2 (a) Given that xy xz z zy x y z a b c 3 3 2 = - ` ` j j , find the exact values of the constants a, b and c. [3] (b) Solve the equation ( ) ( ) 5 2 17 2 3 0 p p 2 1 - + = + . [4]

2(a) 10 1 or 3 3 3 = a B1 7 1 or 2 3 3 = b B1 9 1 or 4 or 4.5 2 2 = c B1 2(b) ( ) ( ) 2 10 2 17 2 3 0 − + = p p ( ) ( ) ( ) ( ) 5 2 1 2 2 3 0 − − = p p 1 3 2 , 2 5 2 = = p p M1 For recognition of a quadratic in 2 p , attempt to factorise and solve for 2 p 1 ln ln1.5 5 or ln 2 ln 2 = = p p oe M1 For correct attempt to deal with 2 = p k 2.32 − A1 0.585 A1 (a) To earn these marks, you must correctly identify the exact values for a, b, and c by comparing the coefficients of the given expression. (b) To earn these marks, you must first recognize the equation as a quadratic in 2^p and attempt to factorize and solve for 2^p. Then, you need to correctly apply logarithms to solve for p, and finally provide both correct numerical answers.

3 (a) Write lg lg a b 3 2 4 + - as a single logarithm to base 10. [4] (b) Solve the equation log log a 3 4 2 7 a 4 + = . [5]

3(a) 2 4 1000 lg a b 4 B1 for 3 lg1000 = B1 for use of power rule once B1 for use of addition or subtraction rule once B1 All correct 3(b) Either 4 3 3log 4 log = a a B1 ( ) 2 4 4 2 log 7log 3 0 − + = a a ( )( ) 4 4 2log 1 log 3 0 − − = a a 4 4 1 log or log 3 2 = = a a M1 For obtaining a quadratic equation and solution 1 3 2 4 or 4 = = a a M1 Dep For dealing with the logarithm correctly once, may be implied by a correct solution 64 A1 2 A1 Or 4 2 2log log 4 = a a (B1) ( ) 2 3 log 4 7log 4 2 0 − + = a a ( )( ) 3log 4 1 log 4 2 0 − − = a a 1 log 4 or log 4 2 3 = = a a (M1) For obtaining a quadratic equation and solution 1 2 3 4 or 4 = = a a (M1) Dep For dealing with the logarithm correctly once, may be implied by a correct solution 64 (A1) 2 (A1) (a) To earn these marks, you need to correctly convert 3 to lg1000, apply the power rule for logarithms, and then correctly apply the addition and subtraction rules to combine the terms into a single logarithm. (b) To earn these marks, you must first correctly rewrite log_a(4) as 2log_a(2) or 3log_a(4) as log_a(64), then form a quadratic equation in terms of log_a(4) or log_a(2), solve this quadratic, and finally correctly solve for 'a' using the definition of logarithms.

5 The functions f and g are defined as follows. ) (f x x x 4 2 = + for x R ! ( ) g e x 1 x 2 = + for x R ! (a) Find the range of f. [2] (b) Write down the range of g. [1] (c) Find the exact solution of the equation ) ( fg x 21 = , giving your answer as a single logarithm. [4]

5 2 5 10 1 1 ... 3 3 9   − = − +     x x x 2 M1 allow one sign error or one arithmetic slip 3 27, 3 = = a a B1 Term in x: 2 3 5 3 3 − = a a b M1 For multiplying their terms, must have sum of 2 relevant products = b 18 = − b A1 Term in 2 : x ( ) 2 3 5 10 3 3 3 9 − + = a a a c M1 For multiplying their terms, must have sum of 3 relevant products = c This question appears to have a mismatch between the question text and the provided mark scheme. The mark scheme seems to correspond to a different question involving polynomial expansion and coefficient matching, not the functions f(x) and g(x) as stated in the question text. Therefore, I cannot provide a relevant note based solely on the given mark scheme for the stated question.

