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S6 Maths: Logarithms and Exponentials: Paper 3 (Challenge)
S6 Maths: Logarithms and Exponentials: Paper 3 (Challenge)
This paper applies the log laws to equations, compound interest and half-life, with domain care.
Direct answers Items 1-3: 4, then 0 or 2, then about 0.6990. Items 4-6: 3, then 3 and 0, then 15 years. Items 7-9: 15 years, ±100, about 1.710. Item 10: about 1.585.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S6 Maths: Logarithms and Exponentials |
| Level | Challenge |
02 Questions (10 items)
- FoundationSolve log₂ x + log₂ (x − 2) = 3 for x > 2.
- FoundationSolve 2^(2x) − 5(2ˣ) + 4 = 0.
- FoundationGiven log 2 ≈ 0.3010, find log 5.
- IntermediateEvaluate log₂ 3 × log₃ 4 × log₄ 8.
- IntermediateEvaluate ln e³ and ln 1.
- IntermediateA sum of $1000 earns 5% compound interest a year. What is the least number of years for it to reach $2000?
- ChallengeA substance halves every 5 years. How long does 80 g take to fall to 10 g?
- ChallengeSolve log₁₀ (x²) = 4.
- ChallengeSolve 3ˣ = 2^(x+1), correct to three decimal places.
- Error redoA student solves 2ˣ = 3 by writing x = 3/2. What is wrong, and what is the correct answer?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | x = 4 | Combine to log₂ x(x − 2) = 3, so x² − 2x − 8 = 0 and the valid root is 4. |
| 2 | x = 0 or x = 2 | Let y = 2ˣ to get y² − 5y + 4 = 0, so y = 1 or 4 and x = 0 or 2. |
| 3 | about 0.6990 | log 5 = 1 − log 2 = 1 − 0.3010 = 0.6990. |
| 4 | 3 | Chaining the factors cancels the middle bases and leaves log₂ 8 = 3. |
| 5 | 3 and 0 | Natural logs have base e, so ln e³ = 3 and ln 1 = 0. |
| 6 | 15 years | Solve 1.05ⁿ ≥ 2, giving n ≥ 14.21, so 15 whole years are needed. |
| 7 | 15 years | 80 → 40 → 20 → 10 is three halvings, so 3 × 5 = 15 years. |
| 8 | x = 100 or x = −100 | Then x² = 10000, so x = 100 or x = −100; the square lets both signs through. |
| 9 | x ≈ 1.710 | Taking logs gives x log 3 = (x + 1) log 2, so x = log 2 / (log 3 − log 2) ≈ 1.710. |
| 10 | The index cannot simply be removed; x ≈ 1.585 | Use logs: x = log 3 ÷ log 2 ≈ 1.585. Dividing 3 by the base 2 has no justification. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
Switch between index and log form first, then apply the log laws.Tutor Registration →
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