The shape of the lesson
By the end of the lesson, learners can
- Take logs of both sides: Taking logarithms turns an index equation into an ordinary one, because \(\ln a^{k} = k\ln a\).
- Inequalities: watch the sign: Solve an inequality exactly like the equation, but every time you divide by a logarithm, ask whether it is negative. \(\ln a > 0\) when \(a > 1\), and \(\ln a < 0\) when \(0 < a < 1\).
- Quadratics in \(a^x\) and moduli: \(4^x = (2^2)^x = (2^x)^2\) and \(2^{x + 1} = 2 \times 2^x\).
- Isolate the power \(a^{f(x)}\) on one side.
- Take ln of both sides: \(f(x)\ln a = \ln b\).
- Both sides: expand, collect \(x\) terms, factorise out \(x\).
- Inequalities: reverse the sign when dividing by \(\ln a\) with \(0 < a < 1\).
- Hidden quadratic: \(u = a^x\), reject \(u \le 0\).
- Integers: for 'smallest \(n\)', round in the direction the inequality demands.
One picture every idea returns to
Take ln (or lg) of both sides: \(a^{f(x)} = b \Rightarrow f(x)\ln a = \ln b\). For an inequality, dividing by \(\ln a\) keeps the sign when \(a > 1\) but REVERSES it when \(0 < a < 1\), because then \(\ln a < 0\).
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, Bring the power down.. 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: Take logs of both sides | Taking logarithms turns an index equation into an ordinary one, because \(\ln a^{k} = k\ln a\). Model the worked example Simple index equation on the board, one line at a time. Your turn: Solve \(6^{x - 2} = 4^x\), giving your answer to 3 significant figures. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Inequalities: watch the sign | Solve an inequality exactly like the equation, but every time you divide by a logarithm, ask whether it is negative. \(\ln a > 0\) when \(a > 1\), and \(\ln a < 0\) when \(0 < a < 1\). Model the worked example Positive base on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: Quadratics in \(a^x\) and moduli | \(4^x = (2^2)^x = (2^x)^2\) and \(2^{x + 1} = 2 \times 2^x\). Model the worked example Hidden quadratic on the board, one line at a time. Your turn: Solve \(4^x - 3(2^{x + 1}) + 8 = 0\). | Mix-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, 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 Quadratics in \(a^x\) and moduli, 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.
Mix-Pair-Share
Students mix around the room until you call pair; they pair with the nearest classmate, discuss the prompt, then mix again. Three prompts of rising difficulty, three different partners.
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 |
|---|---|
| \(\dfrac{\ln 5}{\ln 2}\) is NOT \(\ln \dfrac{5}{2}\) and not \(\ln 5 - \ln 2\). | A quotient of logs does not simplify: work it out as one number divided by another. |
| In \(n\ln 0.8 < \ln 0.1\), both logs are negative. | Forgetting to reverse gives \(n < 10.3\), which says the SMALL values work, the opposite of what happens as \(0.8^n\) shrinks. |
| \(2^{2x + 1}\) is \(2 \times 2^{2x} = 2u^2\), not \(u^2 + 1\) and not \(2u\). | Split the index with \(a^{m + n} = a^m \times a^n\) before you substitute. |
Solving exponential equations and inequalities: worksheet
6 questions, 22 marks. Written in the style of the papers; no past paper question is reproduced.
Solve, giving answers to 3 significant figures: (a) \(3^x = 20\) (b) \(7^x = 0.5\).
Solve \(2^{3x + 1} = 5^x\), giving your answer to 3 significant figures.
Solve the inequality \(4 \times 3^{2x} > 50\).
Solve the equation \(2^{2x + 1} - 7(2^x) + 3 = 0\), giving any non-exact answer to 3 significant figures.
Solve the inequality \(|2^x - 6| < 2\).
Ana invests $2000 at 4% compound interest per year, so after \(n\) years her account holds \(2000 \times 1.04^n\) dollars. Ben invests $2500 at 3% per year, so his holds \(2500 \times 1.03^n\) dollars.
(a) Find the smallest whole number of years after which Ana's account exceeds $3500.
