CAIE P3 2023 March — Question 1 3 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2023
SessionMarch
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLaws of Logarithms
TypeExpress y in terms of x (requires exponentiating both sides)
DifficultyModerate -0.5 This is a straightforward application of logarithm laws (difference of logs becomes division) followed by basic algebraic manipulation to make y the subject. It requires only routine techniques with no problem-solving insight, making it slightly easier than average but not trivial since it involves multiple steps including dealing with a linear equation after exponentiating.
Spec1.06f Laws of logarithms: addition, subtraction, power rules
1 It is given that \(x = \ln ( 2 y - 3 ) - \ln ( y + 4 )\).
Express \(y\) in terms of \(x\).
Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
Use law of logarithm of a quotient or express \(x\) as \(\ln e^x\)M1 \(x = \ln[(2y-3)/(y+4)]\) or \(\ln e^x = \ln(2y-3) - \ln(y+4)\)
Remove logarithms and obtain a correct equation e.g. \(e^x = \dfrac{2y-3}{y+4}\)A1
Obtain answer \(y = \dfrac{3+4e^x}{2-e^x}\)A1 OE ISW
Total: 3 
**Question 1:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Use law of logarithm of a quotient or express $x$ as $\ln e^x$ | **M1** | $x = \ln[(2y-3)/(y+4)]$ or $\ln e^x = \ln(2y-3) - \ln(y+4)$ |
| Remove logarithms and obtain a correct equation e.g. $e^x = \dfrac{2y-3}{y+4}$ | **A1** | |
| Obtain answer $y = \dfrac{3+4e^x}{2-e^x}$ | **A1** | OE ISW |
| **Total: 3** | | |

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1 It is given that $x = \ln ( 2 y - 3 ) - \ln ( y + 4 )$.\\
Express $y$ in terms of $x$.\\

\hfill \mbox{\textit{CAIE P3 2023 Q1 [3]}}
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