DSE Maths Practice Paper: Indices, Logarithms And Sequences

DSE Maths Practice Paper: Indices, Logarithms and Sequences

2026-11-03

DSE Maths: Indices, Logs and Sequences

DSE Maths practice paper on indices, logarithms and sequences: 12 questions, 100 marks, 75 minutes, with full answers and marking points.

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Paper details　Level: S6 | Subject: Mathematics | Time: 75 minutes | Marks: 100 | Questions: 12

Section A: Indices and logarithms (Questions 1-4, 24 marks)

Answer all questions. In calculations, write the formula, substitute the values and state the units, giving 4 decimal places where needed.

#	Question	Answer and marking notes
1	Simplify each of the following. (a) 2³ × 2⁻⁵, giving the answer as a fraction. (b) (3x²)³ ÷ 9x⁴. (c) Find the value of log₂ 32. (6 marks)	(a) 2³ × 2⁻⁵ = 2^(3 − 5) = 2⁻² = 1/4 [1 mark for adding indices; 1 mark for the answer] (b) (3x²)³ = 27x⁶; dividing by 9x⁴ gives 3x² [2 marks] (c) Since 32 = 2⁵, log₂ 32 = 5 [1 mark for the conversion; 1 mark for the answer]
2	Solve the following equations. (a) log₃(x + 2) = 2. (b) 2^(x + 1) = 16. (6 marks)	(a) log₃(x + 2) = 2 means x + 2 = 3² = 9, so x = 7 [1 mark for index form; 1 mark for solving; 1 mark for the answer] (b) Since 16 = 2⁴, x + 1 = 4 and x = 3 [2 marks for equating indices; 1 mark for the answer]
3	Given log 2 = 0.3010 and log 3 = 0.4771 (4 decimal places). (a) Find log 12. (b) Find log 1.5. (c) Find log 8. (6 marks)	(a) log 12 = log(2² × 3) = 2 log 2 + log 3 = 0.6020 + 0.4771 = 1.0791 [1 mark] (b) log 1.5 = log(3 ÷ 2) = 0.4771 − 0.3010 = 0.1761 [1 mark] (c) log 8 = log 2³ = 3 log 2 = 0.9030 [1 mark, 4 decimal places required]
4	(a) Solve 3^(2x) = 27^(x − 1). (b) Find the value of log₅(1/25). (6 marks)	(a) Since 27 = 3³, 3^(2x) = 3^(3x − 3), so 2x = 3x − 3 and x = 3 [2 marks; 1 mark for the answer] (b) 1/25 = 5⁻², so log₅(1/25) = −2 [1 mark; 2 marks for the answer, which must be negative]

Section B: Arithmetic and geometric sequences (Questions 5-8, 32 marks)

In solving equations, show the key steps and check that each answer lies inside the domain.

#	Question	Answer and marking notes
5	An arithmetic sequence has first term 7 and common difference −3. (a) Find the general term T(n). (b) Find T(10). (c) Find the sum of the first 15 terms S(15). (d) Find the first negative term. (8 marks)	(a) T(n) = 7 + (n − 1)(−3) = 10 − 3n [2 marks] (b) T(10) = 10 − 30 = −20 [2 marks] (c) S(15) = 15/2 × (2 × 7 + 14 × (−3)) = 7.5 × (−28) = −210 [1 mark formula; 1 mark substitution; 1 mark answer] (d) 10 − 3n is negative when n is greater than 10/3, so the first negative term is T(4) = −2 [1 mark]
6	A geometric sequence has first term 3 and common ratio 2. (a) Write down T(n). (b) Find T(8). (c) Find S(10). (d) Find the smallest n such that T(n) is greater than 1000. (8 marks)	(a) T(n) = 3 × 2^(n − 1) [2 marks] (b) T(8) = 3 × 2⁷ = 384 [2 marks] (c) S(10) = 3(2¹⁰ − 1)/(2 − 1) = 3 × 1023 = 3069 [2 marks] (d) 3 × 2^(n − 1) greater than 1000 means 2^(n − 1) greater than 333.3, so n − 1 is at least 9 and n = 10, where T(10) = 1536 [2 marks]
7	A geometric sequence has first term 8 and common ratio 3/4. (a) Find the sum to infinity S(infinity). (b) Find T(4). (c) Find S(5). (d) State the condition on the common ratio for the sum to infinity to exist. (8 marks)	(a) S(infinity) = 8 ÷ (1 − 3/4) = 8 ÷ 1/4 = 32 [2 marks] (b) T(4) = 8 × (3/4)³ = 8 × 27/64 = 27/8 = 3.375 [2 marks] (c) S(5) = 8(1 − (3/4)⁵) ÷ (1/4) = 32 × 0.762695… = 781/32, about 24.41 [2 marks] (d) The absolute value of the common ratio must be less than 1, i.e. −1 less than r less than 1 [2 marks]
8	The population of a town is 200 000 and grows by 5% each year. (a) Write an expression for the population after n years. (b) Find the population after 5 years, correct to the nearest integer. (c) Find the smallest n for which the population exceeds 300 000. (8 marks)	(a) P(n) = 200 000 × 1.05^n [2 marks] (b) P(5) = 200 000 × 1.05⁵ = 255 256.3, about 255 256 people [2 marks; 1 mark for rounding] (c) 1.05^n greater than 1.5, so taking logarithms gives n greater than 8.31 and n = 9 [2 marks; 1 mark for the answer]

