Question 4 - A-Level Maths

CAIE P3 2024 March — Question 4 4 marks

Exam Board	CAIE
Module	P3 (Pure Mathematics 3)
Year	2024
Session	March
Marks	4
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Laws of Logarithms
Type	Express log in terms of given variables
Difficulty	Moderate -0.5 This is a straightforward application of logarithm laws (quotient, product, and power rules) requiring algebraic manipulation to solve simultaneous equations. It's slightly easier than average as it follows a standard pattern with clear steps: extract ln(p) and ln(q) from the given equations, then substitute into the target expression.
Spec	1.06f Laws of logarithms: addition, subtraction, power rules

4 The positive numbers $p$ and $q$ are such that $$\ln \left( \frac { p } { q } \right) = a \text { and } \ln \left( q ^ { 2 } p \right) = b .$$ Express $\ln \left( p ^ { 7 } q \right)$ in terms of $a$ and $b$.

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Question 4:

Answer	Marks	Guidance
Answer	Mark	Guidance
Obtain $\ln p - \ln q = a$	B1	$\frac{p}{q} = e^a$
Obtain $\ln p + 2\ln q = b$	B1	$pq^2 = e^b$
Completed method to obtain $\ln(p^7 q)$	M1	E.g. $\ln q = \frac{b-a}{3}$, $\ln p = \frac{2a+b}{3}$ and attempt $7\ln p + \ln q$. All exponentials must be removed to obtain M1
Obtain $\frac{13a + 8b}{3}$	A1
Alternative: State $p^7 q = \left(\frac{p}{q}\right)^x (q^2 p)^y$	B1	Or $\ln p^7 q = x\ln\frac{p}{q} + y\ln q^2 p$
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 $1 = 2y - x$	A1	Leading to $x = \frac{13}{3}$, $y = \frac{8}{3}$
Evaluate $x \times a + y \times b$ to obtain $\frac{13a+8b}{3}$	A1

## Question 4:

| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain $\ln p - \ln q = a$ | B1 | $\frac{p}{q} = e^a$ |
| Obtain $\ln p + 2\ln q = b$ | B1 | $pq^2 = e^b$ |
| Completed method to obtain $\ln(p^7 q)$ | M1 | E.g. $\ln q = \frac{b-a}{3}$, $\ln p = \frac{2a+b}{3}$ and attempt $7\ln p + \ln q$. All exponentials must be removed to obtain M1 |
| Obtain $\frac{13a + 8b}{3}$ | A1 | |
| **Alternative:** State $p^7 q = \left(\frac{p}{q}\right)^x (q^2 p)^y$ | B1 | Or $\ln p^7 q = x\ln\frac{p}{q} + y\ln q^2 p$ |
| 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 $1 = 2y - x$ | A1 | Leading to $x = \frac{13}{3}$, $y = \frac{8}{3}$ |
| Evaluate $x \times a + y \times b$ to obtain $\frac{13a+8b}{3}$ | A1 | |

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4 The positive numbers $p$ and $q$ are such that

$$\ln \left( \frac { p } { q } \right) = a \text { and } \ln \left( q ^ { 2 } p \right) = b .$$

Express $\ln \left( p ^ { 7 } q \right)$ in terms of $a$ and $b$.\\

\hfill \mbox{\textit{CAIE P3 2024 Q4 [4]}}

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