O-LevelAdditional MathematicsLogarithmic and exponential functions4037/21 · May/June 2021 · Q88 Marks

In this question, a, b, c and d are positive constants.

(a)(i)It is given that y = log(x+3)+loga (2x−1). Explain why x must be greater than 1.[1]
(a)(ii)Find the exact solution of the equation log 6 / log (y+3) = 2.[3]
(b)Write the expression log9+(logab) (log√9a) in the form c+dlog 9, where c and d are integers.[4]

Mark scheme

Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.

[B1] Valid explanation e.g. This ensures the argument of both logarithms is greater than 0
[B1] log_a 6 = log_a (y+3)² oe
[M1] (y+3)² = 6
[A1] y = -3+√6 oe only final answer
[M3] Within a complete expression:
[M2] for 2 or 3 correct steps within complete expression
[M1] for 1 correct step within complete expression
[M3 Criterion] Correct change of base to a: log_a 9a / log_a √b
[M3 Criterion] Correct use of power law: log_a √b = 1/2 log_a b
[M3 Criterion] Correct use of addition/multiplication law: log_a 9a = log_a 9 + log_a a
[M3 Criterion] Correct use of log_a a = 1
[A1] leading to 2+3log_a 9. nfww

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