5 (a) Given that log log log log p 5 4 20 a a a a + - = , find the value of p. [2] (b) Solve the equation ( ) 3 8 3 3 0 x x 2 1 + - = + . [3] (c) Solve the equation 2 4log 2 log 4. y y + = [3]

5(a) 16 = p 2 B1 for 5 log log 20 4 = a a p oe B1 for 16, nfww 5(b) ( ) ( )( ) 3 3 1 3 3 0 − + = x x M1 For recognition of a correct quadratic in 3x and attempt to factorise or use quadratic formula 1 3 3 = x 1 = − x 2 M1 dep for a correct attempt to solve 3 , 0 = > x k k A1 for one solution only, must be from a correct solution. 5(c) 2 1 log 2 log = y y or 2 1 log log 2 = y y or log 2 log 2 log = a y a y and 2 log log log 2 = a a y y B1 May be implied ( ) ( ) 2 4 log 2 4 log 2 1 0 − + = y y ( ) 2 2log 2 1 0 − = y , 1 log 2 2 = y or ( ) ( ) 2 2 2 log 4 log 4 0 − + = y y ( ) 2 2 log 2 0 − = y , 2 log 2 = y M1 For obtaining a 3 term quadratic equation in either log 2 y or 2 log y and solving to obtain log 2 = y k or 2 log = y k , may be implied or equivalent using an alternative base. or ( ) ( )( ) 2 log 4 log 2 log 4 log − a a a a y y ( ) 2 4 log 2 0 + = a ( ) 2 log 2log 2 0 − = a a y log 2log 2 = a a y 4 = y A1 nfww

8 Variables x and y are such that y Axb = , where A and b are constants. When lgy is plotted against lgx, a straight line graph passing through the points (0.61, 0.57) and (5.36, 4.37) is obtained. (a) Find the value of A and of b. [5] Using your values of A and b, find (b) the value of y when x 3 = , [2] (c) the value of x when y 3 = . [2]

8(a) lg lg lg = + y b x A B1 May be implied by subsequent work 4.37 5.36 lg = + b A 0.57 0.61 lg = + b A M1 For at least one correct equation 0.8 = b A1 ( ) lg 0.082 = A k 10 = k A M1 Dep for substitution to obtain lg = A k and hence A 1.21 = A A1 Alternative 1 lg lg lg = + y b x A (B1) May be implied by subsequent work Gradient = 4.37 0.57 5.36 0.61 − − (M1) 0.8 = b (A1) ( ) lg 0.082 = A k 10 = k A (M1) Dep for substitution into a correct equation to obtain lg = A k and hence A 1.21 = A (A1) Alternative 2 4.37 5.36 10 10 = × b A or 0.57 0.61 10 10 = × b A (B1) 3.8 4.75 = b (M1) For eliminating A correctly Must have B1. 0.8 = b (A1) ( ) 4.37 5.36 ( ) 10 − × = theirb A oe (M1) For a correct attempt to find A. Must have B1 A = 1.21 (A1) 8(b) ( ) 0.8 1.21 3 or lg 0.8lg3 0.082 = = + y y B1 FT for substitution into their equation y = awrt 2.9 B1 8(c) 0.8 3 1.21 or lg3 0.8lg 0.082 = = + x x B1 FT for substitution into their equation x = awrt 3.1 B1

Cambridge O Level Additional Mathematics · 4037

O-Level Additional Mathematics (4037) Logarithmic and exponential functions past paper questions

Revise O-Level Additional Mathematics (4037) — Logarithmic and exponential functions with real Cambridge past-paper questions grouped by topic, each shown with its marking scheme. Working one topic at a time targets exactly the skill each question tests.

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Logarithmic and exponential functions in O-Level Additional Mathematics 4037 exams, 2020–2025

Across 2020–2025, Logarithmic and exponential functions accounted for 76 questions worth 583 marks in Cambridge O Level Additional Mathematics (4037) papers.