The shape of the lesson
By the end of the lesson, learners can
- Logs and indices are the same fact: \(2^5 = 32\) and \(\log_2 32 = 5\) say the same thing: the power of 2 that gives 32 is 5.
- The three laws: The laws of logarithms are the index laws in disguise: multiplying powers adds the indices, so the log of a product is the sum of the logs.
- Solving log equations: Collect the logs on one side into a single log using the laws, then convert to index form.
- Definition: \(\log_a x = y \iff a^y = x\).
- Power law first: move coefficients inside: \(k\log_a x = \log_a x^k\).
- Combine: sums become products, differences become quotients.
- Undo the log: \(\log_a(\text{stuff}) = c\) means \(\text{stuff} = a^c\).
- Check: reject any root that makes the inside of an original log zero or negative.
One picture every idea returns to
\(\log_a x = y \iff a^y = x\) (for \(a > 0\), \(a \ne 1\), \(x > 0\)). The laws: \(\log_a xy = \log_a x + \log_a y\), \(\log_a \dfrac{x}{y} = \log_a x - \log_a y\), \(\log_a x^k = k\log_a x\). Always check that every logarithm in the original equation is of a positive number.
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, A logarithm is an index.. Papers 2 and 3. \(\log_a x\) answers one question: what power of \(a\) gives \(x\)? 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: Logs and indices are the same fact | \(2^5 = 32\) and \(\log_2 32 = 5\) say the same thing: the power of 2 that gives 32 is 5. Model the worked example Switching forms on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: The three laws | The laws of logarithms are the index laws in disguise: multiplying powers adds the indices, so the log of a product is the sum of the logs. Model the worked example Combining into a single log on the board, one line at a time. Your turn: Without a calculator, show that \(2\lg 5 + \lg 12 - \lg 3 = 2\). | Pairs |
| 21 to 30 min | Teach and check: Solving log equations | Collect the logs on one side into a single log using the laws, then convert to index form. Model the worked example Sum of logs on the board, one line at a time. | Rally Coach |
| 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, 22 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 Solving log equations, 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.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
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 |
|---|---|
| \(\log_2 \dfrac{1}{8}\) is \(-3\), not \(\dfrac{1}{3}\) or \(-\dfrac{1}{3}\). | Negative logs come from fractions; fractional logs come from roots. |
| There is no law for \(\log(x + y)\): \(\log_a(x + y) \ne \log_a x + \log_a y\). | Also \(\dfrac{\log_a x}{\log_a y}\) is NOT \(\log_a \dfrac{x}{y}\). |
| Do not 'cancel the logs' term by term: \(\log_2(x + 3) + \log_2(x - 1) = 5\) does NOT give \((x + 3) + (x - 1) = 5\) or \(= 32\). | Make ONE log first, then undo it. |
Logarithms and the laws of logarithms: worksheet
6 questions, 22 marks. Written in the style of the papers; no past paper question is reproduced.
Find the exact value of (a) \(\log_2 64\) (b) \(\log_9 3\) (c) \(\log_5 \dfrac{1}{125}\) (d) \(\log_8 4\).
Given that \(\log_a x = p\) and \(\log_a y = q\), express in terms of \(p\) and \(q\): (a) \(\log_a(x^2 y)\) (b) \(\log_a \dfrac{\sqrt{x}}{y^3}\) (c) \(\log_a \dfrac{ax}{y}\).
Express \(3\log_a 2 + \log_a 5 - \log_a 4\) as a single logarithm in its simplest form.
Solve the equation \(\log_5(x + 2) + \log_5(x - 2) = 1\).
Solve the equation \(2\log_3 x - \log_3(x - 2) = 2\).
The positive numbers \(x\) and \(y\) satisfy
\(\log_2 x + \log_2 y = 5\) and \(\log_2 x - 2\log_2 y = -1\).
(a) Show that \(xy = 32\) and \(y^2 = 2x\).
(b) Hence find the values of \(x\) and \(y\).
Answer key and marking notes
- (a) \(2^6 = 64\), so \(6\).
- (b) \(9^{1/2} = 3\), so \(\dfrac{1}{2}\).
- (c) \(5^{-3} = \dfrac{1}{125}\), so \(-3\).
