The shape of the lesson
By the end of the lesson, learners can
- Which graph is a straight line?: Take ln of both sides of the model and expand with the laws of logarithms.
- Constants from two points: Exam questions give two points on the straight line (or one point and the gradient).
- From a table of data: Experimental data rarely come labelled with their model.
- \(y = kx^n\): \(\ln y\) against \(\ln x\); gradient \(n\), intercept \(\ln k\).
- \(y = ka^x\): \(\ln y\) against \(x\); gradient \(\ln a\), intercept \(\ln k\).
- Gradient from two points, then intercept from \(Y = mX + c\) at one point.
- Undo the logs: \(k = e^{c}\), \(a = e^{m}\) (or \(10^{(\dots)}\) for lg).
- Answer the question: write the model, then substitute if asked.
One picture every idea returns to
\(y = kx^n \Rightarrow \ln y = \ln k + n\ln x\): plot \(\ln y\) against \(\ln x\); gradient \(n\), intercept \(\ln k\). \(y = ka^x \Rightarrow \ln y = \ln k + x\ln a\): plot \(\ln y\) against \(x\); gradient \(\ln a\), intercept \(\ln k\). Compare with \(Y = mX + c\).
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, Curves made straight.. Papers 2 and 3. 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: Which graph is a straight line? | Take ln of both sides of the model and expand with the laws of logarithms. Model the worked example Choosing the axes on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Constants from two points | Exam questions give two points on the straight line (or one point and the gradient). Model the worked example Exponential model, ln y against x on the board, one line at a time. Your turn: \(y = Ae^{kx}\), and the graph of \(\ln y\) against \(x\) passes through \((0.5, 1.9)\) and \((2.5, 0.7)\). Find \(A\) and \(k\). | Pairs |
| 21 to 30 min | Teach and check: From a table of data | Experimental data rarely come labelled with their model. Model the worked example Confirming an exponential model on the board, one line at a time. | Jigsaw |
| 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, 28 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 From a table of data, 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.
Jigsaw
Split the content into expert pieces. Expert groups master one piece each, then regroup so every home team has one expert per piece, and each expert teaches their piece to the team.
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 |
|---|---|
| For \(y = ka^x\) the horizontal axis is plain \(x\), not \(\ln x\): the \(x\) is already in the index, and taking logs brings it down as it is. | Plot \(\ln x\) only when \(x\) is raised to a constant power. |
| The coordinates in these questions are values of \(\ln y\) (and perhaps \(\ln x\)), not of \(y\). | The intercept of the line is \(\ln k\); answering \(k = 2.6\) instead of \(k = e^{2.6}\) loses the final marks. |
| Undo with the SAME base you took logs in. | If the axis is \(\lg y\), use \(10^{1.2}\); using \(e^{1.2}\) gives \(3.32\), a completely different (wrong) constant. |
Reducing relationships to linear form: worksheet
6 questions, 28 marks. Written in the style of the papers; no past paper question is reproduced.
For each relationship, state which variables should be plotted to give a straight line, and give the gradient and intercept: (a) \(y = 5x^3\) (b) \(y = 2 \times 7^x\).
The variables \(x\) and \(y\) satisfy \(y = kx^n\). The graph of \(\ln y\) against \(\ln x\) is a straight line through \((1.0, 3.2)\) and \((2.5, 2.3)\). Find \(k\) and \(n\).
The variables satisfy \(y = Ae^{kx}\). The graph of \(\ln y\) against \(x\) passes through \((0, 2.4)\) and has gradient \(-0.3\).
(a) Find \(A\) and \(k\). (b) Find \(y\) when \(x = 4\). (c) Find \(x\) when \(y = 5\).
Some experimental values are shown.
| \(x\) | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| \(y\) | 2.50 | 7.07 | 12.99 | 20.00 | 27.95 |
It is thought that \(y = kx^n\). By considering \(\ln y\) and \(\ln x\), show that the data fit this model and find \(k\) and \(n\).
The variables satisfy \(y = Ab^x\). When \(\ln y\) is plotted against \(x\), the line passes through \((1, 2.0)\) and \((3, 3.4)\).
(a) Find \(A\) and \(b\). (b) Hence find the value of \(x\) for which \(y = 100\).
The number of bacteria, \(N\), in a culture \(t\) hours after it is started is modelled by \(N = Ab^t\). A graph of \(\lg N\) against \(t\) is a straight line passing through \((2, 3.10)\) and \((6, 4.30)\).
(a) Find the values of \(A\) and \(b\).
(b) Estimate the number of bacteria after 10 hours.
(c) Find the time at which the number of bacteria first reaches one million.
Answer key and marking notes
- (a) \(\ln y = 3\ln x + \ln 5\): plot \(\ln y\) against \(\ln x\); gradient \(3\), intercept \(\ln 5\).
- (b) \(\ln y = (\ln 7)x + \ln 2\): plot \(\ln y\) against \(x\); gradient \(\ln 7\), intercept \(\ln 2\).
- Gradient: \(\dfrac{2.3 - 3.2}{2.5 - 1.0} = \dfrac{-0.9}{1.5} = -0.6\), so \(n = -0.6\).
