Logarithmic and exponential functions — Paper 3 · A Level Mathematics 9709

2 Two variable quantities x and y are related by the equation y = Axn, where A and n are constants. The diagram shows the result of plotting ln y against ln x for four pairs of values of x and y. Use the diagram to estimate the values of A and n. [5]

5 marks

Mark scheme: 2 State or imply that ln y = ln A + n ln x B1 Equate estimate of ln y -intercept to ln A M1 Obtain value A between 1.97 and 2.03 A1 Calculate the gradient of the line of data points M1 Obtain value n = 0.25, or equivalent A1 [5]

This question in 9709/31 Oct/Nov 2005

2 Solve, correct to 3 significant figures, the equation ex + e2x = e3x. [5]

5 marks

Mark scheme: 2 EITHER State or imply e x + 1 = e 2 x , or 1 + e − x = e x , or equivalent B1 Solve this equation as a quadratic in u = e x , or in e x , obtaining one or two M1 roots Obtain root 1 1 + 5 ) , or decimal in [1.61, 1.62] A1 2 ( Use correct method for finding x from a positive root M1 Obtain x = 0.481 and no other answer A1 [For the solution 0.481 with no working, award B3 (for 0.48 give B2). However a suitable statement can earn the first B1 in addition, giving a maximum of 4/5 (or 3/5) in such cases.] n 1 + e x ) or OR State an appropriate iterative formula, e.g. nx +1 = 12 ln ( n e x n + e 2 x ) B1 nx +1 = 13 ln ( Use the iterative formula correctly at least once M1 Obtain final answer 0.481 A1 Show sufficient iterations to justify its accuracy to 3 d.p., or show there is a sign change in the value of a relevant function in the interval (0.4805, 0.4815) A1 Show that the equation has no other root A1 [5]

This question in 9709/31 May/June 2008

3x 32, giving your answer correct to 3 significant figures. [4]2 Solve the equation 3x+2 = +

4 marks

Mark scheme: 2 EITHER: Use laws of indices correctly and solve a linear equation for 3x, or for 3–x M1 3 2 Obtain 3x, or 3–x in any correct form, e.g. 3x = A1 (3 2 − )1 Use correct method for solving 3±x = a for x, where a > 0 M1 Obtain answer x = 0.107 A1 ln(3 x n + 9 ) OR: State an appropriate iterative formula, e.g. xn+1 = − 2 B1 ln 3 Use the formula correctly at least once M1 Obtain answer x = 0.107 A1 Show that the equation has no other root but 0.107 A1 [4] [For the solution 0.107 with no relevant working, award B1 and a further B1 if 0.107 is shown to be the only root.]

This question in 9709/31 Oct/Nov 2009

3 The variables x and y satisfy the equation xny C, where n and C are constants. When x 1.10, y 5.20, and when x 3.20, y 1.05. = = = = = (i) Find the values of n and C. [5] (ii) Explain why the graph of ln y against ln x is a straight line. [1]

6 marks

Mark scheme: 3 (i) EITHER: State or imply n ln x + ln y = ln C B1 Substitute x- and y-values and solve for n M1 Obtain n = 1.50 A1 Solve for C M1 Obtain C = 6.00 A1 OR: Obtain two correct equations by substituting x- and y-values in x n y = C B1 Solve for n M1 Obtain n = 1.50 A1 Solve for C M1 Obtain C = 6.00 A1 [5] (ii) State that the graph of ln y against ln x has equation nln x + ln y = ln C which is linear in ln y and ln x, or has equation of the form nX + Y = ln C, where X = ln x and Y = ln y, and is thus a straight line B1 [1]

This question in 9709/31 May/June 2010

1 Solve the equation 2x 1 5, + 2x = −1 giving your answer correct to 3 significant figures. [4]

4 marks

Mark scheme: 1 EITHER: Attempt to solve for 2 x M1 Obtain 2x = 6/4, or equivalent A1 Use correct method for solving an equation of the form 2x = a, where a > 0 M1 Obtain answer x = 0.585 A1 n + 6) / 5) / ln 2 B1 OR: State an appropriate iterative formula, e.g. xn + 1 = ln((2 x Use the iterative formula correctly at least once M1 Obtain answer x = 0.585 A1 Show that the equation has no other root but 0.585 A1 [4] [For the solution 0.585 with no relevant working, award B1 and a further B1 if 0.585 is shown to be the only root.] 2 ∫

This question in 9709/32 May/June 2010

2 The variables x and y satisfy the equation y3 Ae2x, where A is a constant. The graph of ln y against = x is a straight line. (i) Find the gradient of this line. [2] (ii) Given that the line intersects the axis of ln y at the point where ln y 0.5, find the value of A = correct to 2 decimal places. [2]

4 marks

Mark scheme: 2 (i) State or imply 3 ln y = ln A + 2 x at any stage B1 2 State gradient is , or equivalent B1 [2] 3 (ii) Substitute x = 0, ln y = 0.5 and solve for A M1 Obtain A = 4.48 A1 [2]

This question in 9709/33 May/June 2010

2 Solve the equation 1 2 ln x, ln(1 + x2) = + giving your answer correct to 3 significant figures. [4]

4 marks

Mark scheme: 2 Use law for the logarithm of a power, a quotient, or a product correctly at least once M1 Use ln e = 1 or e = exp(1) M1 Obtain a correct equation free of logarithms, e.g. 1 + x2 = ex2 A1 Solve and obtain answer x = 0.763 only A1 [4] [For the solution x = 0.763 with no relevant working give B1, and a further B1 if 0.763 is shown to be the only root.] [Treat the use of logarithms to base 10 with answer 0.333 only, as a misread.] [SR: Allow iteration, giving B1 for an appropriate formula, e.g. xn+1 = exp((ln(1 + xn2) – 1)/2), M1 for using it correctly once, A1 for 0.763, and A1 for showing the equation has no other root but 0.763.]

This question in 9709/31 Oct/Nov 2010

3x 8 Let f(x) = (1 + x)(1 + 2x2). (i) Express in partial fractions. [5] f(x) (ii) Hence obtain the expansion of in ascending powers of x, up to and including the term in x3. f(x) [5] [Questions 9 and 10 are printed on the next page.]

10 marks

Mark scheme: A Bx + C 8 (i) State or imply the form + 2 B1 1 + x 1 + 2 x Use any relevant method to evaluate a constant M1 Obtain one of A = –1, B = 2, C = 1 A1 Obtain a second value A1 Obtain the third value A1 [5] (ii) Use correct method to obtain the first two terms of the expansion of (1 + x )−1 or (1 + 2 x 2 )−1 M1 Obtain correct expansion of each partial fraction as far as necessary A1√ + A1√ Multiply out fully by Bx + C, where BC Þ 0 M1 Obtain answer 3x – 3x2 – 3x3 A1 [5] − 1  [Symbolic binomial coefficients, e.g.,   are not sufficient for the first M1. The f.t.  1  is on A, B, C.] [If B or C omitted from the form of fractions, give B0M1A0A0A0 in (i); M1A1√A1√ in (ii), max 4/10.] [If a constant D is added to the correct form, give M1A1A1A1 and B1 if and only if D = 0 is stated.] [If an extra term D/(1 + 2x2) is added, give B1M1A1A1, and A1 if C + D = 1 is resolved to 1/(1 + 2x2).] [In the case of an attempt to expand 3x(1 + x)–1(1 + 2x2)–1, give M1A1A1 for the expansions up to the term in x2, M1 for multiplying out fully, and A1 for the final answer.] [For the identity 3x ≡ (1 + x + 2x2 + 2x3)(a + bx + cx2 + dx3) give M1A1; then M1A1 for using a relevant method to find two of a = 0, b = 3, c = –3 and d = –3; and then A1 for the final answer in series form.]

This question in 9709/31 Oct/Nov 2010

2 Solve the equation 1 2 ln x, ln(1 + x2) = + giving your answer correct to 3 significant figures. [4]

4 marks

Mark scheme: 2 Use law for the logarithm of a power, a quotient, or a product correctly at least once M1 Use ln e = 1 or e = exp(1) M1 Obtain a correct equation free of logarithms, e.g. 1 + x2 = ex2 A1 Solve and obtain answer x = 0.763 only A1 [4] [For the solution x = 0.763 with no relevant working give B1, and a further B1 if 0.763 is shown to be the only root.] [Treat the use of logarithms to base 10 with answer 0.333 only, as a misread.] [SR: Allow iteration, giving B1 for an appropriate formula, e.g. xn+1 = exp((ln(1 + xn2) – 1)/2), M1 for using it correctly once, A1 for 0.763, and A1 for showing the equation has no other root but 0.763.]

