The shape of the lesson
By the end of the lesson, learners can
- The graph of \(y = e^{kx}\): growth and decay: In the lab, set a positive percentage rate and watch the curve climb faster and faster; then make the rate negative and watch it decay towards zero.
- \(\ln x\): the inverse of \(e^x\): \(\ln x\) means \(\log_e x\).
- Solving equations with \(e\) and ln: Isolate the exponential (or the log) first, then apply the inverse function to both sides.
- Inverse pair: \(\ln(e^x) = x\) and \(e^{\ln x} = x\) (\(x > 0\)).
- Isolate first: get \(e^{(\dots)}\) or \(\ln(\dots)\) alone on one side.
- Quadratic in \(e^x\): let \(u = e^x\); reject \(u \le 0\).
- Sketch: \(y\)-intercept, asymptote \(y = c\) for \(Ae^{kx} + c\), and any \(x\)-intercept.
- Exact or 3 s.f.: give \(\ln\) forms when asked for exact answers.
One picture every idea returns to
\(\ln x = \log_e x\), so \(y = e^x \iff x = \ln y\). They undo each other: \(\ln(e^x) = x\) for all \(x\), and \(e^{\ln x} = x\) for \(x > 0\). \(y = e^{kx}\) always passes through \((0, 1)\), stays above the \(x\)-axis, grows for \(k > 0\) and decays towards \(y = 0\) for \(k < 0\).
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, Two functions, one undoes the other.. 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: The graph of \(y = e^{kx}\): growth and decay | In the lab, set a positive percentage rate and watch the curve climb faster and faster; then make the rate negative and watch it decay towards zero. Project tab 4 of the lab (Graphs of functions lab, growth and decay) and let a student drive it while the class predicts. Model the worked example Sketching a transformed exponential on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: \(\ln x\): the inverse of \(e^x\) | \(\ln x\) means \(\log_e x\). Model the worked example Exact simplification on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: Solving equations with \(e\) and ln | Isolate the exponential (or the log) first, then apply the inverse function to both sides. Model the worked example Exponential to ln on the board, one line at a time. Your turn: Solve \(e^x - 6e^{-x} = 1\). | Think-Pair-Share |
| 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, 25 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 equations with \(e\) and ln, 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.
Think-Pair-Share
Pose the question, give thirty seconds of silent thinking, then a minute to compare with a partner before a few pairs share with the class. The silent minute is the part that cannot be cut.
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 |
|---|---|
| \(e^{-x}\) is never negative: \(y = e^{-x}\) is a decreasing curve ABOVE the \(x\)-axis, not the curve \(y = e^x\) flipped below it. | The minus sign in the power reflects in the \(y\)-axis; a minus sign in front, \(-e^x\), reflects in the \(x\)-axis. |
| \(\ln(e^x + 1)\) is NOT \(x + 1\), and \(\ln(2e^x)\) is \(\ln 2 + x\), not \(2x\). ln only undoes \(e\) when the whole inside of the log is a single power of \(e\). | Return to \(\ln x\): the inverse of \(e^x\): ask a pair to state the correct version and show it on the lab before moving on. |
| Take ln only after isolating the exponential: from \(3e^{x} + 1 = 10\), first get \(e^x = 3\), then \(x = \ln 3\). | Writing \(\ln 3 + x + \ln 1 = \ln 10\) is not a valid step, because there is no law for the log of a sum. |
The exponential function and natural logarithms: worksheet
6 questions, 25 marks. Written in the style of the papers; no past paper question is reproduced.
Simplify exactly: (a) \(e^{3\ln 2}\) (b) \(\ln \dfrac{1}{e^2}\) (c) \(e^{\ln 6 - \ln 3}\) (d) \(\ln\left(e^{2x} \times e^3\right)\).
Solve, giving answers exactly and to 3 significant figures: (a) \(e^{3x} = 20\) (b) \(\ln(2x - 1) = 3\).
Sketch the graph of \(y = 2e^{-x} - 1\), stating the exact coordinates of the points where it meets the axes and the equation of the asymptote.
Solve the equation \(3e^{2x} + e^x - 2 = 0\), giving your answer in exact form.
Solve the equation \(2\ln(x + 1) - \ln x = \ln 6\), giving your answers in exact form.
A cup of tea cools so that its temperature \(T\,\degree\text{C}\) after \(t\) minutes is \(T = 20 + 70e^{-0.05t}\).