- (d) \(8^y = 4\) means \(2^{3y} = 2^2\), so \(y = \dfrac{2}{3}\).
- (a) \(2p + q\)
- (b) \(\dfrac{1}{2}p - 3q\)
- (c) \(1 + p - q\)
- \(3\log_a 2 = \log_a 8\).
- \(\log_a \dfrac{8 \times 5}{4} = \log_a 10\).
- \(\log_5\left[(x + 2)(x - 2)\right] = 1\), so \(x^2 - 4 = 5\).
- \(x^2 = 9\), so \(x = 3\) or \(x = -3\).
- \(x = -3\) makes \(x + 2 = -1\) negative: reject. So \(x = 3\).
- \(\log_3 \dfrac{x^2}{x - 2} = 2\), so \(\dfrac{x^2}{x - 2} = 9\).
- \(x^2 = 9x - 18\), so \(x^2 - 9x + 18 = 0\).
- \((x - 3)(x - 6) = 0\), so \(x = 3\) or \(x = 6\).
- Both exceed 2, so every log is defined: \(x = 3\) or \(x = 6\).
- (a) \(\log_2(xy) = 5\), so \(xy = 2^5 = 32\).
- \(\log_2 \dfrac{x}{y^2} = -1\), so \(\dfrac{x}{y^2} = 2^{-1} = \dfrac{1}{2}\), i.e. \(y^2 = 2x\).
- (b) From \(y^2 = 2x\): \(x = \dfrac{y^2}{2}\). Substitute: \(\dfrac{y^3}{2} = 32\), so \(y^3 = 64\).
- \(y = 4\) and \(x = \dfrac{16}{2} = 8\). (Check: \(\log_2 8 + \log_2 4 = 3 + 2 = 5\); \(3 - 2(2) = -1\).)
8 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.
Find the exact value of \(\log_8 4\).
- A\(\dfrac{2}{3}\)
- B\(\dfrac{3}{2}\)
- C\(\dfrac{1}{2}\)
- D\(2\)
Express \(3\log_a 2 + \log_a 5 - \log_a 4\) as a single logarithm.
- A\(\log_a 10\)
- B\(\log_a 7.5\)
- C\(\log_a 7\)
- D\(\log_a 160\)
Given that \(\log_a x = p\) and \(\log_a y = q\), express \(\log_a \dfrac{ax^2}{y}\) in terms of \(p\) and \(q\).
- A\(1 + 2p - q\)
- B\(a + 2p - q\)
- C\(2p - q\)
- D\(1 + p^2 - q\)
Solve the equation \(\log_3(2x - 1) = 2\).
- A\(x = 5\)
- B\(x = 4\)
- C\(x = \dfrac{7}{2}\)
- D\(x = \dfrac{3}{2}\)
Solve the equation \(\log_3 x + \log_3(x + 6) = 3\).
- A\(x = 3\)
- B\(x = 3\) or \(x = -9\)
- C\(x = \dfrac{21}{2}\)
- D\(x = -\dfrac{3}{2}\)
Solve the equation \(2\log_2 x - \log_2(x + 3) = 2\).
- A\(x = 6\)
- B\(x = 6\) or \(x = -2\)
- C\(x = \dfrac{1 + \sqrt{29}}{2}\)
- D\(x = -6\)
Solve the simultaneous equations \(\log_3(xy) = 4\) and \(\log_3\left(\dfrac{x}{y}\right) = 2\), where \(x > 0\) and \(y > 0\).
- A\(x = 27\), \(y = 3\)
- B\(x = 3\), \(y = 27\)
- C\(x = 9\), \(y = 9\)
- D\(x = 3\), \(y = 1\)
Solve the equation \(\lg(x + 1) = 1 + \lg(x - 2)\), giving \(x\) as an exact fraction.
- A\(x = \dfrac{7}{3}\)
- B\(x = \dfrac{1}{3}\)
- C\(x = \dfrac{19}{9}\)
- D\(x = \dfrac{1 + 2e}{e - 1}\)
Quiz key
If \(\log_8 4 = y\) then \(8^y = 4\), so \(2^{3y} = 2^2\), \(3y = 2\), \(y = \dfrac{2}{3}\). Check: \(8^{2/3} = (\sqrt[3]{8})^2 = 4\). The tempting \(\dfrac{3}{2}\) is \(\log_4 8\), the base and the number swapped. \(\dfrac{1}{2}\) is \(4 \div 8\), and 2 is \(\log_2 4\).