This question in 9709/32 Oct/Nov 2010

4 The polynomial is defined by f(x) 12x3 25x2 f(x) = + −4x −12. (i) Show that 0 and factorise completely. [4] f(−2) = f(x) (ii) Given that 12 27y 25 9y 3y 0, × + × −4 × −12 = state the value of 3y and hence find y correct to 3 significant figures. [3]

7 marks

Mark scheme: 4 (i) Verify that –96 + 100 + 8 – 12 = 0 B1 Attempt to find quadratic factor by division by (x + 2), reaching a partial quotient 12x2 + kx, inspection or use of an identity M1 Obtain 12x2 + x – 6 A1 State (x + 2)(4x + 3)(3x – 2) A1 [4] [The M1 can be earned if inspection has unknown factor Ax2 + Bx – 6 and an equation in A and/or B or equation 12x2 + Bx + C and an equation in B and/or C.] (ii) State 3y = 23 and no other value B1 Use correct method for finding y from equation of form 3y = k, where k > 0 M1 Obtain –0.369 and no other value A1 [3] 2 2

This question in 9709/31 May/June 2011

5 The curve with equation 6e2x key e2y c, + + = where k and c are constants, passes through the point P with coordinates 3, ln (ln 2). (i) Show that 58 2k c. [2] + = (ii) Given also that the gradient of the curve at P is find the values of k and c. [5] −6,

7 marks

Mark scheme: 5 (i) Use at least one of e2x = 9, e y = 2 and e2y = 4 B1 Obtain given result 58 + 2k = c AG B1 [2] dy dy y (ii) Differentiate left-hand side term by term, reaching ae2x + be + ce2y M1 dx dx dy dy y Obtain 12e2x + ke + 2e2y A1 dx dx Substitute (ln 3, ln 2) in an attempt involving implicit differentiation at least once, where RHS = 0 M1 Obtain 108 – 12k – 48 = 0 or equivalent A1 Obtain k = 5 and c = 68 A1 [5] 2 2

This question in 9709/31 May/June 2011

2 (i) Show that the equation log2(x + 5) = 5 −log2 x can be written as a quadratic equation in x. [3] (ii) Hence solve the equation log2(x + 5) = 5 −log2 x. [2]

5 marks

Mark scheme: 2 (i) Use law for the logarithm of a product or quotient M1 Use log232 = 5 or 25 = 32 M1 Obtain x2 + 5x – 32 = 0, or horizontal equivalent A1 [3] (ii) Solve a 3-term quadratic equation M1 153 − 5 Obtain answer x = 3.68 only, or exact equivalent, e.g. A1 [2] 2

This question in 9709/32 May/June 2011

giving your answer correct to 3 significant figures.1 Use logarithms to solve the equation 52x−1 = 2(3x), [4]

4 marks

Mark scheme: 1 Use law for the logarithm of a product, power or quotient M1* Obtain a correct linear equation, e.g. (2 x − 1) ln 5 = ln 2 + x ln 3 A1 Solve a linear equation for x M1(dep*) Obtain answer x = 1.09 A1 [4] x x  25  [SR: Reduce equation to the form a = b M1*, obtain   = 10 Al, use correct method to  3  calculate value of x M1(dep*), obtain answer 1.09 A1.]

This question in 9709/33 May/June 2011

1 Using the substitution u ex, or otherwise, solve the equation = 1 ex = + 6e−x, giving your answer correct to 3 significant figures. [4]

4 marks

Mark scheme: 1 Rearrange as e2x – ex – 6 = 0, or u2 – u – 6 = 0, or equivalent B1 Solve a 3-term quadratic for ex or for u M1 Obtain simplified solution ex = 3 or u = 3 A1 Obtain final answer x = 1.10 and no other A1 [4]

This question in 9709/31 Oct/Nov 2011

1 Using the substitution u ex, or otherwise, solve the equation = 1 ex = + 6e−x, giving your answer correct to 3 significant figures. [4]

4 marks

Mark scheme: 1 Rearrange as e2x – ex – 6 = 0, or u2 – u – 6 = 0, or equivalent B1 Solve a 3-term quadratic for ex or for u M1 Obtain simplified solution ex = 3 or u = 3 A1 Obtain final answer x = 1.10 and no other A1 [4]

This question in 9709/32 Oct/Nov 2011

1 Solve the equation 2 ln(3x + 4) = ln(x + 1), giving your answer correct to 3 significant figures. [4]

4 marks

Mark scheme: 1 EITHER: Use law of the logarithm of a power or quotient and remove logarithms M1 Obtain a 3-term quadratic equation x 2 −x − 3 = 0 , or equivalent A1 Solve 3-term quadratic obtaining 1 or 2 roots M1 Obtain answer 2.30 only A1  1  OR1: Use an appropriate iterative formula, e.g. x n +1 = exp ln (3 x n + 4 )  − 1 correctly at  2  least once M1 Obtain answer 2.30 A1 Show sufficient iterations to at least 3 d.p. to justify 2.30 to 2 d.p., or show there is a sign change in the interval (2.295, 2.305) A1 Show there is no other root A1 OR2: Use calculated values to obtain at least one interval containing the root M1 Obtain answer 2.30 A1 Show sufficient calculations to justify 2.30 to 3 s.f., e.g. show it lies in (2.295, 2.305) A1 Show there is no other root A1 [4] 1 2 1

This question in 9709/32 May/June 2012

2 Solve the equation 2 ln x ln 3, giving your answer correct to 3 significant figures. [4] ln(2x + 3) = +

4 marks

Mark scheme: 2 Use law of the logarithm of a power and a product or quotient and remove logarithms M1 2 x + 3 Obtain a correct equation in any form, e.g. = 3 A1 x 2 Solve 3-term quadratic obtaining at least one root M1 Obtain final answer 1.39 only A1 [4] dx dy

This question in 9709/33 May/June 2012

1 Solve the equation 1 ln x, ln(x + 5) = + giving your answer in terms of e. [3]

3 marks

Mark scheme: 1 State or imply 1n e = 1 B1 Apply at least one logarithm law for product or quotient correctly M1 (or exponential equivalent) 5 Obtain x + 5= ex or equivalent and hence A1 [3] e − 1

This question in 9709/33 Oct/Nov 2012

4 (i) Solve the equation 4x x . [3] −1 = −3 4y correct to 3 significant figures. [3] −3 (ii) Hence solve the equation 4y+1 −1 =

6 marks

Mark scheme: 4 (i) Either State or imply non-modular equation (4 x − 1) 2 = ( x − 3) 2 or pair of linear equations 4 x −=1 ± ( x − 3) B1 Solve a three-term quadratic equation or two linear equations M1 2 4 Obtain − and A1 3 5 2 Or Obtain value − from inspection or solving linear equation B1 3 4 Obtain value similarly B2 [3] 5 y 4 (ii) State or imply at least 4 = , following a positive answer from part (i) B1√ 5 Apply logarithms and use log a b = b log a property M1 Obtain –0.161 and no other answer A1 [3]

This question in 9709/31 May/June 2013

2 Solve the equation 2 3x 3x, giving your answers correct to 3 significant figures. [4] −1 =

4 marks

Mark scheme: x2 EITHER: State or imply non-modular equation 2 2 ( 3 x − 1) 2 = ( 3 ) 2 , or pair of equations x M1 2 ( 3 x − 1) = ±3 x 2 Obtain 3x = 2 and 3 = (or 3x+1 = 2) A1 3 OR: Obtain 3x = 2 by solving an equation or by inspection B1 x 2 Obtain 3 = (or 3x+1 = 2) by solving an equation or by inspection B1 3 Use correct method for solving an equation of the form 3x = a (or 3x+1 = a), where a > 0 M1 Obtain final answers 0.631 and –0.369 A1 [4] 1 1 1 ∫

