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S5 Mathematics Practice Paper: Indices and Logarithms (12 Questions, 100 Marks, 75 Minutes, with Answers)
12 questions, 100 marks and 75 minutes on the laws of indices, exponential and logarithmic equations, with growth, decay and log-graph applications and full marking points.
Instructions: This paper has three sections, A, B and C. Answer all questions. Total marks: 100. Time allowed: 75 minutes.
State the law of indices or logarithms that you use. After solving a logarithmic equation, check that the number is positive and reject any invalid root.
Give numerical answers to three significant figures unless stated otherwise.
Spend about 35 minutes on Section A, 25 minutes on Section B and the last 15 minutes on Section C.
Section A: Laws of Indices and Exponential Equations
Simplifying, evaluating and exponential equations
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | Simplify (a) (2x³)² × 3x⁻⁴ ÷ 6x² and (b) evaluate 27^(2/3) + 16^(−1/2), stating the laws of indices you use. | (a) (2x³)² = 4x⁶ (2 marks); times 3x⁻⁴ gives 12x² (2 marks); divided by 6x² gives 2 (1 mark). (b) 27^(2/3) = 9 (2 marks) and 16^(−1/2) = 1/4, so the total is 37/4 = 9.25 (1 mark). |
| 2 | Solve the exponential equation 3^(2x−1) = 27^(x−2). | (a) Write 27 as 3³, so the right side becomes 3^(3x−6) (4 marks). (b) Equating indices gives 2x − 1 = 3x − 6, hence x = 5 (4 marks). |
| 3 | Given log₂ 3 = 1.585, find (a) log₂ 48 and (b) log₂ 0.75, correct to 3 decimal places. | (a) 48 = 16 × 3, so log₂ 48 = 4 + 1.585 = 5.585 (4 marks). (b) 0.75 = 3 ÷ 4, so log₂ 0.75 = 1.585 − 2 = −0.415 (4 marks). |
| 4 | Solve the logarithmic equation log₂ (x + 3) + log₂ (x − 3) = 4. | (a) Combining gives log₂ (x² − 9) = 4 (3 marks). (b) Hence x² − 9 = 16 and x² = 25, so x = 5 or x = −5 (3 marks). (c) The number must be positive, so x = 5 and −5 is rejected (3 marks). |
Section B: Laws of Logarithms and Logarithmic Equations
Evaluating, simplifying and solving logarithmic equations
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | A bacterial population grows as N = 500 × 2^(t/3), where t is in hours. (a) Find N when t = 9. (b) Find the time for the population to reach 16000. | (a) N = 500 × 2³ = 4000 (4 marks). (b) 2^(t/3) = 16000 ÷ 500 = 32 = 2⁵ (2 marks), so t/3 = 5 and t = 15 hours (2 marks). |
| 6 | Without a calculator, evaluate (a) log₅ 125 + log₄ (1/16) and (b) log₇ 49³. | (a) log₅ 125 = 3 (2 marks) and log₄ (1/16) = −2 (1 mark), giving 1 (1 mark). (b) log₇ 49³ = 3 × log₇ 49 = 3 × 2 = 6 (4 marks). |
| 7 | Given log 2 = 0.3010 and log 3 = 0.4771, find (a) log 12 and (b) log 0.15. | (a) log 12 = 2 log 2 + log 3 = 0.6020 + 0.4771 = 1.0791 (4 marks). (b) 0.15 = 3 ÷ 20, so log 0.15 = 0.4771 − 1.3010 = −0.8239 (4 marks). |
| 8 | Simplify (a) logₐ 8 ÷ logₐ 2 and (b) logₐ a³, where a is positive and a is not 1. | (a) logₐ 8 = 3 logₐ 2, so the quotient is 3 (4 marks). (b) logₐ a³ = 3 logₐ a = 3 (4 marks). |
Section C: Graphs and Integrated Applications
Growth, decay and straight-line log graphs
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | Solve log (x − 1) + log (x + 2) = 1. | (a) Combining gives log[(x − 1)(x + 2)] = 1 (3 marks). (b) Hence (x − 1)(x + 2) = 10, which gives x² + x − 12 = 0 (3 marks). (c) The roots are x = 3 and x = −4; the number must be positive, so x = 3 (3 marks). |
| 10 | Solve 4^x − 5 × 2^x + 4 = 0. | (a) Let u = 2^x, so the equation becomes u² − 5u + 4 = 0 (3 marks). (b) (u − 1)(u − 4) = 0, so u = 1 or u = 4 (3 marks). (c) 2^x = 1 gives x = 0 and 2^x = 4 gives x = 2 (2 marks). |
| 11 | A radioactive sample of mass m = 80 × (1/2)^(t/5) grams decays over t years. (a) Find the mass when t = 10. (b) Find the time for the mass to fall to 10 grams. | (a) m = 80 × (1/2)² = 20 grams (4 marks). (b) (1/2)^(t/5) = 10 ÷ 80 = 1/8 = (1/2)³ (3 marks), so t/5 = 3 and t = 15 years (2 marks). |
| 12 | Given log y = 2 log x + log 3 with x positive, (a) express y in terms of x and (b) find y when x = 10. | (a) log y = log x² + log 3 = log(3x²) (4 marks), so y = 3x². (b) When x = 10, y = 3 × 10² = 300 (3 marks), and y increases as x increases (2 marks). |
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A TutorZone practice paper; original questions and passages, not publisher material.