S6 Maths: Logarithms and Exponentials: Paper 2 (Intermediate)
This paper focuses on combining and splitting logs, with change of base and two-step index equations.
| 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 | Intermediate |
| # | Answer | One-line note |
|---|---|---|
| 1 | log₂ 16 = 4 | Subtracting logs of the same base gives the log of the quotient; 48 ÷ 3 = 16, so the value is 4. |
| 2 | log₃ 27 = 3 | 54 ÷ 2 = 27 and 3³ = 27, so the value is 3. |
| 3 | log₄ 16 = 2 | Adding gives log₄ 16, and 4² = 16, so the answer is 2. |
| 4 | 3/2 | 8 = 4^(3/2) because the square root of 4 is 2 and 2³ = 8. |
| 5 | 3/2 | 27 = 9^(3/2), so the value is 3/2. |
| 6 | log₂ 8 = 3 | When the middle bases match the factors chain together, leaving log₂ 8 = 3. |
| 7 | x = 2 | Since 81 = 3⁴ we get 2x = 4, so x = 2. |
| 8 | x = 4 | 128 = 2⁷, so x + 3 = 7 and x = 4. |
| 9 | x = 0.01 and x = 64 | log₁₀ x = −2 means x = 10⁻²; log₂ x = 6 means x = 2⁶ = 64. |
| 10 | 3 | log 8 ÷ log 2 equals log₂ 8 = 3; dividing two logs is not the log of the divided numbers. |
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