This question in 9709/31 Oct/Nov 2013

2 Solve the equation 2 3x −1 = 3x, giving your answers correct to 3 significant figures. [4]

4 marks

Mark scheme: x2 EITHER: State or imply non-modular equation 2 2 ( 3 x − 1) 2 = ( 3 ) 2 , or pair of equations x M1 2 ( 3 x − 1) = ±3 x 2 Obtain 3x = 2 and 3 = (or 3x+1 = 2) A1 3 OR: Obtain 3x = 2 by solving an equation or by inspection B1 x 2 Obtain 3 = (or 3x+1 = 2) by solving an equation or by inspection B1 3 Use correct method for solving an equation of the form 3x = a (or 3x+1 = a), where a > 0 M1 Obtain final answers 0.631 and –0.369 A1 [4] 1 1 1 ∫

This question in 9709/32 Oct/Nov 2013

1 Use logarithms to solve the equation ex giving your answer correct to 3 decimal places. [3] = 3x−2,

3 marks

Mark scheme: 1 Use law of the logarithm of a power M1 Obtain a correct linear equation in any form, e.g. x = ( x − 2)ln 3 A1 Obtain answer x = 22.281 A1 [3]

This question in 9709/32 Oct/Nov 2014

1 Use logarithms to solve the equation 25x giving the answer correct to 3 significant figures. = 32x+1, [4]

4 marks

Mark scheme: 1 Use law for the logarithm of a power at least once *M1 Obtain correct linear equation, e.g. 5 xIn 2 = ( 2 x + )1 In3 A1 Solve a linear equation for x M1 dep *M Obtain x = 0.866 A1 [4]

This question in 9709/31 May/June 2015

2 Using the substitution giving your answer correct to 4x, solve the equation 4x 42 u = + = 4x+2, 3 significant figures. [4]

4 marks

Mark scheme: 2 Use laws of indices correctly and solve for u M1 16 Obtain u in any correct form, e.g. u = A1 16 − 1 Use correct method for solving an equation of the form 4 x = a , where a > 0 M1 Obtain answer x = 0.0466 A1 [4]

This question in 9709/32 May/June 2015

2 Using the substitution u = 3x, solve the equation 3x + 32x = 33x giving your answer correct to 3 significant figures. [5]

5 marks

Mark scheme: 2 State or imply 1 + u = u 2 B1 Solve for u M1 1 1( + 5 ) , or decimal in [1.61, 1.62] A1 Obtain root 2 Use correct method for finding x from a positive root M1 Obtain x = 0.438 and no other answer A1 [5]

This question in 9709/31 Oct/Nov 2015

2 Using the substitution u = 3x, solve the equation 3x + 32x = 33x giving your answer correct to 3 significant figures. [5]

5 marks

Mark scheme: 2 State or imply 1 + u = u 2 B1 Solve for u M1 1 1( + 5 ) , or decimal in [1.61, 1.62] A1 Obtain root 2 Use correct method for finding x from a positive root M1 Obtain x = 0.438 and no other answer A1 [5]

This question in 9709/32 Oct/Nov 2015

1 Sketch the graph of y eax where a is a positive constant. [2] = −1

2 marks

Mark scheme: 1 Draw curve with increasing gradient existing for negative and positive values of x M1 Draw correct curve passing through the origin A1 [2]

This question in 9709/33 Oct/Nov 2015

1 (i) Solve the equation 2 x 3 x . [3] −1 = (ii) Hence solve the equation 2 5x 3 5x , giving your answer correct to 3 significant figures. −1 = [2] 1

5 marks

Mark scheme: 1 (i) EITHER: State or imply non-modular equation (2( x − 1)) 2 = (3 x ) 2 , or pair of linear equations 2( x − 1) = ±3 x B1 Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations M1 Obtain answers x = −and2 x = 52 A1 OR: Obtain answer x = −by2 inspection or by solving a linear equation (B1 Obtain answer x = 52 similarly B2) [3] (ii) Use correct method for solving an equation of the form 5x = a or 5x +1 = a , where a > 0 M1 Obtain answer x =– 0.569 only A1 [2] 2 ∫ 2

This question in 9709/31 May/June 2016

2 The variables x and y satisfy the relation 3y = 42−x. (i) By taking logarithms, show that the graph of y against x is a straight line. State the exact value of the gradient of this line. [3] (ii) Calculate the exact x-coordinate of the point of intersection of this line with the line with equation y 2x, simplifying your answer. [2] =

5 marks

Mark scheme: 2 (i) State or imply y ln3 = (2 − x )ln 4 B1 State that this is of the form ay = bx + c and thus a straight line, or equivalent B1 ln4 State gradient is − , or exact equivalent B1 ln3 [3] (ii) Substitute y = 2x and solve for x, using a log law correctly at least once M1 Obtain answer x = ln 4 / ln6 , or exact equivalent A1 [2]

This question in 9709/33 May/June 2016

3x 21 Solve the equation + 8, giving your answer correct to 3 decimal places. [3] 3x = −2

3 marks

Mark scheme: 1 Solve for3x and obtain 3x = 187 B1 Use correct method for solving an equation of the form 3x = a , where a > 0 M1 Obtain answer x = 0.860 3 d.p. only A1 [3] 3

This question in 9709/31 Oct/Nov 2016

3x 21 Solve the equation + 8, giving your answer correct to 3 decimal places. [3] 3x = −2

3 marks

Mark scheme: 1 Solve for3x and obtain 3x = 187 B1 Use correct method for solving an equation of the form 3x = a , where a > 0 M1 Obtain answer x = 0.860 3 d.p. only A1 [3] 3

This question in 9709/32 Oct/Nov 2016

1 Solve the equation ln 1 2x 2, giving your answer correct to 3 decimal places. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: Question Answer Marks 1 Remove logarithm and obtain 1 + 2 x = e 2 B1 Use correct method to solve an equation of the form 2 x = a , where a > 0 M1 Obtain answer x = 2.676 A1 Total: 3

This question in 9709/32 Feb/March 2017

3 It is given that x ln 1 y, where 0 y 1. = −y −ln < < e−x (i) Show that y . [2] = 1 e−x + … … … … … … … … … … … … … … … … … … … … … … … … 1 @ A 2e (ii) Hence show that y dx ln . [4] Ó 0 = e 1 + … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: Question Answer Marks 3(i) 1 − y B1 Remove logarithms correctly and obtain ex = y e − x B1 Obtain the given answer y = − x following full working 1 + e Total: 2 3(ii) State integral k ln(1 + e − x ) where k = ± 1 *M1 State correct integral − ln(1 + e − x ) A1 Use limits correctly DM1  2e  A1 Obtain the given answer ln   following full working  e + 1  Total: 4

This question in 9709/31 May/June 2017

3 Using the substitution u ex, solve the equation 3ex 4. Give your answer correct to 3 significant figures. = 4e−x = + [4] … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 3 Rearrange as 3u 2 + 4u − 4 = 0 , or 3e 2 x + 4e x − 4 = 0 , or equivalent B1 Solve a 3-term quadratic for ex or for u M1 Obtain e x = 23 or u = 23 A1 Obtain answer x = –0.405 and no other A1 Total: 4

This question in 9709/33 May/June 2017

4 The variables x and y satisfy the equation yn Ax3, where n and A are constants. It is given that = y 2.58 when x 1.20, and y 9.49 when x 2.51. = = = = (i) Explain why the graph of ln y against ln x is a straight line. [2] … … … … … … … … … … … … … (ii) Find the values of n and A, giving your answers correct to 2 decimal places. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 4(i) State or imply n ln y = ln A + 3 ln x B1 State that the graph of ln y against ln x has an equation which is linear in ln y and ln B1 x, or has equation of the form nY = ln A + 3X, where Y = ln y and X = ln x, and is thus a straight line. 2 4(ii) Substitute x- and y-values in n ln y = ln A + 3 ln x or in the given equation and solve M1 for one of the constants Obtain a correct constant, e.g. n = 1.70 A1 Solve for a second constant M1 Obtain the other constant, e.g. A = 2.90 A1 4

This question in 9709/32 Feb/March 2018

7 (i) By sketching suitable graphs, show that the equation e2x 6 has exactly one real root. [2] = + e−x (ii) Verify by calculation that this root lies between 0.5 and 1. [2] … … … … … … … … … … … … (iii) Show that if a sequence of values given by the iterative formula 3 + xn+1 = 1 ln 1 6exn converges, then it converges to the root of the equation in part (i). [2] … … … … … … … … … … (iv) Use this iterative formula to calculate the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3] … … … … … … … … … … … …