(a) Find the initial temperature.
(b) Find the temperature after 10 minutes.
(c) Find the time taken for the temperature to fall to \(40\,\degree\text{C}\).
(d) State the temperature that the tea approaches in the long term.
Answer key and marking notes
- (a) \(e^{\ln 8} = 8\)
- (b) \(\ln e^{-2} = -2\)
- (c) \(e^{\ln 2} = 2\)
- (d) \(\ln e^{2x + 3} = 2x + 3\)
- (a) \(3x = \ln 20\), so \(x = \dfrac{\ln 20}{3} = 0.999\) (3 s.f.).
- (b) \(2x - 1 = e^3\), so \(x = \dfrac{e^3 + 1}{2} = 10.5\) (3 s.f.).
- \(x = 0\): \(y = 2 - 1 = 1\), so \((0, 1)\).
- \(y = 0\): \(e^{-x} = \dfrac{1}{2}\), so \(x = \ln 2\): \((\ln 2, 0)\).
- As \(x \to \infty\), \(y \to -1\): asymptote \(y = -1\). Decreasing curve, approaching \(y = -1\) from above.
- Let \(u = e^x\): \(3u^2 + u - 2 = 0\).
- \((3u - 2)(u + 1) = 0\), so \(u = \dfrac{2}{3}\) or \(u = -1\).
- \(e^x = -1\) is impossible, so \(e^x = \dfrac{2}{3}\).
- \(x = \ln \dfrac{2}{3}\) \((\approx -0.405)\).
- \(\ln \dfrac{(x + 1)^2}{x} = \ln 6\), so \((x + 1)^2 = 6x\).
- \(x^2 + 2x + 1 = 6x\), so \(x^2 - 4x + 1 = 0\).
- \(x = \dfrac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}\).
- Both are positive (\(2 - \sqrt{3} \approx 0.27\)), so both logs are defined: \(x = 2 + \sqrt{3}\) or \(x = 2 - \sqrt{3}\).
- (a) \(t = 0\): \(T = 20 + 70 = 90\,\degree\text{C}\).
- (b) \(T = 20 + 70e^{-0.5} = 62.5\,\degree\text{C}\) (3 s.f.).
- (c) \(70e^{-0.05t} = 20\), so \(e^{-0.05t} = \dfrac{2}{7}\) and \(-0.05t = \ln \dfrac{2}{7}\).
- \(t = 20\ln \dfrac{7}{2} = 25.1\) minutes (3 s.f.).
- (d) As \(t \to \infty\), \(e^{-0.05t} \to 0\), so \(T \to 20\,\degree\text{C}\) (room temperature).
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 \(e^{3\ln 2}\).
- A\(8\)
- B\(6\)
- C\(9\)
- D\(e^{6}\)
Solve \(e^{3x + 1} = 5\), giving your answer correct to 3 significant figures.
- A\(0.203\)
- B\(0.870\)
- C\(-0.464\)
- D\(0.462\)
Solve the equation \(\ln(2x - 3) = 1\), giving the exact value of \(x\).
- A\(x = \dfrac{e + 3}{2}\)
- B\(x = 2\)
- C\(x = \dfrac{13}{2}\)
- D\(x = \dfrac{3}{2} + e\)
For the curve \(y = 5 - 2e^{-x}\), state the \(y\)-intercept and the equation of the horizontal asymptote.
- A\((0, 3)\); \(y = 5\)
- B\((0, 5)\); \(y = 3\)
- C\((0, 3)\); \(y = 0\)
- D\((0, 7)\); \(y = 5\)
Solve the equation \(e^{2x} - 4e^{x} - 5 = 0\), giving exact answers.
- A\(x = \ln 5\)
- B\(x = 5\) or \(x = -1\)
- C\(x = 0\)
- D\(x = \dfrac{1}{2}\ln 5\)
The function f is defined by \(f(x) = \ln(2x + 1)\) for \(x > -\dfrac{1}{2}\). Find an expression for \(f^{-1}(x)\).
- A\(f^{-1}(x) = \dfrac{e^{x} - 1}{2}\)
- B\(f^{-1}(x) = \dfrac{e^{x} + 1}{2}\)
- C\(f^{-1}(x) = \dfrac{1}{2}e^{x} - 1\)
- D\(f^{-1}(x) = \dfrac{1}{\ln(2x + 1)}\)
The number of bacteria in a culture after \(t\) hours is \(N = 200e^{0.04t}\). Find the time taken for the number to reach 500, giving your answer correct to 3 significant figures.