Section C: Equations, applications and explanations (Questions 9-12, 44 marks)

In application questions, name the sequence or function model used and explain the conclusion in full sentences.

#	Question	Answer and marking notes
9	(a) Solve 4^x − 5(2^x) + 4 = 0. (b) Solve log x + log(x − 3) = 1. (10 marks)	(a) Let y = 2^x; then y² − 5y + 4 = 0, giving y = 1 or 4, so x = 0 or 2 [2 marks substitution; 2 marks quadratic; 1 mark answers] (b) log[x(x − 3)] = 1 gives x² − 3x − 10 = 0, so x = 5 or −2. Since the logarithm requires x greater than 3, x = 5 and −2 is rejected [2 marks equation; 1 mark solving; 1 mark rejection; 1 mark answer]
10	A hall has 15 seats in the first row and each later row has 2 more seats than the row before, with 20 rows in total. (a) Find the number of seats in row 20. (b) Find the total number of seats. (c) If the hall may hold at most 700 people, are 20 rows enough? Find the maximum number of rows. (10 marks)	(a) Row 20 has 15 + 19 × 2 = 53 seats [3 marks] (b) S(20) = 20/2 × (15 + 53) = 10 × 68 = 680 seats [4 marks] (c) 20 rows give 680 seats, which is below 700, so they are enough; S(21) = 21/2 × (15 + 55) = 735 exceeds the limit, so at most 20 rows [1 mark judgement; 2 marks calculation]
11	The sequence is 5, 15, 45, … (a) Show that it is geometric and find T(6). (b) Find S(6). (c) Find the smallest n such that S(n) is greater than 5000. (12 marks)	(a) 15 ÷ 5 = 3 and 45 ÷ 15 = 3, so the common ratio is 3 throughout and the sequence is geometric; T(6) = 5 × 3⁵ = 1215 [2 marks proof; 2 marks answer] (b) S(6) = 5(3⁶ − 1)/(3 − 1) = 5 × 728 ÷ 2 = 1820 [3 marks] (c) 5(3^n − 1)/2 greater than 5000 means 3^n greater than 2001; since 3⁷ = 2187, n = 7 and S(7) = 5465 [2 marks inequality; 2 marks trial; 1 mark answer]
12	A radioactive sample has mass 80 mg and a half-life of 6 hours. (a) Write an expression for the mass m(t) after t hours. (b) Find the mass after 18 hours. (c) Find the least whole number of hours after which the mass falls below 5 mg. (12 marks)	(a) m(t) = 80 × (1/2)^(t/6) [3 marks] (b) m(18) = 80 × (1/2)³ = 80 ÷ 8 = 10 mg [4 marks] (c) 80 × (1/2)^(t/6) less than 5 means (1/2)^(t/6) less than 1/16, so t/6 is greater than 4 and t is greater than 24; the least whole number of hours is 25 [2 marks inequality; 2 marks solving; 1 mark answer]

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