9 marks

Mark scheme: 7(i) Sketch a relevant graph, e.g. y = 2e x B1 Sketch a second relevant graph, e.g. y = 6 + e− x , and justify the given statement B1 2 7(ii) Calculate the value of a relevant expression or values of a pair of relevant M1 expressions at x = 0.5 and x = 1 Complete the argument correctly with correct calculated values A1 2 7(iii) 1 x B1 State a suitable equation, e.g. x = ln 1 + 6e ( ) 3 Rearrange this as 2e x = 6 + e− x , or commence working vice versa B1 2 7(iv) Use the iterative formula correctly at least once M1 Obtain final answer 0.928 A1 Show sufficient iterations to 5 d.p. to justify 0.928 to 3 d.p., or show there is a sign A1 change in the interval (0.9275, 0.9285) 3

This question in 9709/32 Feb/March 2018

1 Showing all necessary working, solve the equation 3 2x 2x, giving your answers correct to 3 significant figures. −1 = [4] … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 1 EITHER: State or imply non-modular equation ( ) ( ) 2 2 2 3 2 1 2 − = x x , or pair of equations ( ) 3 2 1 2 − =± x x M1 2 8 2 18 2 9 0 − + = x x Obtain 2x = 3 2 and 2x = 3 4 or equivalent A1 OR: Obtain 2x = 3 2 by solving an equation B1 Obtain 2x = 3 4 by solving an equation B1 Use correct method for solving an equation of the form 2 = x a , where a > 0 M1 Obtain final answers x = 0.585 and x = – 0.415 only A1 The question requires 3 s.f. Do not ISW if they go on to reject one value 4

This question in 9709/32 May/June 2018

2ex 2 Showing all necessary working, solve the equation + e−x 4, giving your answer correct to ex = −e−x 2 decimal places. [4] … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2 Rearrange the equation in the form 2e x a b = or e e x x a b − = M1 Obtain correct equation in either form with a = 2 and b = 5 A1 Use correct method to solve for x M1 Obtain answer x = 0.46 A1 4

This question in 9709/31 Oct/Nov 2018

4 Showing all necessary working, solve the equation ex e−x 4, + ex 1 = + giving your answer correct to 3 decimal places. [5] … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 4 Substitute and obtain 3-term quadratic 2 3 4 1 0 + −= u u , or equivalent 2 3 e 4e 1 0 + −= x x Solve a 3 term quadratic for u M1 Must be an equation with real roots Obtain root ( ) 7 2 / 3 − , or decimal in [0.21, 0.22] A1 Or equivalent. Ignore second root (even if incorrect) Use correct method for finding x from a positive value of ex M1 Must see some indication of method: use of ln = x u Obtain answer x = – 1.536 only A1 CAO. Must be 3 dp 5

This question in 9709/32 Oct/Nov 2018

2ex 2 Showing all necessary working, solve the equation + e−x 4, giving your answer correct to ex = −e−x 2 decimal places. [4] … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2 Rearrange the equation in the form 2e x a b = or e e x x a b − = M1 Obtain correct equation in either form with a = 2 and b = 5 A1 Use correct method to solve for x M1 Obtain answer x = 0.46 A1 4

This question in 9709/33 Oct/Nov 2018

2 Showing all necessary working, solve the equation 9x 3x 12. Give your answer correct to 2 decimal = + places. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2 State or imply 2 12 0 − − = u u B1 Need to be convinced they know 2 23 3 = x x Solve for u, or for 3x , and obtain root 4 B1 Use a correct method to solve an equation of the form 3 = x a where a >0 M1 Need to see evidence of method. Do not penalise an attempt to use the negative root as well. e.g. 3 ln3 ln , log = = x a x a If seen, accept solution of straight forward cases such as 3x = 3 , x = 1 without working Obtain final answer x = 1.26 only A1 The Q asks for 2 dp 4

This question in 9709/32 May/June 2019

1 Given that ln 1 e2y x, express y in terms of x. [3] + = … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 1 State 2 1 e e y x + = B1 Make y the subject M1 Rearrange to 2 e ... y = and use logs Obtain answer ( ) 1 ln e 1 2 x y = − A1 OE 3

This question in 9709/31 Oct/Nov 2019

1 Solve the equation 5 ln 4 6. Show all necessary working and give the answer correct to −3x = 3 decimal places. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 1 Remove logarithms and state 1.2 4 3 e x − = , or equivalent B1 Accept 4 3 3.32 01169 … x − = 3 s.f. or better Use correct method to solve an equation of the form 3x a = , where a > 0. M1 ( ) 3 0.67988.. x = Complete method to x =… If using log3 the subscript can be implied Obtain answer x = – 0.351 only A1 CAO must be to 3 d.p. 3

This question in 9709/32 Oct/Nov 2019

32x 3 Showing all necessary working, solve the equation + 3−x 4. Give your answer correct to 32x = −3−x 3 decimal places. [4] … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 3 Reduce the equation to a horizontal equation in M1 Simplify and reach 3( 33 ) 5 = x , ( ) 3 27 5 = x , or equivalent A1 Use correct method for finding x from a positive value of 33 x , 3 1 3 + x or 27x M1 Obtain answer x = 0.155 A1 4

This question in 9709/33 Oct/Nov 2019

2 (a) Expand 2 in ascending powers of x, up to and including the term in x2, simplifying the −3x −2 coefficients. [4] … … … … … … … … … … … … … … … … … … (b) State the set of values of x for which the expansion is valid. [1] … … … … …

5 marks

Mark scheme: 2(a) State a correct unsimplified version of the x or x2 term of the expansion of (2 – 3x)–2 or 2 3 1 2 −   −     x State correct first term 1 4 B1 Obtain the next two terms 2 3 27 4 16 + x x A1 + A1 4 2(b) State answer 2 3 < x , or equivalent B1 1

This question in 9709/31 May/June 2020

3 The variables x and y satisfy the relation 2y = 31−2x. (a) By taking logarithms, show that the graph of y against x is a straight line. State the exact value of the gradient of this line. [3] … … … … … … … … … … … … (b) Find the exact x-coordinate of the point of intersection of this line with the line y 3x. Give your = ln a answer in the form where a and b are integers. [2] ln b, … … … … … … … … …

5 marks

Mark scheme: 3(a) State or imply log2 log3 2 log3 y x = − B1 Accept ( ) ln 2 1 2 ln3 y x = − State that the graph of y against x has an equation which is linear in x and y, or is of the form ay = bx + c B1 Correct equation. Need a clear statement/comparison with matching linear form. Clear indication that the gradient is – 2ln3 ln2 B1 Must be exact. Any equivalent e.g. 3 2 2log log k k − , 2 1 log 9 3 3(b) Substitute y = 3x in an equation involving logarithms and solve for x M1 Obtain answer x = ln3 ln72 A1 Allow M1A1 for the correct answer following decimals 2

This question in 9709/32 Oct/Nov 2020

2ex e−x2 Find the real root of the equation 3, giving your answer correct to 3 decimal places. + 2 ex = Your working should show clearly that +the equation has only one real root. [5] … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 Reduce to a 3-term quadratic 2 6 1 0 + −= u u OE B1 Allow ‘= 0’ implied Solve a 3-term quadratic for u M1 Obtain root 10 3 − A1 Obtain answer x = – 1.818 only A1 The question asks for 3 d.p. Reject 10 3 − − correctly B1 e.g. by stating that e 0 > x or ( ) ln 10 3 − − is impossible Not "math error". Alternative method for Question 2 Rearrange to obtain a correct iterative formula B1 e.g. ( ) 1 ln 6 e + = − + n x nx Use the iterative process at least twice M1 Obtain answer x = – 1.818 A1 Show sufficient iterations to at least 4 d.p. to justify x = – 1.818 A1 1, 2.165..., 1.811..., 1.819..., 1.818..., 1.818... − − − − − Clear explanation of why there is only one real root B1 5

This question in 9709/31 May/June 2021

3 The variables x and y satisfy the equation x A 3−y , where A is a constant. = (a) Explain why the graph of y against ln x is a straight line and state the exact value of the gradient of the line. [3] … … … … … … … … … … … It is given that the line intersects the y-axis at the point where y 1.3. = (b) Calculate the value of A, giving your answer correct to 2 decimal places. [2] … … … … … … … … … …

