S6 Maths: Logarithms and Exponentials: Paper 1 (Basics)
Ten items starting from switching between index and log form, then the basic log laws.
| 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 | Basics |
| # | Answer | One-line note |
|---|---|---|
| 1 | log₂ 32 = 5 | Since 2⁵ = 32, the logarithm is 5. |
| 2 | log₁₀ 10000 = 4 | 10⁴ = 10000, so the value is 4. |
| 3 | log₃ 81 = 4 | 3⁴ = 81, so the value is 4. |
| 4 | 3 + 4 = 7 | The two parts are 3 and 4, so the sum is 7; log₂ 128 gives the same answer. |
| 5 | 3 − 2 = 1 | The values are 3 and 2, so the difference is 1, which is log₅ 5. |
| 6 | log₂ 1 = 0 and log₇ 7 = 1 | The log of 1 is always 0, and the log of the base itself is 1. |
| 7 | x = 6 | Write 64 as 2⁶ and compare the indices to get x = 6. |
| 8 | x = −3 | Since 0.001 = 10⁻³, the index is −3. |
| 9 | 3 + (−1) = 2 | The first term is 3 and the second is −1, giving 2. |
| 10 | The log law applies to products only; the correct value is 3 | log₂ 8 = 3. A sum inside a log cannot be split, so log(a + b) is not log a + log b. |
Switch between index and log form first, then apply the log laws.Tutor Registration →
Questions and answers are written originally by the TutorZone team and are not taken from any school, publisher or examination authority paper. Every value is checked by computer.

