Question 9 - A-Level Maths

AQA Paper 2 2022 June — Question 9 4 marks

Exam Board	AQA
Module	Paper 2 (Paper 2)
Year	2022
Session	June
Marks	4
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Laws of Logarithms
Type	Simplify or prove logarithmic identity
Difficulty	Moderate -0.8 This is a straightforward application of logarithm laws (power rule and quotient rule) requiring only 2-3 algebraic steps to reach the answer. The question explicitly tells students what form to reach and involves routine manipulation with no problem-solving insight needed, making it easier than average.
Spec	1.06f Laws of logarithms: addition, subtraction, power rules

9 Given that $$\log _ { 2 } x ^ { 3 } - \log _ { 2 } y ^ { 2 } = 9$$ show that $$x = A y ^ { p }$$ where $A$ is an integer and $p$ is a rational number. \includegraphics[max width=\textwidth, alt={}, center]{ad6590e8-6673-45ca-bef3-a14716978827-15_2488_1716_219_153}

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Question 9:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Uses a log (or index) law correctly on an algebraic term: $\log A \pm \log B$ or $n\log A$	B1 (AO 1.1b)
$\log_2 x^3 - \log_2 y^2 = 9 \Rightarrow \log_2 \frac{x^3}{y^2} = 9 \Rightarrow \frac{x^3}{y^2} = 2^9$	M1 (AO 1.1a)	Raises 2 to the power of both sides (removal of $\log_2$), or writes 9 as $9\log_2 2$ or $\log_2 512$
$x^3 = 2^9 y^2$	A1 (AO 1.1b)	Obtains correct equation without logs, or obtains $\log_2(x) = \log_2(8y^{2/3})$
$x = 8y^{\frac{2}{3}}$	R1 (AO 2.1)	Completes a reasoned argument to obtain $x = 8y^{\frac{2}{3}}$

Question 9 Total: 4 marks

## Question 9:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses a log (or index) law correctly on an algebraic term: $\log A \pm \log B$ or $n\log A$ | B1 (AO 1.1b) | |
| $\log_2 x^3 - \log_2 y^2 = 9 \Rightarrow \log_2 \frac{x^3}{y^2} = 9 \Rightarrow \frac{x^3}{y^2} = 2^9$ | M1 (AO 1.1a) | Raises 2 to the power of both sides (removal of $\log_2$), or writes 9 as $9\log_2 2$ or $\log_2 512$ |
| $x^3 = 2^9 y^2$ | A1 (AO 1.1b) | Obtains correct equation without logs, or obtains $\log_2(x) = \log_2(8y^{2/3})$ |
| $x = 8y^{\frac{2}{3}}$ | R1 (AO 2.1) | Completes a reasoned argument to obtain $x = 8y^{\frac{2}{3}}$ |

**Question 9 Total: 4 marks**

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9 Given that

$$\log _ { 2 } x ^ { 3 } - \log _ { 2 } y ^ { 2 } = 9$$

show that

$$x = A y ^ { p }$$

where $A$ is an integer and $p$ is a rational number.\\

\includegraphics[max width=\textwidth, alt={}, center]{ad6590e8-6673-45ca-bef3-a14716978827-15_2488_1716_219_153}

\hfill \mbox{\textit{AQA Paper 2 2022 Q9 [4]}}

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