5 marks

Mark scheme: 3(a) = − B1 1 ln ln ln3 ln3   = − +     A y x State that the graph of y against ln x has an equation that is linear in y and ln x, or has an equation of the standard form ‘y = mx + c’ and is thus a straight line B1 Must be a correct statement. Accept if the 2 equations are written side by side with no comment. An equation with ln3 y should be compared with the form py + q ln x = c. State that the gradient is – 1 ln3 B1 OE. Exact answer required. ISW after a correct statement. 3 Question Answer Marks Guidance 3(b) Substitute ln x = 0, y = 1.3 and use correct method to solve for A M1 ( ) ln 1.3ln3 A = Follow their equation in y and ln x . Must be substituting lnx = 0, not x = 0. ln 0 ‘used’ in the solution scores M0A0. Obtain answer A = 4.17 only A1 Must be 2 d.p. as specified in question 2

This question in 9709/32 May/June 2021

2 Solve the equation 4x 3 Give your answer correct to 3 decimal places. [5] = + 4−x. … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 State or imply 2 3 1 0 − −= u u B1 Solve for u or 4x M1 Obtain root ( ) 1 3 13 2 + , or decimal in [3.30, 3.31] A1 Use correct method for finding x from a positive root M1 Obtain answer x = 0.862 and no other A1 5

This question in 9709/33 May/June 2021

1 Solve the equation 4 5x 5x, giving your answers correct to 3 decimal places. [4] −1 = … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 1 State or imply non-modular equation ( ) ( ) 2 2 2 4 5 1 5 − = x x or pair of equations ( ) 4 5 1 5 − = ± x x M1 Obtain 4 5 3 = x and 4 5 5 = x (or 1 5 4 + = x ) A1 Use correct method for solving an equation of the form 5 = x a , or 1 5 + = x b where a > 0, or b > 0 M1 Obtain answers x = 0.179 and x = – 0.139 A1 Alternative method for question 1 Obtain 4 5 3 = x by solving an equation B1 Obtain 4 5 5 = x (or 1 5 4) + = x by solving an equation B1 Use correct method for solving an equation of the form 5 = x a , or 1 5 + = x b where a > 0, or b > 0 M1 Obtain answers x = 0.179 and x = – 0.139 A1 4

This question in 9709/31 Oct/Nov 2021

4x giving your answer correct to 3 decimal places. [4]3 Solve the equation 4x−2 = −42, … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 3 M1 Obtain correct solution in any form, e.g. 256 4 15 x = A1 Use a correct method for solving an equation of the form 4x a = , where a > 0 M1 Obtain answer 2.047 A1 4

This question in 9709/33 Oct/Nov 2021

5 (a) By sketching a suitable pair of graphs, show that the equation ln x 3x has one real root. = −x2 [2] (b) Verify by calculation that the root lies between 2 and 2.8. [2] … … … … … … … (c) Use the iterative formula /3xn xn xn+1 = −ln to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … …

7 marks

Mark scheme: 5(a) Sketch a relevant graph, e.g. y = ln x B1 ln( ) x : sketch should imply y-axis is an asymptote. Through (1, 0) if marked. Correct shape. 2 3x x  : Symmetrical. Through (0, 0) and (3, 0) if marked. If ln(x) correct accept parabola for +ve y only. If ln(x) incorrect then need parabola in 3 quadrants. Sketch a second relevant graph, e.g. 2 3 y x x   , and justify the given statement by marking the root on the sketch or by use of a suitable comment B1 2 5(b) Calculate the values of a relevant expression or pair of expressions at x = 2 and x = 2.8 M1 Allow for a smaller interval. At least one value correct if comparing with 0. If using pairs then the pairing must be clear. Complete the argument correctly with correct calculated values A1 e.g. 0.693 2 and1.03 0.56   or 1.307 0, 0.47 0    using 3 ln x x  0.304 0, 0.085 0    . Need to have calculated values to at least 2 sf. 2 2 2 Question Answer Marks Guidance 5(c) Use the iterative process correctly at least once M1 Obtain final answer 2.63 A1 Show sufficient iterations to at least 4 dp to justify 2.63 to 2 dp or show there is a sign change in the interval (2.625, 2.635) A1 SC Allow M1 A1 A0 to a candidate who starts at a point in the interval and reaches a premature conclusion 3

This question in 9709/32 May/June 2022

1 Solve the equation 3e2x 5. −4e−2x = Give the answer correct to 3 decimal places. [3] … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 1 2 2 2 3 e 5 e 4 0 x x    2 2 5 73 1 5 73 e , ln 6 2 6 x x             M1 Use correct method to solve for x. x = 0.407 A1 Only 3

This question in 9709/31 May/June 2023

8 (a) By sketching a suitable pair of graphs, show that the equation x = ex −3 has only one root. [2] (b) Show by calculation that this root lies between 1 and 2. [2] … … … … … … … … … … … (c) Show that, if a sequence of values given by the iterative formula xn+1 = ln 3 + / xn ! converges, then it converges to the root of the equation in (a). [1] … … … … … … … (d) Use the iterative formula to calculate the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … … … … … … …

8 marks

Mark scheme: 8(a) Sketch a relevant graph, e.g. y = ex – 3 B1 y = ex – 3 Should cut vertical axis at (0, –2) and have increasing gradient. 2 y= x Sketch a second relevant graph, e.g. y = x and justify the given statement B1 y = x should start at (0, 0) and have reducing grading y=ex - 3 2 Ignore anything outside 1st and 4th quadrants. For second B1 need to mark intersection with a dot, a cross, or say root at point of intersection, or equivalent. 2 8(b) Calculate the values of a relevant expression or pair of expressions at x = 1 M1 and x = 2 Complete the argument correctly with correct calculated values A1 e.g.1 −0.28..,1.41  4.39.. 1.28 > 0, –2.98 < 0. 2 8(c) State x = ln 3 + x and rearrange to the given equation x = e x − 3 B1 Or rearrange x = e x −to3 x = ln(3 + x ) and state ( ) iterative formula of xn +1 = ln(3 + xn ) . AG 1 8(d) Use the iterative process correctly at least once M1 Obtain final answer 1.43 A1 Show sufficient iterations to at least 4 d.p. to justify 1.43 to 2 d.p. or show A1 e.g. 1, 1.3864, 1.4297, 1.4341, … there is a sign change in the interval (1.425, 1.435) 1.5, 1.4210, 1.4332, 14.344, 1.4345, … Condone recovery and small differences in the final figure in the iteration 2, 1.4848, 1.4395, 1.4350, 1.4346, 1.4345, … 3

This question in 9709/31 Oct/Nov 2023

1 Find the set of values of x satisfying the inequality 2x+1 −2 < 0.5, giving your answer to 3 significant figures. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: Question Answer Marks Guidance 1 State or imply non-modular inequality −0.5  2 x+1 − 2  0.5 , can be in two B1 −0.25  2 x −1 0.25 , can be in two separate statements, separate statements, x 2 2 or 2 − 1  0.25 or corresponding pair of linear 2 ( ) or 2 x+1 − 2  0.5 2 ( ) equations 0.25 = 2x – 1 and −0.25 = 2x – 1 or quadratic or corresponding pair of linear equations 0.5 = 2x+1 – 2 and − 0.5 = 2x+1 – 2 x 2 2 equation 2 − 1 = 0.25 . 2 ( ) or quadratic equation 2 x+1 − 2 = 0.52 ( ) Incorrect inequality mark recoverable by correct final answer or x < 0.32 and x > –0.42 . Use correct method for solving an equation or inequality of the form M1 Reach (x + 1)ln2 = lna or equivalent, do not need to reach 2x +1 = a or 2x = b where a, b > 0 x = … Obtain critical values x = 0.322 and –0.415 A1 ln 2.5 ln1.5 e.g. − 1 and −.1 or awrt x = 0.32 and –0.42 ln 2 ln2 or exact equivalents State final answer –0.415 < x < 0.322 or (–0.415, 0.322) A1 Need 3 significant figures. Need combined result, not x < 0.32 and x >–0.42 . Must be strict inequalities. No working, 0/4. Alternative method for Question 1 Use correct method for solving an equation or inequality of the form M1 May see 2x+1 = 1.5 and 2x+1 = 2.5 . 2x +1 = a or 2x = b where a, b > 0 Reach (x + 1)ln2 = lna or equivalent, don’t need to reach x = … Obtain one critical value, e.g. 0.322 A1 ln 2.5 e.g. −.1 or awrt x = 0.32 ln 2 or exact equivalent Obtain the other critical value e.g. –0.415 or awrt x = –0.42 or exact A1 ln1.5 e.g. −.1 equivalent ln2 1 State final answer –0.415 < x < 0.322 or (–0.415, 0.322) A1 Need 3 significant figures. Need combined result, not x < 0.32 and x > – 0.42 . Must be strict inequalities. No working, 0/4. 4

