The shape of the lesson
By the end of the lesson, learners can
- The three laws: In the lab pick a rule and move the sliders for \(m\) and \(n\).
- Zero and negative powers: Divide \(x^{3}\) by \(x^{3}\): the answer is 1, and the law gives \(x^{0}\).
- Fractional powers (Extended): Since \(x^{\frac{1}{2}} \times x^{\frac{1}{2}} = x^{1}\), the power \(\frac{1}{2}\) must mean square root.
- The unknown in the power: Write both sides as powers of the same base.
- Multiply: add the powers; multiply the numbers in front.
- Divide: subtract the powers; divide the numbers in front.
- Bracket to a power: every factor inside gets the power.
- Negative and zero: \(x^{-n} = \dfrac{1}{x^{n}}\), \(x^{0} = 1\).
- Fractional (Extended): bottom is the root, top is the power.
- Equations: same base on both sides, then equate the powers.
One picture every idea returns to
Multiply: add the powers. Divide: subtract the powers. Power of a power: multiply the powers. \(x^{0} = 1\), \(x^{-n} = \dfrac{1}{x^{n}}\), and (Extended) \(x^{\frac{m}{n}} = (\sqrt[n]{x})^{m}\). Numbers in front are multiplied or divided as normal.
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, Powers that obey rules.. Apply the laws of indices to letters as well as numbers: positive, zero and negative powers, then (Extended) fractional powers, and use matching bases to solve equations such as \(2^{x} = 32\). | Whole class, then pairs |
| 5 to 11 min | Teach and check: The three laws | In the lab pick a rule and move the sliders for \(m\) and \(n\). Project tab 5 of the lab (Algebra lab, indices) and let a student drive it while the class predicts. Model the worked example Power of a product on the board, one line at a time. | Whole class, then pairs |
| 11 to 17 min | Teach and check: Zero and negative powers | Divide \(x^{3}\) by \(x^{3}\): the answer is 1, and the law gives \(x^{0}\). Model the worked example Writing without negative powers on the board, one line at a time. Your turn: Without a calculator, find the value of \(2^{-3} \times 4^{0}\). | Pairs |
| 17 to 23 min | Teach and check: Fractional powers (Extended) | Since \(x^{\frac{1}{2}} \times x^{\frac{1}{2}} = x^{1}\), the power \(\frac{1}{2}\) must mean square root. Model the worked example Fractional power of a term (Extended) on the board, one line at a time. Your turn: (Extended) Simplify \(\left(\dfrac{16x^{8}}{81}\right)^{\frac{3}{4}}\). | Pairs |
| 23 to 30 min | Teach and check: The unknown in the power | Write both sides as powers of the same base. Model the worked example Same base on the board, one line at a time. Your turn: (Extended) Solve \(4^{x} = 8\). | 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, 18 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 The unknown in the power, 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 |
|---|---|
| \((5x^{3})^{2}\) is not \(5x^{6}\): the 5 is inside the bracket, so it is squared too. | And the numbers in front are multiplied, never added: \(3x^{2} \times 4x^{3} = 12x^{5}\). |
| \(3x^{-2}\) is \(\dfrac{3}{x^{2}}\), not \(\dfrac{1}{3x^{2}}\). | The power belongs only to the letter it touches. |
| \(8^{\frac{2}{3}}\) is not \(8 \times \frac{2}{3}\). | The fraction is an instruction: cube root, then square. |
| \(\dfrac{1}{27}\) is \(3^{-3}\), not \(3^{3}\). | A fraction 'one over' needs a negative power. |
Indices II: algebraic and fractional indices: worksheet
6 questions, 18 marks. Written in the style of the papers; no past paper question is reproduced.
Simplify (a) \(a^{4} \times a^{6}\) (b) \(b^{9} \div b^{3}\) (c) \((c^{2})^{5}\).
Simplify \(6x^{7}y^{4} \times 5x^{-5}y\).
Find the value of \(x\) when (a) \(2^{x} = 32\) (b) \(3^{x} = \dfrac{1}{9}\).
(Extended) Without a calculator, find the value of (a) \(64^{\frac{1}{3}}\) (b) \(9^{-\frac{3}{2}}\) (c) \(\left(\dfrac{8}{27}\right)^{\frac{2}{3}}\).
(Extended) Simplify (a) \((27x^{6})^{\frac{2}{3}}\) (b) \(4x^{\frac{1}{2}} \times 3x^{-\frac{3}{2}}\).
(Extended)
(a) Solve \(9^{x + 1} = 27^{x}\).
(b) Solve \(4^{x} \times 2^{x + 3} = 64\).
