Logarithmic and exponential functions — Paper 2 · A Level Mathematics 9709

3 Solve the equation 2 x [5] ln(x + 3) −ln = ln(2x −2).

5 marks

Mark scheme: 3 Use 2 ln(x + 3) = ln(x + 3)2 M1 Use law for addition or subtraction of logarithms M1 Obtain correct quadratic expression in x A1 Make reasonable solution attempt at a 3-term quadratic M1 State x = 9 and no other solutions (condone x = –1 not deleted) A1 [5] 1 1

This question in 9709/23 Oct/Nov 2011

giving your answer correct to 3 significant figures.2 Use logarithms to solve the equation 5x = 32x−1, [4]

4 marks

Mark scheme: 2 Use law for the logarithm of a product, a quotient or a power M1* Obtain x log 5 = (2x – 1) log 3 or equivalent A1 Solve for x M1(dep*) Obtain answer x = 1.87 A1 [4]

This question in 9709/21 Oct/Nov 2012

giving your answer correct to 3 significant figures.2 Use logarithms to solve the equation 5x = 32x−1, [4]

4 marks

Mark scheme: 2 Use law for the logarithm of a product, a quotient or a power M1* Obtain x log 5 = (2x – 1) log 3 or equivalent A1 Solve for x M1(dep*) Obtain answer x = 1.87 A1 [4]

This question in 9709/23 Oct/Nov 2012

1 Solve the equation 3e2x 27 0, giving your answers in the form k ln 3. [5] −82ex + = … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 1 M1 Either directly or using substitution e = x u Obtain 1 3 e , e 27 = = x x A1 1 3 e , e 27 = = x x may be implied if e = x u is stated Use correct process at least once for solving e = x c where 0 > c M1 Obtain ln3 − from a correct solution A1 Condone use of e = x x Obtain 3ln3 from a correct solution A1 5 Question Answer Marks Guidance 2 Either State or imply equation ln ln ln ln = + y A B x B1 Equate gradient of line to ln B M1 Obtain ln 1.6486... = B and hence 5.2 = B A1 Substitute appropriate values to find ln A M1 Obtain ln 1.2809... = A and hence 3.6 = A A1 Or State or imply equation ln ln ln ln = + y A B x B1 Use given coordinates to obtain a correct equation B1 Equations are 4.908 ln 2.2ln = + A B and 11.008 ln 5.9ln A B = + Use given coordinates to obtain a second correct equation and attempt to solve both equations simultaneously to obtain at least one of the unknowns ln A or ln B M1 Obtain ln 1.6486... = B and hence 5.2 = B A1 Obtain ln 1.2809... = A and hence 3.6 = A A1

This question in 9709/21 May/June 2018

7 (a) Find the quotient when 9x3 1 is divided by 3x 2 , and show that the remainder is 9. −6x2 −20x + + [3] … … … … … … … … … … … … 6 9x3 1 (b) Hence find dx, giving the answer in the form a ln b where a and b are −6x2 −20x + 3x 2 + Ô1 + integers. [5] … … … … … … … … … … … … … … … … … (c) Find the exact root of the equation 9e9y 0. [4] −6e6y −20e3y −8 = … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 7(a) Carry out division at least as far as 2 3 + x kx M1 Obtain quotient 2 3 4 4 − − x x A1 Confirm remainder is 9 AG A1 3 7(b) Integrate to obtain at least 3 1k x and 2 ln(3 2) + k x terms *M1 Obtain 3 2 2 4 3ln(3 2) − − + + x x x x (FT from quotient in part (a)) A1FT Apply limits correctly DM1 Apply appropriate logarithm properties correctly M1 Obtain 125 ln64 + A1 5 7(c) State or imply 3 2 2 9 6 20 8 (3 2)(3 4 4) − − − = + − − x x x x x x (FT from quotient in part (a)) B1FT Attempt to solve cubic eqn to find positive value of x (or of 3e y ) M1 Use logarithms to solve equation of form 3e = y k where 0 > k M1 Obtain 1 ln 2 3 or exact equivalent A1 4

This question in 9709/21 May/June 2020

8 23x+22 Given that 5, find the value of 23x and hence, using logarithms, find the value of x correct + 23x = to 4 significant figures.−7 [5] … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 Use 3 2 3 2 4 2 x x × Solve equation for 3 2 x M1 Obtain 3 2 43 x = A1 Apply logarithms and use power law for 3 2 x k = where 0 k > M1 Obtain 1.809 A1 AWRT 5

This question in 9709/22 Oct/Nov 2020

2 Use logarithms to solve the equation giving your answer correct to 3 significant figures. 14e−2x = 5x+1, [4] … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2 Apply logarithms correctly to both sides and apply power law at least once *M1 Obtain ln14 − 2 x = ( x + 1)ln5 A1 OE with x no longer part of a power. Attempt solution of linear equation DM1 Must have ln14 − ln5 = x ( 2 + ln5 ) . Obtain 0.285 A1 4