- Intercept: \(3.2 = -0.6(1.0) + c\), so \(c = 3.8 = \ln k\).
- \(k = e^{3.8} = 44.7\) (3 s.f.), so \(y = 44.7x^{-0.6}\).
- (a) \(k = -0.3\) and \(\ln A = 2.4\), so \(A = e^{2.4} = 11.0\) (3 s.f.).
- (b) \(\ln y = 2.4 - 0.3(4) = 1.2\), so \(y = e^{1.2} = 3.32\).
- (c) \(\ln 5 = 2.4 - 0.3x\), so \(x = \dfrac{2.4 - \ln 5}{0.3} = 2.64\).
- \(\ln x = 0,\ 0.693,\ 1.099,\ 1.386,\ 1.609\); \(\ln y = 0.916,\ 1.956,\ 2.564,\ 2.996,\ 3.330\).
- Gradients between successive points: \(\dfrac{1.040}{0.693} = 1.50\), \(\dfrac{0.608}{0.406} = 1.50\), \(\dfrac{0.432}{0.287} = 1.50\), \(\dfrac{0.334}{0.223} = 1.50\): constant, so the points lie on a straight line.
- So \(n = 1.5\), and the intercept (at \(\ln x = 0\)) is \(0.916 = \ln k\).
- \(k = e^{0.916} = 2.50\), so \(y = 2.5x^{1.5}\).
- (a) Gradient \(\dfrac{3.4 - 2.0}{3 - 1} = 0.7 = \ln b\), so \(b = e^{0.7} = 2.01\). Intercept \(2.0 - 0.7 = 1.3 = \ln A\), so \(A = e^{1.3} = 3.67\) (3 s.f.).
- (b) \(\ln 100 = 1.3 + 0.7x\), so \(x = \dfrac{\ln 100 - 1.3}{0.7} = \dfrac{4.6052 - 1.3}{0.7} = 4.72\) (3 s.f.).
- (a) \(\lg N = \lg A + t\lg b\). Gradient \(\dfrac{4.30 - 3.10}{6 - 2} = 0.3 = \lg b\), so \(b = 10^{0.3} = 2.00\) (3 s.f.).
- Intercept: \(3.10 = 0.3(2) + \lg A\), so \(\lg A = 2.5\) and \(A = 10^{2.5} = 316\) (3 s.f.).
- (b) \(\lg N = 2.5 + 0.3(10) = 5.5\), so \(N = 10^{5.5} \approx 316\,000\).
- (c) One million is \(10^6\), so \(\lg N = 6\): \(6 = 2.5 + 0.3t\).
- \(t = \dfrac{3.5}{0.3} = 11.7\) hours (3 s.f.).
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.
The variables \(x\) and \(y\) satisfy \(y = 3x^4\). The graph of \(\ln y\) against \(\ln x\) is a straight line. State its gradient and the intercept on the \(\ln y\) axis.
- AGradient \(4\), intercept \(\ln 3\)
- BGradient \(\ln 4\), intercept \(3\)
- CGradient \(4\), intercept \(3\)
- DGradient \(3\), intercept \(\ln 4\)
The variables satisfy \(y = 2 \times 5^x\). Which graph is a straight line, and what is its gradient?
- A\(\ln y\) against \(x\), gradient \(\ln 5\)
- B\(\ln y\) against \(\ln x\), gradient \(5\)
- C\(\ln y\) against \(x\), gradient \(\ln 2\)
- D\(y\) against \(x\), gradient \(5\)
The variables satisfy \(y = Ab^x\). The graph of \(\ln y\) against \(x\) is the straight line \(\ln y = 1.2x + 0.7\). Find \(A\) and \(b\), correct to 3 significant figures.
- A\(A = 2.01\), \(b = 3.32\)
- B\(A = 0.7\), \(b = 1.2\)
- C\(A = 3.32\), \(b = 2.01\)
- D\(A = 2.01\), \(b = 1.2\)
The graph of \(\ln y\) against \(\ln x\) is a straight line with gradient \(-0.5\) and intercept 2 on the \(\ln y\) axis. Express \(y\) in terms of \(x\), with the constant correct to 3 significant figures.
- A\(y = 7.39x^{-0.5}\)
- B\(y = 2x^{-0.5}\)
- C\(y = 7.39e^{-0.5x}\)
- D\(y = 0.607x^{2}\)
The variables satisfy \(y = kx^n\). The diagram shows the graph of \(\ln y\) against \(\ln x\): a straight line through \((1.0, 3.2)\) and \((2.5, 2.3)\). Find \(k\) and \(n\), correct to 3 significant figures.
- A\(k = 44.7\), \(n = -0.6\)
- B\(k = 3.8\), \(n = -0.6\)
- C\(k = 44.7\), \(n = 0.6\)
- D\(k = 24.5\), \(n = -0.6\)
The variables satisfy \(y = Ab^x\). The graph of \(\lg y\) against \(x\) is a straight line with gradient 0.25 that passes through the point \((2, 1.1)\). Find \(A\) and \(b\), correct to 3 significant figures.