This question in 9709/33 Oct/Nov 2023

4 The positive numbers p and q are such that p 2 ln = a and ln q p = b . e q o ` j Express ln p 7 q in terms of a and b. [4] ` j … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 4 Obtain ln p − ln q = a B1 p = ea. q Obtain ln p + 2ln q = b B1 pq2 = eb. Completed method to obtain ln ( p 7 q ) M1 E.g. ln q = b − a , ln p = 2 a + b 3 3 and attempt 7ln p + ln q. All exponentials must be removed to obtain M1. 13a + 8b A1 Obtain 3 Alternative solution for Question 4 x p y B1 7  p  2 Or ln p 7 q = x ln + y ln q 2 p . State p q =   ( q p ) q  q  Equate indices to form simultaneous equations in x and y, can have errors M1 x + y = 7 and − x + 2y = 1. Obtain 7 = x + y and 1 = 2 y − x A1 Leading to x = 133 , y = 83 . 13a + 8b A1 Evaluate x×a + y×b to obtain 3 4

This question in 9709/32 Feb/March 2024

2 Solve the equation ln ( x - 5) = 7 - ln x . Give your answer correct to 2 decimal places. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2 Use law of logarithm of a product (or quotient) on correct terms *M1 Use correct method to eliminate logarithm DM1 Obtain a correct quadratic in x, e.g. 2 7 5 e 0    x x (allow decimals) A1 Obtain answer x = 35.71 only A1 4

This question in 9709/31 May/June 2024

3 y (1.31, 1.50) (0.336, 1.00) O ln x The variables x and y satisfy the equation a y = bx , where a and b are constants. The graph of y against lnx is a straight line passing through the points (0.336, 1.00) and (1.31, 1.50), as shown in the diagram. Find the values of a and b. Give each value correct to the nearest integer. [4] … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 3 State or imply that ln ln ln y a b x Carry out a completely correct method for finding lna or lnb M1 E.g., from ln ln 0.336 a b   1.5ln ln 1.31. a b   Obtain value a = 7 A1 Obtain value b = 5 A1 4

This question in 9709/31 May/June 2024

3 The variables x and y satisfy the equation a 2 y - 1 = b x - y , where a and b are constants. (a) Show that the graph of y against x is a straight line. [3] … … … … … … … … … … … … … (b) Given that a = b3 , state the equation of the straight line in the form y = px + q , where p and q are rational numbers in their simplest form. [2] … … … … … … … … … … … …

5 marks

Mark scheme: 3(a) Use logarithms to obtain a correct expression without powers e.g.     2 1 ln ln y a x y b    2 1 log . a y x y b   Do not condone missing brackets unless recovered later. Separate terms and factorise to obtain   2ln ln ln ln y a b x b a    B1 Or equivalent, e.g. 2 2 ln ln ln ln b a y x a b a b   or   2 log log 1. a a y b x b    Clear explanation of linear form. From correct work only. B1 E.g. equation matches the linear form y mx c   or . py qx r   Condone if they compare with , y mx c   but do not actually state that it must therefore be a straight line. Stating “this is a linear equation” without comparing to a relevant standard form scores B0. B0 if they have m = … and c = … correct but never actually mention y = mx + c. 3 3(b) Use 3 a b  and log laws to simplify their equation M1 3 7 7 ln ln ln ln b b y x b b         Denominator reduced to a single log term. Obtain 1 3 7 7 y x   A1 Accept 3 7 7 x y   but not 3. 7 x y   Alternative method for Question 3(b) Use 3 a b  to obtain   3 2 1 y x y b b    or equivalent (M1) Or 1 3 log . a b  Obtain 1 3 7 7 y x   (A1) Accept 3 7 7 x y   but not 3. 7 x y   2

This question in 9709/32 May/June 2024

5 (a) It is given that the equation e 2 x = 5 + cos 3x has only one root. Show by calculation that this root lies in the interval 0.7 1 x 1 0.8 . [2] … … … … … … (b) Show that if a sequence of values in the interval 0.7 1 x 1 0.8 given by the iterative formula 1 x n + 1 = ln 5 + cos 3x n 2 ` j converges then it converges to the root of the equation in part (a). [1] … … … … … … (c) Use this iterative formula to determine the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3] … … … … … … … … … …

6 marks

Mark scheme: 5(a) Calculate the value of a relevant expression or values of a pair of expressions at 0.7 x  and 0.8 x  Need all relevant values but condone one error. Pairings must be clear for solutions involving four values (do not accept embedded values). M0 if working in degrees e.g -1.94…, 1.04… Complete the argument correctly with correct calculated values. (can be using the equation in the rubric or the equation in (b) or equivalent) A1 E.g. 4.95... 4.26...  and 4.06... 4.49...  -0.439…< 0, 0.690…> 0 1.4 < 1.5029…, 1.6 > 1.449… 1.1 > 1, 0.86 < 1 0.0515 > 0, -0.075 < 0. Allow values rounded or truncated to 2sf. 2 5(b) State   2 ln 5 cos3 x x   and take exponential of both sides to obtain 2e 5 cos3 x x   B1 Given answer requires fully correct working or work vice versa. If working in reverse, must get to the iterative formula, including subscripts. 1 5(c) Use the iterative process correctly at least once M1 M0 if working in degrees (e.g. values heading for 0.89….). Obtain final answer 0.740 A1 Show sufficient iterations to at least 5dp to justify 0.740 to 3dp, or show that there is a sign change in the interval   0.7395, 0.7405 A1 E.g. 0.7,0.75150,0.73719,0.74105,0.74000,0.74028 0.75,0.73759,0.74094,0.74003,0.74028 0.8,0.72494,0.74443,0.73909,0.74053,0.74014,0.74025 Allow recovery. Allow truncation or rounding and condone small differences in the final decimal place. 3

This question in 9709/32 May/June 2024

1 Solve the equation 8 3 - 6 x = 4 # 5 -2 x . Give your answer correct to 3 decimal places. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 1 Use law of the logarithm of product or quotient on each side *B1 Allow logs to any base, as well as decimals, throughout. ln 83 + ln 8−6x and ln 4 + ln 5−2x. Allow for ln 3 8 4 and ln 86x – ln 52x. (3 − 6x) ln 8 and ln 4 + ln 5−2x gains next DB1 as well. Use law of logarithm of a power involving x on ONE side, e.g. ln 83 + (−)6x ln 8 or (3 − 6x) ln 8 or   9 18x  ln 2 or ln 4 − 2x ln 5 DB1 SC If *B0 DB0, then allow B1 (1/4) for a correct logarithm law seen anywhere. Obtain a correct linear equation in x, e.g.     3 6 ln8 9 18 ln2 ln 4 2 ln5 x x x      B1 If in decimals, allow small errors in 2nd and 3rd dp. Obtain answer x = 0.524 B1 3dp required. No working scores 0/4 marks. After *B1 DB1 to correct answer with no more log working seen, then SC B1 for x = 0.524. Maximum 3/4 possible. Alternative Method for Question 1 Use laws of indices to get to a = b ±2x or c ± x in a correct form so now only ONE log power law required (B2) (83/4) and (5/83)−2x or (52/86)−x opposite sides or (4/83) and (83/5)–2x or (86/52)–x opposite sides Obtain a correct linear equation in x, e.g. 3 3 8 8 ln 2 ln 4 5 x  (B1) −2x ln (5/83) or 2x ln (83/5) or x ln (86/52) or – x ln (52/86). SC: If B0 then allow B1 (1/4) for a correct term seen anywhere. If in decimals, allow small errors in 2nd and 3rd dp. Obtain answer x = 0.524 (B1) 3dp required. No working scores 0/4 marks. From the first line to correct answer with no log working seen, then B2 and SC B1 for x = 0.524. Maximum 3/4 possible. Question Answer Marks Guidance 1 Alternative Method 2 for Question 1 Use laws of indices to get to any correct form with indices combined so now TWO log power laws are required (*B1) Allow 27 – 18x and 5–2x on opposite sides or 29 – 18x and 22 – 4.64x on opposite sides. Use law of logarithm of a power involving x on ONE side, e.g. (7 – 18x) ln 2 = ln 5–2x or ln 27 – 18x = − 2x ln 5 or … Allow 7 – 18x ln 2 or 9 18 ln2 x  (DB1) e.g. (7 – 18x) ln 2 or   9 18 ln2 x  or –2x ln 5 or (2 – 4.64x) ln 2 SC: If *B0 DB0 then allow B1 (1/4) for a correct term seen anywhere. E.g. any term in *B1 shown above. Obtain a correct linear equation in x, e.g. (7 – 18x) ln 2 = − 2x ln 5 or (9 – 18x) ln 2 = (2 – 4.64x) ln 2 (B1) If in decimals, allow small errors in 2nd and 3rd dp. Obtain answer x = 0.524 (B1) 3dp required. No working scores 0/4 marks. From the first line to correct answer with no log working seen, then *B1 and SC B1 for x = 0.524. Maximum 2/4 possible. 4

