = power: root first.
- Equations: write both sides as powers of the same base, then compare indices.
- Check: a negative index never makes the answer negative.
One picture every idea returns to
With the same base: \(a^{m} \times a^{n} = a^{m+n}\), \(a^{m} \div a^{n} = a^{m-n}\) and \((a^{m})^{n} = a^{mn}\). Also \(a^{0} = 1\), \(a^{-n} = \dfrac{1}{a^{n}}\) and (Extended) \(a^{\frac{1}{n}} = \sqrt[n]{a}\).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Count the factors.. An index counts how many times a number is multiplied by itself. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: The three laws | In the lab, choose a base \(a\) and the powers \(m\) and \(n\), and watch the factors line up: multiplying joins the two strings of factors, dividing cancels them in pairs. Project tab 2 of the lab (Powers and accuracy lab, indices) and let a student drive it while the class predicts. Model the worked example Using the laws on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Zero and negative indices | Dividing \(5^{3}\) by itself gives 1, but the law gives \(5^{3-3} = 5^{0}\), so \(5^{0} = 1\). Model the worked example Negative indices on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: Fractional indices | Extended. Model the worked example Root, then power (Extended) on the board, one line at a time. Your turn: (Extended) Find the value of \(32^{-\frac{3}{5}}\). | Round Table |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 19 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Fractional indices, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| The laws only work with the SAME base, and the base never changes: \(3^{4} \times 3^{2} = 3^{6}\), not \(9^{6}\). | And \(2^{3} \times 5^{2}\) cannot be combined at all. |
| A negative index does NOT make the number negative: \(7^{-2} = \dfrac{1}{49}\), not \(-49\) and not \(-14\). | Return to Zero and negative indices: ask a pair to state the correct version and show it on the lab before moving on. |
| \(8^{-\frac{2}{3}}\) is not \(8 \times -\dfrac{2}{3}\). | An index is never multiplied by the base; it tells you to take roots, powers and reciprocals. |
Indices: the laws of indices: worksheet
6 questions, 19 marks. Written in the style of the papers; no past paper question is reproduced.
Find the value of (a) \(6^{0}\) (b) \(4^{-2}\) (c) \(10^{-3}\).
Simplify, leaving each answer in index form: (a) \(7^{5} \times 7^{-2}\) (b) \((5^{-2})^{3}\) (c) \(9^{4} \div 9^{6}\).
Find the value of (a) \(2^{-3} \times 2^{4}\) (b) \((2^{3})^{2}\) (c) \(2^{3} \div 2^{4}\).
(Extended) Find the value of (a) \(81^{\frac{1}{2}}\) (b) \(125^{\frac{2}{3}}\) (c) \(16^{-\frac{1}{2}}\).
Find \(x\) when (a) \(3^{x} = \dfrac{1}{81}\) (b) \(5^{x} = 1\) (c) (Extended) \(4^{x} = 32\).
A digital image is \(2^{11}\) pixels wide and \(2^{10}\) pixels high. Each pixel uses \(2^{5}\) bits of memory. There are \(2^{3}\) bits in a byte, and \(2^{20}\) bytes in a megabyte (MB).
(a) Find the number of bits in the image, as a power of 2. (b) Find the size of the image in megabytes.
Answer key and marking notes
- (a) 1
- (b) \(\dfrac{1}{4^{2}} = \dfrac{1}{16}\)
- (c) \(\dfrac{1}{1000} = 0.001\)
- (a) \(7^{5 + (-2)} = 7^{3}\)
- (b) \(5^{-2 \times 3} = 5^{-6}\)
- (c) \(9^{4-6} = 9^{-2}\)
- (a) \(2^{1} = 2\)
- (b) \(2^{6} = 64\)
- (c) \(2^{-1} = \dfrac{1}{2}\)
- (a) \(\sqrt{81} = 9\)
- (b) \(\sqrt[3]{125} = 5\), \(5^{2} = 25\)
- (c) \(\sqrt{16} = 4\), so \(\dfrac{1}{4}\)
- (a) \(\dfrac{1}{81} = 3^{-4}\), so \(x = -4\)
- (b) \(5^{0} = 1\), so \(x = 0\)
- (c) \(2^{2x} = 2^{5}\), so \(2x = 5\) and \(x = \dfrac{5}{2}\)
- (a) Pixels: \(2^{11} \times 2^{10} = 2^{21}\). Bits: \(2^{21} \times 2^{5} = 2^{26}\)
- (b) Bytes: \(2^{26} \div 2^{3} = 2^{23}\)
- Megabytes: \(2^{23} \div 2^{20} = 2^{3} = 8\) MB
6 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Simplify \(5^{7} \times 5^{3} \div 5^{4}\), leaving your answer as a power of 5.
