AQA AS Paper 2 2023 June — Question 4 5 marks

Exam BoardAQA
ModuleAS Paper 2 (AS Paper 2)
Year2023
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLaws of Logarithms
TypeSolve log equation with domain restrictions
DifficultyModerate -0.3 This is a straightforward logarithm equation requiring standard log laws (sum and multiple rules), leading to a quadratic equation. The 'fully justify' requirement adds minimal difficulty as students just need to check their solution is in the domain (x > 1). Slightly easier than average due to being a routine application of well-practiced techniques with no conceptual surprises.
Spec1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b

Find the exact solution of the equation \(\ln(x + 1) + \ln(x - 1) = \ln 15 - 2\ln 7\) Fully justify your answer. [5 marks]

Question 4:
AnswerMarks
4Obtains ln(x + 1)(x – 1)
Or
ln(x2 – 1)
PI by correct equation in x2
AnswerMarks Guidance
Condone missing brackets1.1b B1
= ln15 – ln49
15
= ln
49
15
x2 – 1 =
49
64
x2 =
49
8
x =
7
8
x cannot be – because the ln
7
functions would not exist with this
value
Obtains ln 49 or ln72 for 2 ln 7
AnswerMarks Guidance
PI by correct equation in x21.1b B1
Applies subtraction rule for ln to
right-hand side
AnswerMarks Guidance
PI by correct equation in x21.1a M1
Obtains correct exact value for
x2
AnswerMarks Guidance
PI1.1b A1
8
Explains why x = - is not a
7
valid solution.
AnswerMarks Guidance
Must refer to ln(-ve)2.4 E1
Question 4 Total5
QMarking instructions AO
Question 4:
4 | Obtains ln(x + 1)(x – 1)
Or
ln(x2 – 1)
PI by correct equation in x2
Condone missing brackets | 1.1b | B1 | ln(x + 1)(x – 1)
= ln15 – ln49
15
= ln
49
15
x2 – 1 =
49
64
x2 =
49
8
x =
7
8
x cannot be – because the ln
7
functions would not exist with this
value
Obtains ln 49 or ln72 for 2 ln 7
PI by correct equation in x2 | 1.1b | B1
Applies subtraction rule for ln to
right-hand side
PI by correct equation in x2 | 1.1a | M1
Obtains correct exact value for
x2
PI | 1.1b | A1
8
Explains why x = - is not a
7
valid solution.
Must refer to ln(-ve) | 2.4 | E1
Question 4 Total | 5
Q | Marking instructions | AO | Marks | Typical solution
Find the exact solution of the equation $\ln(x + 1) + \ln(x - 1) = \ln 15 - 2\ln 7$

Fully justify your answer.
[5 marks]

\hfill \mbox{\textit{AQA AS Paper 2 2023 Q4 [5]}}