S5 Maths Indices and Logarithms Paper 2: Intermediate
Ten original logarithm items, each with an answer and a one-line explanation.
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S5 Maths: Indices and Logarithms |
| Level | S5 |
| # | Answer | One-line note |
|---|---|---|
| 1 | 6 | Adding logarithms of the same base multiplies the numbers: 16 x 4 = 64 and log base 2 of 64 is 6. |
| 2 | 3 | Subtracting logarithms of the same base divides the numbers: 54 / 2 = 27 and log base 3 of 27 is 3. |
| 3 | 4 | 5 to the 4th is 625, so the logarithm is 4; dividing 625 by 5 once only reaches 125. |
| 4 | x = 2 | 125 is 5 cubed, so 2x – 1 = 3 and x = 2. |
| 5 | 0.9030 | 8 is 2 cubed, so the logarithm is 3 x 0.3010 = 0.9030; multiplying 0.3010 by 8 is wrong. |
| 6 | x = 13 | Rewrite in index form: x + 3 = 16, so x = 13, and the argument stays positive. |
| 7 | 3/2, that is 1.5 | 4 to the power 3/2 is the cube of its square root, which is 8; change of base gives 3 divided by 2. |
| 8 | 1 | The numbers multiply to 2, so the sum is log base 2 of 2, which is 1. |
| 9 | 3 log base a of x | A power moves to the front as a coefficient: log base a of x cubed is 3 log base a of x. |
| 10 | 3 | A logarithm does not distribute over addition: the sum is 8 and its logarithm is 3, while 2 + 2 = 4 misuses the product law. |
Book a Maths tutorTutor 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.