(b) Find the smallest whole number of years after which Ana's account holds more than Ben's.
Answer key and marking notes
- (a) \(x = \dfrac{\ln 20}{\ln 3} = 2.73\)
- (b) \(x = \dfrac{\ln 0.5}{\ln 7} = -0.356\)
- \((3x + 1)\ln 2 = x\ln 5\), so \(3x\ln 2 + \ln 2 = x\ln 5\).
- \(\ln 2 = x(\ln 5 - 3\ln 2)\), so \(x = \dfrac{\ln 2}{\ln 5 - \ln 8} = \dfrac{\ln 2}{\ln 0.625}\).
- \(x = -1.47\) (3 s.f.).
- \(3^{2x} > 12.5\), so \(2x\ln 3 > \ln 12.5\).
- \(\ln 3 > 0\), so the sign stays: \(x > \dfrac{\ln 12.5}{2\ln 3}\).
- \(x > 1.15\) (3 s.f.).
- Let \(u = 2^x\): \(2u^2 - 7u + 3 = 0\), so \((2u - 1)(u - 3) = 0\).
- \(2^x = \dfrac{1}{2}\) gives \(x = -1\).
- \(2^x = 3\) gives \(x = \dfrac{\ln 3}{\ln 2} = 1.58\).
- \(-2 < 2^x - 6 < 2\), so \(4 < 2^x < 8\).
- \(2^x\) is increasing, so \(2^2 < 2^x < 2^3\) gives \(2 < x < 3\).
- (a) \(1.04^n > 1.75\), so \(n > \dfrac{\ln 1.75}{\ln 1.04} = 14.27\dots\) Smallest \(n = 15\). (Check: \(2000 \times 1.04^{14} = 3463\), \(2000 \times 1.04^{15} = 3602\).)
- (b) \(2000 \times 1.04^n > 2500 \times 1.03^n\), so \(\left(\dfrac{1.04}{1.03}\right)^n > 1.25\).
- \(n\ln \dfrac{1.04}{1.03} > \ln 1.25\), and \(\ln \dfrac{1.04}{1.03} > 0\), so \(n > \dfrac{\ln 1.25}{\ln (1.04/1.03)} = 23.09\dots\)
- Smallest \(n = 24\). (Check: after 23 years 4929 < 4934; after 24 years 5127 > 5082.)
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.
Solve \(5^x = 12\), giving your answer correct to 3 significant figures.
- A\(1.54\)
- B\(0.875\)
- C\(2.4\)
- D\(4.00\)
Solve \(2^{x - 1} = 7\), giving your answer correct to 3 significant figures.
- A\(3.81\)
- B\(1.81\)
- C\(2.81\)
- D\(2.95\)
Find the smallest integer \(n\) for which \(0.7^n < 0.01\).
- A\(n = 13\)
- B\(n = 12\)
- C\(n = 12.9\)
- DEvery integer \(n \le 12\)
Solve the equation \(4^x = 8^{x - 1}\), giving the exact value of \(x\).
- A\(x = 3\)
- B\(x = 1\)
- C\(x = 2\)
- D\(x = \dfrac{1}{\ln 2}\)
Solve \(5^{2x - 1} = 3^{x + 2}\), giving your answer correct to 3 significant figures.
- A\(1.80\)
- B\(0.277\)
- C\(0.882\)
- D\(1.28\)
Solve the equation \(3^{2x + 1} - 10(3^x) + 3 = 0\).
- A\(x = 1\) or \(x = -1\)
- B\(x = 1\) only
- C\(x = 3\) or \(x = \dfrac{1}{3}\)
- D\(x = 2.07\) or \(x = -1.07\)
Solve the inequality \(2 \times 0.6^{x} < 0.5\), giving the boundary value correct to 3 significant figures.
- A\(x > 2.71\)
- B\(x < 2.71\)
- C\(x > 1.36\)
- D\(x < -3.80\)
Solve the inequality \(|2^x - 6| < 2\).
