S6 Maths: Logarithms and Exponentials: Paper 4 (Error clinic)
Ten items on the usual slips: splitting a sum, dividing logs, mixing up added and multiplied indices, and dropping a coefficient.
| 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 | Error clinic |
| # | Answer | One-line note |
|---|---|---|
| 1 | Undefined | The number inside a log must be positive, so 0 has no logarithm. |
| 2 | 0 | Since 2⁰ = 1, the value is 0. |
| 3 | 6 and 9 | The first gives 2 log 1000 = 6; the second squares the answer 3 to get 9. |
| 4 | 1/2 | Since 16^(1/2) = 4 the value is 1/2, not 2. |
| 5 | 128; the indices add, not multiply | Multiplying powers of the same base adds the indices: 2⁷ = 128, while 2¹² = 4096. |
| 6 | 64; different from 512 | A power of a power multiplies the indices: 2⁶ = 64, whereas 2⁹ = 512. |
| 7 | 6 and 9 | The sum is log₂ 64 = 6; squaring the answer 3 gives 9. |
| 8 | x = 3/2 | 125 = 5³ so 2x = 3 and x = 3/2; answering 3 forgets the factor of 2. |
| 9 | 1; log_a a = 1 | For any a > 0 with a ≠ 1, a¹ = a, so the log of the base is 1. |
| 10 | 1/2 | Since 4^(1/2) = 2 the value is 1/2; writing 2 swaps the base and the number. |
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.