This question in 9709/33 May/June 2024

4 1n y (5.10, 2.21) (2.80, 0.372) O x The variables x and y satisfy the equation ky = ecx , where k and c are constants. The graph of ln y against x is a straight line passing through the points (2.80, 0.372) and (5.10, 2.21), as shown in the diagram. Find the values of k and c. Give each value correct to 2 significant figures. [4] … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 4 State or imply that ln k + ln y = cx or ln y = cx + ln 1 k etc. B1 Allow ln k + ln y = cx lne Carry out a completely correct method for finding ln k or c M1 Equations must have been formulated correctly. Obtain value c = 0.80 A1 AWRT Allow 0.8 for 0.80. Not a fraction. Accept in the equation ky = ecx. Obtain value k = 6.5 A1 AWRT Not a fraction. Accept in the equation ky = ecx.

This question in 9709/33 May/June 2024

4 Solve the equation 5 x = 5 x + 2 - 10 . Give your answer correct to 3 decimal places. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 4 Use laws of indices correctly and solve for 5x *M1 x 5 E.g. obtain 5 = OE. 12 Allow for y = … if they have previously stated y = 5 x . Could be implied if they have a correct simplified equation in 5x, e.g. 12  5 x = 5. Use a correct method for solving an equation of the form 5 x = a , where a > 0 DM1  10  Allow x ln5 = ln   .  24  Obtain answer – 0.544 A1 CWO. If no working shown, 0/3. Note: 3 dp required. 3

This question in 9709/32 Oct/Nov 2024

6 ln y (3.40, 8.27) (0.50, 2.24) O x The variables x and y satisfy the equation ay = bx , where a and b are constants. The graph of lny against x is a straight line passing through the points (0.50, 2.24) and (3.40, 8.27), as shown in the diagram. Find the values of a and b. Give each value correct to 1 significant figure. [4] … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 6 Form a pair of equations in a and b *M1 Condone sign slips but must be using the given coordinates correctly. ln a + 8.27 = 3.4ln b e.g.  ln a + 2.24 = 0.5ln b ae 2.24 = b 0.5 or  8.27 3.4  ae = b Carry out a correct method for finding lna or lnb or a or b DM1 Condone sign slip. Obtain value a = 0.3 A1 (0.30109…) Obtain value b = 8 A1 (7.99895…) Allow A0A1 if both values ‘correct’ but not rounded to 1 sf. Allow 4/4 for 0.3 y = 8 x with correct working shown. 6 Alternative Method for Question 6: Carry out a correct method for finding lnb or b *M1 Condone sign slips but must be using the given (Need to link the gradient to lnb at some point) coordinates correctly. 8.27 − 2.24 ln b = ( = 2.079 … ) 3.4 − 0.5 Obtain value b = 8 A1 Correct method to find  ln aor a DM1 Condone sign slip ( ln a = − 1.200... ) . Obtain value a = 0.3 A1 Allow A0A1 if both values ‘correct’ but not rounded to 1 sf. Allow 4/4 for 0.3 y = 8 x with correct working shown. 4

This question in 9709/32 Oct/Nov 2024

3 lnP 3 O t The number of bacteria in a population, P, at time t hours is modelled by the equation P = aekt , where a and k are constants. The graph of lnP against t, shown in the diagram, has gradient 1 and intersects 20 the vertical axis at ( 0, 3) . (a) State the value of k and find the value of a correct to 2 significant figures. [3] … … … … … … … (b) Find the time taken for P to double. Give your answer correct to the nearest hour. [2] … … … … … … … … … …

5 marks

Mark scheme: 3(a) State or imply that ln P = ln a + kt or ln P = ln a + k(ln e)t B1 Can be implied by both a and k correct. 1 t P = e 3 e 20 gets B1B1. 1 ln P = t + 3 B0 until associated with a and /or k 20 1 dP B1 OE. Can be embedded in P = ae kt . State k = , not from = k 20 dt ln a = 3  a = 20 to 2 sf B1 Must be 2 sf, can be embedded in P = ae kt . 3 3(b) Form a correct equation in t using a and k, or their a and k where a will cancel (or M1 E.g. 2 a = ae kt , 2 = e kt , kt = ln 2. are both numerical) Obtain t = 14  hours  A1 Allow 13.75 [hrs] (13 hrs 45 min) to 14 [hrs]. ISW 2

This question in 9709/33 Oct/Nov 2024

1 Solve the equation ln `1 - e -2 xj + 3 = 0 . Give your final answer correct to 4 decimal places. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: Question Answer Marks Guidance 1 State that 1 – e–2x = e–3 B1 OE, with ln removed. Use correct method to solve an equation of the form e±2x = a, where a > 0, and a M1 E.g. [e–2x = 1 − e–3] reasonable attempt at the B1, for ± 2x ln e or ± x ln e −2x = ln(1 − e–3) … OE. Can be numerical ln (1 − 0.049787) = ln 0.9502. Evidence of method must be seen. Obtain answer 0.0255 A1 CAO Must be 4 decimal places. No working seen scores 0. Alternative Method for Question 1 State that 1 – e–2x = e–3 B1 OE, without ln. Rearrange to obtain an expression for ex and solve an equation of the form e±x = a, M1  3  x 1 x e where a > 0, and a reasonable attempt at the B1, for x E.g.  e = −3 , e = 3  ,  1 − e e − 1    1 x = ln 3 1 − e− Can be numerical. Evidence of method must be seen. Obtain answer 0.0255 A1 CAO Must be 4 decimal places. No working seen scores 0. 3

This question in 9709/32 Feb/March 2025

2 It is given that 2 ln p + ln ( p - 1 ) - 1 ln ( q + 1) = 3 . 2 Find q in terms of p. [3] … … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 2 Use logarithm of a root or a power M1 E.g. 2ln p = ln p 2 , 12 ln ( q + 1) = ln q + 1 or 3 = ln e 3 . Obtain p 2 ( p − 1) = e 3 q + 1 A1 OE without logs. 2 2 A1 OE, ISW  p ( p − 1)  Obtain q =  3  − 1  e  3

This question in 9709/31 May/June 2025

e x + 2e -x 1 Solve the equation x = 4 . Give your answer correct to 3 decimal places. [5] e - 3 … … … … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: Question Answer Marks Guidance 1 Obtain a 3-term quadratic in ex *M1 Obtain e.g. 3e2x – 12ex – 2 = 0 or 3-term equivalent A1 2 E.g. 3m − 12m − 2 = 0. ‘= 0’ could be implied by subsequent working. Solve a 3-term quadratic to obtain a value for x or ex DM1 Need to get as far as a value for x or ex. 6 + 42 A1 OE Obtain root or 4.16… Ignore second root if seen. 3 Obtain answer 1.426 only A1 CAO, must be 3 d.p. 0/5 for answer with no working. Second root must be rejected if seen. 5