This question in 9709/23 Oct/Nov 2022

7 The curve with equation e2x 18x y3 y 11 has a stationary point at p, q . (a) Find the exact value of p. [4] … … … … … … … … … … … … … … … … … … … … … … … … (b) Show that q 3 2 18 ln 3 [2] = + −q. … … … … … … … … (c) Show by calculation that the value of q lies between 2.5 and 3.0. [2] … … … … … (d) Use an iterative formula, based on the equation in (b), to find the value of q correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3] … … … … … … … … … Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(a) 3 2 d y B1 Differentiate y to obtain 3 y d x Differentiate complete equation to produce at least one term involving M1 d y using implicit differentiation. d x 2 x 2 dy dy A1 Obtain 2e − 18 + 3 y + = 0 dx dx dy 1 A1 Substitute = 0 to obtain either p = 2 ln9 or p = ln3 dx 4 7(b) Substitute value of p in original equation and rearrange as far as y 3 = ... M1 Allow in terms of ln9 . or q3 = … Obtain given result q = 3 2 + 18ln3 − q or y = 3 2 + 18ln3 − y with A1 AG sufficient detail 2 7(c) Consider sign of q − 3 2 + 18ln3 − q or equivalent for 2.5 and 3.0 M1 Obtain −0.18... and 0.34... with sufficient detail and justify A1 OE conclusion 2 7(d) Use iteration process correctly at least once M1 Obtain final answer q = 2.673 A1 Answer required to exactly 4 s.f. Show sufficient iterations to 6 sf to justify answer or show sign change A1 in the interval [2.6725, 2.6735] 3

This question in 9709/23 Oct/Nov 2023

3 (a) Sketch on the same diagram the graphs of y = 3x - 8 and y = 5 - x . [2] (b) Solve the inequality 3x - 8 1 5 - x . [4] … … … … … … … … … … … … … … … (c) Hence determine the largest integer N satisfying the inequality 3 e 0 .1 N - 8 1 5 - e 0 .1 N . [2] … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 3(a) Draw V-shaped graph with vertex on positive x-axis in the first quadrant. B1 Draw correct graph of 5 y x   correctly positioned with respect to modulus graph. B1 Two points of intersection. 2 3(b) Solve 3 8 5 x x    to obtain 13 4 B1 Or inequality. Solve linear equation or inequality with signs of 3x and x the same M1 Obtain 3 2 A1 Conclude 3 13 2 4 x   or 3 2 x  and 13 4 x  A1 Allow alternative notation e.g.   3 13 2 4 , . Alternative Method for Question 3(b) State or imply non-modulus equation (or inequality) 2 2 (3 8) (5 ) x x    (B1) Attempt solution of three-term equation (or inequality) (M1) Obtain 3 2 and 13 4 (A1) Conclude 3 13 2 4 x   or 3 2 x  and 13 4 x  (A1) Allow alternative notation e.g.   3 13 2 4 , . 4 3(c) Attempt value of N (maybe non-integer at this stage) for 0.1 13 4 e N their  M1 Allow 0.1 13 4 e N their  (or inequality). Conclude with single integer 11 A1 2

This question in 9709/21 May/June 2024

2 (a) Use logarithms to solve the inequality 4 x 1 0.05 . Give your answer in the form x 1 a , where the value of a is correct to 3 significant figures. [2] … … … … … … … … (b) Solve the inequality 3x + 8 1 9 . [3] … … … … … … … … … … … … (c) Hence state the integers that satisfy both of the inequalities in parts (a) and (b). [1] … … … … …

6 marks

Mark scheme: 2(a) Apply logarithms to both sides and use relevant logarithm property M1 Allow for x  log 4 0.05. ln0.05 A1 Allow greater accuracy. Obtain x  or equivalent and hence x −2.16 ln 4 2 2(b) Attempt solution of equations 3x + 8 = 9 or of quadratic equation (3 x + 8) 2 = 9 2 M1 Need a complete method to obtain 2 values. Obtain values − 173 and 13 A1 Conclude − 173  x  13 A1 OE 3 2(c) State −5, − 4, − 3 only B1 1

This question in 9709/21 May/June 2025

1 Solve the equation ln ( 3x + 5) - ln ( x - 2) = 4 . Give your answer in an exact form. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: Question Answer Marks Guidance 1 Apply relevant logarithm property *M1 3 x + 5 4 A1 Allow recovery from an incorrect log statement. Obtain correct equation without logarithms = e or equivalent x − 2 Solve equation to find exact value of x DM1 Must be in correct form, condone one sign slip. 2e 4 + 5 A1 Or exact equivalent. Obtain e 4 − 3 4

This question in 9709/22 Oct/Nov 2025