\(3\log_a 2 = \log_a 8\). Then \(\log_a 8 + \log_a 5 - \log_a 4 = \log_a \dfrac{8 \times 5}{4} = \log_a 10\). The tempting \(\log_a 7.5\) uses \(3 \times 2 = 6\) instead of \(2^3 = 8\). \(\log_a 7\) adds and subtracts the numbers \(6 + 5 - 4\) as if logs were numbers, and \(\log_a 160\) multiplies by 4 instead of dividing.
\(\log_a a + 2\log_a x - \log_a y = 1 + 2p - q\), since \(\log_a a = 1\). The tempting \(a + 2p - q\) takes \(\log_a a\) to be \(a\). \(2p - q\) leaves out the \(\log_a a\) term altogether, and \(p^2\) treats \(\log_a x^2\) as \((\log_a x)^2\).
\(2x - 1 = 3^2 = 9\), so \(2x = 10\) and \(x = 5\). Check: \(\log_3 9 = 2\). The tempting \(x = 4\) comes from \(2x = 9 - 1\), a sign slip when rearranging. \(\dfrac{7}{2}\) uses \(3 \times 2 = 6\) instead of \(3^2\), and \(\dfrac{3}{2}\) ignores the log and solves \(2x - 1 = 2\).
\(\log_3 x(x + 6) = 3\), so \(x^2 + 6x = 27\), \((x + 9)(x - 3) = 0\), \(x = 3\) or \(x = -9\). But \(\log_3(-9)\) is undefined, so \(x = 3\) only. Check: \(\log_3 3 + \log_3 9 = 1 + 2 = 3\). The tempting "\(x = 3\) or \(x = -9\)" skips the check. \(\dfrac{21}{2}\) turns the sum of logs into the log of a sum, \(x + (x + 6) = 27\), and \(-\dfrac{3}{2}\) also adds instead of multiplying, using \(3\) instead of \(3^3\).
\(\log_2 \dfrac{x^2}{x + 3} = 2\), so \(x^2 = 4(x + 3)\), \(x^2 - 4x - 12 = 0\), \((x - 6)(x + 2) = 0\). \(x = -2\) is rejected because \(\log_2(-2)\) is undefined, so \(x = 6\). Check: \(2\log_2 6 - \log_2 9 = \log_2 \dfrac{36}{9} = \log_2 4 = 2\). The tempting "6 or \(-2\)" does not check. \(\dfrac{1 + \sqrt{29}}{2}\) comes from \(x^2 - (x + 3) = 4\), treating the difference of logs as a difference of numbers, and \(-6\) from writing \(2\log_2 x\) as \(\log_2 2x\).
\(xy = 81\) and \(\dfrac{x}{y} = 9\). Multiplying: \(x^2 = 729\), \(x = 27\) (positive), and \(y = \dfrac{81}{27} = 3\). Check: \(\log_3 81 = 4\), \(\log_3 9 = 2\). The tempting \(x = 3\), \(y = 27\) swaps the roles of \(x\) and \(y\), giving \(\dfrac{x}{y} = \dfrac{1}{9}\). \(x = 9\), \(y = 9\) gives \(\dfrac{x}{y} = 1\), and \(x = 3\), \(y = 1\) solves \(x + y = 4\), \(x - y = 2\), ignoring the logs.
\(\lg(x + 1) = \lg 10 + \lg(x - 2) = \lg 10(x - 2)\), so \(x + 1 = 10x - 20\), \(9x = 21\), \(x = \dfrac{7}{3}\). Check: \(x - 2 = \dfrac{1}{3} > 0\), and \(\dfrac{x + 1}{x - 2} = \dfrac{10/3}{1/3} = 10\). The tempting \(\dfrac{1}{3}\) expands \(10(x - 2)\) as \(10x - 2\), and it is invalid anyway since \(\lg(x - 2)\) needs \(x > 2\). \(\dfrac{19}{9}\) is a sign slip giving \(9x = 19\), and the \(e\) option treats \(\lg\) as \(\ln\) (writing 1 as \(\ln e\)).
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.