AQA AS Paper 2 2019 June — Question 4 4 marks

Exam BoardAQA
ModuleAS Paper 2 (AS Paper 2)
Year2019
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLaws of Logarithms
TypeSimplify or prove logarithmic identity
DifficultyModerate -0.8 This is a straightforward application of standard logarithm laws (product, quotient, and power rules) with no conceptual challenges. The algebraic manipulation is routine and mechanical, requiring only direct application of log₁₀(a/b) = log₁₀a - log₁₀b, log₁₀(ab) = log₁₀a + log₁₀b, and log₁₀(xⁿ) = n log₁₀x, making it easier than a typical A-level question.
Spec1.01a Proof: structure of mathematical proof and logical steps1.06f Laws of logarithms: addition, subtraction, power rules

4 Show that, for \(x > 0\) $$\log _ { 10 } \frac { x ^ { 4 } } { 100 } + \log _ { 10 } 9 x - \log _ { 10 } x ^ { 3 } \equiv 2 \left( - 1 + \log _ { 10 } 3 x \right)$$

Question 4:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Uses a law of logarithms correctly (Multiplication / Division / Power) e.g. \(= 4\log_{10} x - \log_{10} 100 + \log_{10} 9 + \log_{10} x - 3\log_{10} x\)M1 (AO 1.1a) NB any attempt to show result with numerical values scores 0/4
Uses a different law of logarithms correctly e.g. \(= 2(-\log_{10} 10 + \log_{10} 3 + \log_{10} x)\)M1 (AO 1.1a) NB \(\log_{10}\frac{9x^2}{100}\) OE scores M1M1
Obtains at least two terms equivalent to \(-2\log_{10} 10 + 2\log_{10} 3 + 2\log_{10} x\)A1 (AO 1.1b)
Completes rigorous argument to obtain \(2(-1 + \log_{10} 3x)\) correctly with Base 10 identified in final answerR1 (AO 2.1) AG
## Question 4:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses a law of logarithms correctly (Multiplication / Division / Power) e.g. $= 4\log_{10} x - \log_{10} 100 + \log_{10} 9 + \log_{10} x - 3\log_{10} x$ | M1 (AO 1.1a) | NB any attempt to show result with numerical values scores 0/4 |
| Uses a different law of logarithms correctly e.g. $= 2(-\log_{10} 10 + \log_{10} 3 + \log_{10} x)$ | M1 (AO 1.1a) | NB $\log_{10}\frac{9x^2}{100}$ OE scores M1M1 |
| Obtains at least two terms equivalent to $-2\log_{10} 10 + 2\log_{10} 3 + 2\log_{10} x$ | A1 (AO 1.1b) | |
| Completes rigorous argument to obtain $2(-1 + \log_{10} 3x)$ correctly with Base 10 identified in final answer | R1 (AO 2.1) | AG |

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4 Show that, for $x > 0$

$$\log _ { 10 } \frac { x ^ { 4 } } { 100 } + \log _ { 10 } 9 x - \log _ { 10 } x ^ { 3 } \equiv 2 \left( - 1 + \log _ { 10 } 3 x \right)$$

\hfill \mbox{\textit{AQA AS Paper 2 2019 Q4 [4]}}