- A\(22.9\) hours
- B\(155\) hours
- C\(10.1\) hours
- D\(62.5\) hours
Solve the equation \(3e^{x} = 2e^{-x} + 5\), giving your answer in exact form.
- A\(x = \ln 2\)
- B\(x = 2\)
- C\(x = \ln 2\) or \(x = -\ln 3\)
- D\(x = \ln 5\)
Quiz key
\(e^{3\ln 2} = e^{\ln 8} = 8\), because \(e^x\) and \(\ln x\) undo each other. The tempting 6 multiplies \(3 \times 2\) instead of finding \(2^3\). 9 is \(3^2\), the power and the number swapped, and \(e^6\) ignores the ln altogether.
\(3x + 1 = \ln 5\), so \(x = \dfrac{\ln 5 - 1}{3} = \dfrac{1.6094 - 1}{3} = 0.203\). The tempting 0.870 adds the 1 instead of subtracting it. \(-0.464\) divides only \(\ln 5\) by 3 before subtracting 1, and 0.462 subtracts the 1 before taking logs, \(\dfrac{\ln 4}{3}\).
\(2x - 3 = e\), so \(x = \dfrac{e + 3}{2}\) (about 2.86). The tempting \(x = 2\) solves \(2x - 3 = 1\), ignoring the ln. \(\dfrac{13}{2}\) treats ln as \(\log_{10}\) (\(2x - 3 = 10\)), and \(\dfrac{3}{2} + e\) forgets to halve the \(e\).
At \(x = 0\): \(y = 5 - 2(1) = 3\). As \(x \to \infty\), \(e^{-x} \to 0\), so \(y \to 5\): the asymptote is \(y = 5\). The tempting \((0, 5)\), \(y = 3\) swaps the two. \(y = 0\) is the asymptote of \(y = e^{-x}\) before it is transformed, and \((0, 7)\) adds the 2 instead of subtracting it.
\(u^2 - 4u - 5 = 0\), \((u - 5)(u + 1) = 0\), so \(e^x = 5\) or \(e^x = -1\). \(e^x = -1\) is impossible, so \(x = \ln 5\) only. The tempting "\(x = 5\) or \(x = -1\)" stops at the values of \(u\). \(x = 0\) comes from factorising as \((u + 5)(u - 1)\), and \(\dfrac{1}{2}\ln 5\) sets \(e^{2x} = 5\) instead of \(e^x = 5\).
\(y = \ln(2x + 1)\) gives \(e^y = 2x + 1\), so \(x = \dfrac{e^y - 1}{2}\) and \(f^{-1}(x) = \dfrac{e^x - 1}{2}\), defined for all real \(x\) (the range of f). Check: \(f^{-1}(f(1)) = \dfrac{3 - 1}{2} = 1\). The tempting \(\dfrac{e^x + 1}{2}\) is a sign slip when subtracting 1. \(\dfrac{1}{2}e^x - 1\) divides before subtracting, and \(\dfrac{1}{\ln(2x + 1)}\) is the reciprocal, not the inverse.
\(e^{0.04t} = \dfrac{500}{200} = 2.5\), so \(0.04t = \ln 2.5\) and \(t = \dfrac{\ln 2.5}{0.04} = 22.9\) hours. The tempting 155 takes ln of 500 without dividing by 200. 10.1 uses \(\dfrac{300}{200}\) (the increase, not the total), and 62.5 forgets to take ln: \(\dfrac{2.5}{0.04}\).
\(3e^{2x} = 2 + 5e^x\), so \(3e^{2x} - 5e^x - 2 = 0\), \((3e^x + 1)(e^x - 2) = 0\). \(e^x = -\dfrac{1}{3}\) is impossible, so \(e^x = 2\) and \(x = \ln 2\). Check: \(3(2) = 2\left(\dfrac{1}{2}\right) + 5\). The tempting \(x = 2\) stops at \(e^x = 2\). "\(\ln 2\) or \(-\ln 3\)" wrongly accepts \(e^x = \dfrac{1}{3}\) after a sign slip in the factor, and \(\ln 5\) treats \(e^{-x}\) as \(e^x\), giving \(e^x = 5\).
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.