- A\(5^{6}\)
- B\(25^{6}\)
- C\(5^{17}\)
- D\(5^{14}\)
Simplify \(\left(x^{4}\right)^{3}\).
- A\(x^{12}\)
- B\(x^{7}\)
- C\(x^{64}\)
- D\(3x^{4}\)
Find the value of \(4^{-2}\).
- A\(\dfrac{1}{16}\)
- B\(-16\)
- C\(-8\)
- D\(\dfrac{1}{8}\)
Find the value of \(7^{0} + 2^{-3}\).
- A\(\dfrac{9}{8}\)
- B\(\dfrac{1}{8}\)
- C\(-6\)
- D\(7\dfrac{1}{8}\)
Find the value of \(16^{-\frac{3}{4}}\).
- A\(\dfrac{1}{8}\)
- B\(-8\)
- C\(-12\)
- D\(\dfrac{1}{12}\)
Find \(x\) when \(9^{x} = 27\).
- A\(\dfrac{3}{2}\)
- B\(3\)
- C\(\dfrac{2}{3}\)
- D\(\dfrac{1}{3}\)
Quiz key
\(5^{7} \times 5^{3} = 5^{7+3} = 5^{10}\), then \(5^{10} \div 5^{4} = 5^{10-4} = 5^{6}\). \(25^{6}\) multiplies the bases, but the base never changes. \(5^{17}\) multiplies the indices \(7 \times 3\) before subtracting 4, and \(5^{14}\) adds all three indices instead of subtracting for the division.
\(\left(x^{4}\right)^{3} = x^{4} \times x^{4} \times x^{4} = x^{4+4+4} = x^{12}\): a power of a power multiplies the indices. The tempting \(x^{7}\) adds the indices, which is the rule for multiplying, not for a power of a power. \(x^{64}\) works out \(4^{3}\), and \(3x^{4}\) is \(x^{4} + x^{4} + x^{4}\).
\(4^{-2} = \dfrac{1}{4^{2}} = \dfrac{1}{16}\). The tempting \(-16\) treats the negative index as making the answer negative. \(-8\) multiplies \(4 \times (-2)\), and \(\dfrac{1}{8}\) takes the reciprocal of \(4 \times 2\) instead of \(4^{2}\).
\(7^{0} = 1\) and \(2^{-3} = \dfrac{1}{2^{3}} = \dfrac{1}{8}\), so the total is \(1 + \dfrac{1}{8} = \dfrac{9}{8}\). \(\dfrac{1}{8}\) treats \(7^{0}\) as 0, but any non-zero number to the power 0 is 1. \(7\dfrac{1}{8}\) treats \(7^{0}\) as 7, and \(-6\) uses \(7^{0} = 0\) and \(2^{-3} = -6\).
The bottom of the index is a root: \(\sqrt[4]{16} = 2\). The top is a power: \(2^{3} = 8\). The minus sign means a reciprocal: \(16^{-\frac{3}{4}} = \dfrac{1}{8}\). \(-8\) treats the minus sign as making the answer negative. \(-12\) and \(\dfrac{1}{12}\) multiply \(16 \times \dfrac{3}{4}\), but an index is never multiplied by the base.
\(9 = 3^{2}\) and \(27 = 3^{3}\), so \(\left(3^{2}\right)^{x} = 3^{3}\), which gives \(3^{2x} = 3^{3}\). Then \(2x = 3\), so \(x = \dfrac{3}{2}\). Check: \(9^{\frac{3}{2}} = \left(\sqrt{9}\right)^{3} = 27\). \(\dfrac{2}{3}\) solves \(3x = 2\), the indices the wrong way round. \(x = 3\) gives \(9^{3} = 729\), and \(\dfrac{1}{3}\) gives \(\sqrt[3]{9}\), about 2.08.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.