- A\(A = 3.98\), \(b = 1.78\)
- B\(A = 1.82\), \(b = 1.28\)
- C\(A = 0.6\), \(b = 0.25\)
- D\(A = 12.6\), \(b = 1.78\)
The variables satisfy \(y = Ab^{-x}\), where \(A\) and \(b\) are constants. The graph of \(\ln y\) against \(x\) is a straight line with gradient \(-0.4\) and intercept 1.5 on the \(\ln y\) axis. Find \(A\) and \(b\), correct to 3 significant figures.
- A\(A = 4.48\), \(b = 1.49\)
- B\(A = 4.48\), \(b = 0.670\)
- C\(A = 1.5\), \(b = 0.4\)
- D\(A = 1.49\), \(b = 4.48\)
The variables satisfy \(y = kx^n\). When \(x = 4\), \(y = 40\), and when \(x = 9\), \(y = 135\). Find \(k\) and \(n\).
- A\(k = 5\), \(n = 1.5\)
- B\(k = 1.61\), \(n = 1.5\)
- C\(k = 5\), \(n = \dfrac{2}{3}\)
- D\(k = 1.5\), \(n = 5\)
Quiz key
\(\ln y = \ln 3 + 4\ln x\), which is \(Y = 4X + \ln 3\) with \(Y = \ln y\), \(X = \ln x\). Gradient 4, intercept \(\ln 3\). The tempting "gradient \(\ln 4\), intercept 3" puts the log on the wrong constant. "Intercept 3" forgets that the intercept is \(\ln k\), and "gradient 3" swaps the roles of \(k\) and \(n\).
\(\ln y = \ln 2 + x\ln 5\): this is linear in \(x\) and \(\ln y\), with gradient \(\ln 5\) and intercept \(\ln 2\). The tempting "\(\ln y\) against \(\ln x\)" is the right plot for \(y = kx^n\), not for an exponential. Gradient \(\ln 2\) mixes up the gradient and the intercept, and \(y\) against \(x\) is a curve, not a line.
\(\ln A = 0.7\), so \(A = e^{0.7} = 2.01\); \(\ln b = 1.2\), so \(b = e^{1.2} = 3.32\). The tempting \(A = 0.7\), \(b = 1.2\) forgets that the gradient and intercept are logarithms. \(A = 3.32\), \(b = 2.01\) swaps them, and \(b = 1.2\) undoes only one of the two logs.
\(\ln y = \ln x^{-0.5} + \ln e^2 = \ln\left(e^2x^{-0.5}\right)\), so \(y = e^2x^{-0.5} = 7.39x^{-0.5}\). The tempting \(y = 2x^{-0.5}\) forgets to exponentiate the intercept. \(7.39e^{-0.5x}\) is the exponential model, which belongs with a graph of \(\ln y\) against \(x\), and \(0.607x^2\) swaps the gradient and the intercept.
Gradient \(= \dfrac{2.3 - 3.2}{2.5 - 1.0} = -0.6 = n\). Intercept: \(3.2 = -0.6(1.0) + c\), \(c = 3.8 = \ln k\), so \(k = e^{3.8} = 44.7\). The tempting \(k = 3.8\) forgets that the intercept is \(\ln k\). \(n = 0.6\) loses the sign of a falling line, and \(k = 24.5\) is \(e^{3.2}\), taking the first point as the intercept although it is at \(\ln x = 1\), not 0.
Intercept: \(1.1 = 0.25(2) + c\), so \(c = 0.6 = \lg A\) and \(A = 10^{0.6} = 3.98\). \(\lg b = 0.25\), so \(b = 10^{0.25} = 1.78\). The tempting \(A = 1.82\), \(b = 1.28\) uses \(e\) instead of 10 to undo \(\lg\). \(A = 0.6\), \(b = 0.25\) forgets to undo the logs, and \(A = 12.6 = 10^{1.1}\) treats the given point as the intercept.
\(\ln y = \ln A - (\ln b)x\). Intercept \(\ln A = 1.5\), so \(A = e^{1.5} = 4.48\). Gradient \(-\ln b = -0.4\), so \(\ln b = 0.4\) and \(b = e^{0.4} = 1.49\). The tempting \(b = 0.670 = e^{-0.4}\) ignores the minus sign in \(b^{-x}\). \(A = 1.5\), \(b = 0.4\) forgets to exponentiate, and the last option swaps \(A\) and \(b\).
\(n = \dfrac{\ln 135 - \ln 40}{\ln 9 - \ln 4} = \dfrac{\ln 3.375}{\ln 2.25} = 1.5\) (indeed \(3.375 = 1.5^3\) and \(2.25 = 1.5^2\)). Then \(40 = k \times 4^{1.5} = 8k\), so \(k = 5\). Check: \(5 \times 9^{1.5} = 5 \times 27 = 135\). The tempting \(k = 1.61\) gives \(\ln k\) instead of \(k\) (\(\ln 5 = 1.61\)). \(n = \dfrac{2}{3}\) inverts the gradient (change in \(\ln x\) over change in \(\ln y\)), and the last option swaps \(k\) and \(n\).
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.