Answer key and marking notes
- (a) \(a^{10}\)
- (b) \(b^{6}\)
- (c) \(c^{10}\)
- \(6 \times 5 = 30\); \(x^{7 + (-5)} = x^{2}\); \(y^{4 + 1} = y^{5}\)
- \(30x^{2}y^{5}\)
- (a) \(32 = 2^{5}\), so \(x = 5\)
- (b) \(\dfrac{1}{9} = 3^{-2}\), so \(x = -2\)
- (a) \(\sqrt[3]{64} = 4\)
- (b) \(9^{\frac{3}{2}} = 3^{3} = 27\), so \(9^{-\frac{3}{2}} = \dfrac{1}{27}\)
- (c) \(\left(\dfrac{2}{3}\right)^{2} = \dfrac{4}{9}\)
- (a) \(27^{\frac{2}{3}} = 9\) and \(x^{6 \times \frac{2}{3}} = x^{4}\), so \(9x^{4}\)
- (b) \(12x^{\frac{1}{2} - \frac{3}{2}} = 12x^{-1} = \dfrac{12}{x}\)
- (a) \(9^{x + 1} = 3^{2x + 2}\) and \(27^{x} = 3^{3x}\)
- \(2x + 2 = 3x\), so \(x = 2\)
- (b) \(4^{x} = 2^{2x}\), so the left side is \(2^{2x + x + 3} = 2^{3x + 3}\)
- \(64 = 2^{6}\), so \(3x + 3 = 6\)
- \(x = 1\)
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 \(2a^{4}b^{3} \times 7a^{2}b^{-5}\).
- A\(14a^{6}b^{-2}\)
- B\(9a^{6}b^{-2}\)
- C\(14a^{8}b^{-15}\)
- D\(14a^{6}b^{8}\)
Simplify \((3x^{4})^{3}\).
- A\(27x^{12}\)
- B\(3x^{12}\)
- C\(9x^{12}\)
- D\(27x^{7}\)
Simplify \(20x^{6} \div 4x^{-3}\).
- A\(5x^{9}\)
- B\(5x^{3}\)
- C\(16x^{9}\)
- D\(5x^{-2}\)
Solve \(2^{x} = \dfrac{1}{16}\).
- A\(x = -4\)
- B\(x = 4\)
- C\(x = -8\)
- D\(x = \dfrac{1}{4}\)
Without a calculator, find the value of \(27^{-\frac{2}{3}}\).
- A\(\dfrac{1}{9}\)
- B\(-9\)
- C\(9\)
- D\(-18\)
Solve \(9^{x} = 27^{x - 1}\).
- A\(x = 3\)
- B\(x = 1\)
- C\(x = -3\)
- D\(x = \dfrac{3}{5}\)
Quiz key
\(2 \times 7 = 14\), \(a^{4 + 2} = a^{6}\), \(b^{3 + (-5)} = b^{-2}\). Answer \(14a^{6}b^{-2}\). \(9a^{6}b^{-2}\) adds the numbers in front instead of multiplying. \(14a^{8}b^{-15}\) multiplies the powers. \(b^{8}\) treats \(-5\) as \(+5\).
\(3^{3} = 27\) and \((x^{4})^{3} = x^{4 \times 3} = x^{12}\), so \(27x^{12}\). \(3x^{12}\) forgets to cube the 3; \(9x^{12}\) multiplies 3 by 3 instead of cubing; \(x^{7}\) adds the powers instead of multiplying.
\(20 \div 4 = 5\) and \(x^{6 - (-3)} = x^{9}\), so \(5x^{9}\). \(5x^{3}\) does \(6 - 3\) instead of \(6 - (-3)\). \(16x^{9}\) subtracts the numbers in front. \(x^{-2}\) divides the powers.
\(16 = 2^{4}\), so \(\dfrac{1}{16} = 2^{-4}\). Then \(2^{x} = 2^{-4}\) gives \(x = -4\). \(x = 4\) gives \(2^{4} = 16\), not \(\dfrac{1}{16}\). \(x = -8\) halves 16 instead of finding the power.
\(\sqrt[3]{27} = 3\), \(3^{2} = 9\), and the negative power gives the reciprocal: \(\dfrac{1}{9}\). A negative power never makes the answer negative, so \(-9\) is wrong. 9 ignores the minus sign. \(-18\) multiplies 27 by \(-\dfrac{2}{3}\).
\(9^{x} = 3^{2x}\) and \(27^{x - 1} = 3^{3(x - 1)} = 3^{3x - 3}\). So \(2x = 3x - 3\), giving \(x = 3\). Check: \(9^{3} = 729 = 27^{2}\). \(x = 1\) comes from \(2x = 3x - 1\), forgetting to multiply the \(-1\) by 3.
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.