- A\(2 < x < 3\)
- B\(x < 3\)
- C\(x < 2\) or \(x > 3\)
- D\(4 < x < 8\)
Quiz key
\(x = \dfrac{\ln 12}{\ln 5} = \dfrac{2.4849}{1.6094} = 1.54\). Check: \(5^{1.54} \approx 11.9\). The tempting 0.875 is \(\ln \dfrac{12}{5}\): a quotient of logs is NOT the log of a quotient. 2.4 is \(12 \div 5\) with no logs, and 4.00 multiplies \(\ln 12 \times \ln 5\).
\(x - 1 = \dfrac{\ln 7}{\ln 2} = 2.807\), so \(x = 3.81\). Check: \(2^{2.81} \approx 7\). The tempting 1.81 subtracts the 1 instead of adding it. 2.81 forgets the 1 altogether, and 2.95 is \(\ln 7 + 1\), forgetting to divide by \(\ln 2\).
\(\ln 0.7 < 0\), so dividing by it REVERSES the inequality: \(n > \dfrac{\ln 0.01}{\ln 0.7} = 12.9\), so the smallest integer is 13. Check: \(0.7^{12} = 0.0138\), \(0.7^{13} = 0.0097\). The tempting 12 rounds 12.9 down, but \(0.7^{12}\) is still above 0.01. 12.9 is not an integer, and "every \(n \le 12\)" forgets to reverse the inequality.
\(2^{2x} = 2^{3(x - 1)} = 2^{3x - 3}\), so \(2x = 3x - 3\) and \(x = 3\). Check: \(4^3 = 64 = 8^2\). The tempting \(x = 1\) expands \(3(x - 1)\) as \(3x - 1\). \(x = 2\) treats \(4^x\) and \(8^{x - 1}\) as \(4x\) and \(8(x - 1)\), and \(\dfrac{1}{\ln 2}\) comes from taking logs without the bracket: \(x\ln 4 = x\ln 8 - 1\).
\((2x - 1)\ln 5 = (x + 2)\ln 3\), so \(2x\ln 5 - x\ln 3 = 2\ln 3 + \ln 5\) and \(x = \dfrac{2\ln 3 + \ln 5}{2\ln 5 - \ln 3} = \dfrac{3.8067}{2.1203} = 1.80\). The tempting 0.277 moves \(\ln 5\) across without changing its sign (\(2\ln 3 - \ln 5\) on top). 0.882 makes the same slip with \(\ln 3\) in the denominator, and 1.28 forgets to multiply \(\ln 3\) by 2 when expanding \((x + 2)\ln 3\).
\(3y^2 - 10y + 3 = 0\), \((3y - 1)(y - 3) = 0\), so \(3^x = \dfrac{1}{3}\) or \(3^x = 3\), giving \(x = -1\) or \(x = 1\). Both are valid because both values of \(y\) are positive. The tempting "\(x = 1\) only" discards \(3^x = \dfrac{1}{3}\), which is perfectly possible. "3 or \(\dfrac{1}{3}\)" stops at the values of \(y\), and \(2.07\), \(-1.07\) treat \(3^{2x + 1}\) as \(y^2\), solving \(y^2 - 10y + 3 = 0\).
\(0.6^x < 0.25\), so \(x\ln 0.6 < \ln 0.25\). Dividing by \(\ln 0.6 < 0\) reverses the sign: \(x > \dfrac{\ln 0.25}{\ln 0.6} = 2.71\). Check: \(2 \times 0.6^3 = 0.432 < 0.5\). The tempting \(x < 2.71\) forgets to reverse the inequality. \(x > 1.36\) forgets to divide by 2 (\(0.6^x < 0.5\)), and \(x < -3.80\) multiplies \(2 \times 0.6 = 1.2\) first, as if \(2 \times 0.6^x = 1.2^x\).
\(-2 < 2^x - 6 < 2\) gives \(4 < 2^x < 8\), i.e. \(2^2 < 2^x < 2^3\). Since \(2^x\) is increasing, \(2 < x < 3\). The tempting \(x < 3\) uses only \(2^x - 6 < 2\) and loses the lower bound. "\(x < 2\) or \(x > 3\)" solves \(|2^x - 6| > 2\), and \(4 < x < 8\) stops at the bounds for \(2^x\).
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.