This question in 9709/32 May/June 2025

2 Solve the equation 2 ln ( 2x + 3) - ln ( 2x + 5) = ln ( 3x) . [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2 Use the correct rule for logarithm of a power, product or quotient or an equivalent *M1 Use of any correct law applied to original terms. method using exponentials Obtain an equation free of logarithms A1 ( 2 x + 3 ) 2 E.g. = 3 x. 2 x + 5 Form a 3-term quadratic from completely correct use of logarithms and solve for x DM1 2 x 2 + 3x −=9 0 State final answer 32 only A1 OE Rejection of negative value if given must be clear. 4

This question in 9709/33 May/June 2025

1 Solve the equation 3 4 - 2x = 5 ( 6x - 1 ) . Give your answer correct to 3 significant figures. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: Question Answer Marks Guidance 1 Use law of logarithm of a product or quotient M1 ln 5 + ln 6x−1 or ln 34−2x – ln 5 or ln 34−2x − ln 6x−1 Allow logs to any base but must be consistent throughout the equation. Use law of logarithm of a power twice M1 (4 – 2x)ln 3 and (x – 1)ln 6. Omission of bracket(s) is an accuracy error if not corrected later. Can have M0M1, i.e. go wrong with product but powers dealt with correctly. E.g. (4 – 2x)ln 3 = ln 5×(x – 1)ln 6, or (4 – 2x)ln 3 = (x – 1)ln 30 or 33x×2x written as 3xln 3×xln 2. Obtain a correct equation in any form, A1 May see (4 – 2x)ln 3 written as (2 – x)ln 9. e.g. (4 – 2x)ln 3 = ln 5 + (x – 1)ln 6 or 1.10(4 – 2x) = 1.61 + 1.79(x – 1) Obtain x = 1.15 A1 Must be 3 s.f. If no working seen, no marks available. 4

This question in 9709/35 May/June 2025

2 (a) Show that the equation log ( 2x + 1 ) = 2 log ( 3 x - 1 ) - 2 can be written as a quadratic equation 4 4 in x. [3] … … … … … … … … … … … … (b) Hence solve the equation log ( 2x + 1 ) = 2 log ( 3 x - 1 ) - 2 . [2] 4 4 … … … … … … … … … … … … … …

5 marks

Mark scheme: 2(a) State or imply 2 = log4 42 B1 Use log power, product or quotient law e.g. 2log4 (3x – 1) = log4 (3x – 1)2 M1 ( 3 x − 1) 2 E.g. log4 (3x – 1)2 – log4 42 = log4 4 2 Obtain a correct equation in any form, free of logs A1 E.g. 16(2x + 1) = (3x – 1)2 or 9x2 – 38x – 15 = 0 3 2(b) Solve 3-term quadratic M1 Obtain answer 4.59 only A1 19 + 4 31 OE. 9 Accept AWRT 4.59. 2

This question in 9709/31 Oct/Nov 2025

x 3 + 2x - 11 10 Let f ( x) = . ( 3 + x) `2 + x 2j (a) Express f ( x) in partial fractions. [6] … … … … … … … … … … … … … … … … … … … … … … … … … (b) Hence obtain the expansion of f ( x) in ascending powers of x, up to and including the term in x2. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 10(a) B Cx + D B1 State or imply the form A + + 3 + x 2 + x 2 Use a correct method for finding a constant M1 −3 x 2 − 17 Might be working from 1 + . (3 + x )(2 + x 2 ) Obtain one of A = 1, B = –4, C = 1 and D = –3 A1 SC: If B0 scored due to missing term(s), then a maximum of M1A1 is available for a correct method leading to a correct value. Obtain a second value A1 SC if obtaining A = 1 and then scoring B0, they can score maximum M1A1 + A1 for a correct value for one other constant. Obtain a third value A1 Obtain all four correct values A1 6 10(b) Use a correct method to find the first two terms in the expansion of (3 + x) –1, M1 Symbolic binomial coefficients not sufficient for the −1 2 −1 M1.  x  –1  x   1 +  , (2 + x2) or  1 +   3   2  Obtain correct unsimplified expansions up to the term in x2 of each partial A1 FT FT B, C and D. fraction A1 FT B  x x 2  Cx + D  x 2  E.g.  1 − +  and  1 −  . 3  3 9  2  2  −1 −1 2 M1 2  x   x  Multiply (Cx + D) by the expansion of  1 +  up to the term in x2 where Expansion of  1 +  .  2   2  CD ≠ 0 Must be of the form 1 + Qx². Multiplication must include 3 relevant terms. 11 17 65 2 A1 Or exact equivalent. Obtain final answer − + x + x Do not ISW attempts to multiply expression by, e.g., 6 18 108 108. 5

This question in 9709/31 Oct/Nov 2025

2 Solve the equation 3 # 2 x + 1 = 4 # 3 2 x - 3 . Give your answer correct to 3 significant figures. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2 Use a correct law for the logarithm of a product or the logarithm of a power *M1 Not available after incorrect manipulation, e.g. 6 x +1 = 12 2 x − 3 Obtain ln3 + ( x + 1) ln 2 = ln 4 + ( 2 x − 3 ) ln3 A1 Or equivalent with no powers Solve for x DM1 ln 812 E.g. x = ln 92 Allow with logarithms evaluated. Obtain 2.46 only A1 Alternative Method for Question 2 Use a correct law of indices for a product or a power *M1 x +1 x E.g. 2 = 2  2 Not available after incorrect manipulation, e.g. 6 x +1 = 12 2 x − 3. 4 = 2 2 on its own is not enough. 2 4  9 x = 2 A1 OE with powers of x only, e.g. 6  2 x = 27 ) x Obtain ( 9 81 2 Note: 3x is not far enough. ( ) Solve for x DM1 ln 812 2 or x = ln E.g. x = 2 81 . 2 9 ln 9 Allow with logarithms evaluated. Obtain 2.46 only A1 4

This question in 9709/32 Oct/Nov 2025

3 x - 4 3 ln3 Solve the equation 2 = x . Give your answer in the form m, where m and n are integers. [4] 5 ln n … … … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 3 3 M1 May work in log to any base for first 3 marks. Use log quotient law, e.g. ln = ln 3 – ln 5x 5x or log product law, e.g. ln5x + ln(23x – 4) Use log power law, e.g. ln 5x = x ln5 or ln (23x – 4) = (3x – 4) ln 2 M1 Condone missing brackets if recovered at some point. ln3 + 4ln 2 A1 Obtain correct expression for x in any exact form, e.g. 3ln 2 + ln5 ln 48 A1 Do not ISW. Obtain final answer Final answer of log40 48 scores 3 marks only. ln 40 No working award 0 marks. Alternative Method for Question 3 Rearrange to obtain (23 × 5)x = 3 × 24 B1 OE Use log power law on an equation of the form abx = c M1 May work in log to any base for first 3 marks. e.g. x ln (23 × 5) = ln (3 × 24) 4 A1 ln ( 3 × 2 ) Obtain correct expression for x in any exact form, e.g. x = ln ( 2 3 × 5 ) ln 48 A1 Do not ISW. Obtain final answer Final answer of log40 48 scores 3 marks only. ln 40 No working award 0 marks. 4

This question in 9709/33 Oct/Nov 2025

3 The variables x and y satisfy the equation Ay = bx , where A and b are constants. (a) Show that the graph of lny against x is a straight line. [2] … … … … … … … … (b) When x = 3 .4 , ln y = 0.86 and when x = 5. 7, ln y = 2.56 . Find the value of A and the value of b. Give your answers correct to 2 significant figures. [3] … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3(a) Take logarithms of both sides to obtain ln A + ln y = x ln b M1 Or three-term equivalent with no indices. Compare with Y = mX + c , pY + qX + r = 0 A1 AG Condone use of y = mx + c in comparison. 2 3(b) Form equations in A and b and solve for A or b M1 ln A + 0.86 = 3.4ln b and ln A + 2.56 = 5.7ln b. Or Ae 0.86 = b 3.4 and Ae 2.56 = b 5.7 . Obtain b = 2.1 or A = 5.2 A1 Allow for 2 sf or more. Obtain b = 2.1 and A = 5.2 A1 Accept A = 5.3. 3

This question in 9709/35 Oct/Nov 2025