Question 5 - A-Level Maths

AQA C2 2005 January — Question 5 7 marks

Exam Board	AQA
Module	C2 (Core Mathematics 2)
Year	2005
Session	January
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Laws of Logarithms
Type	Evaluate log expression using laws
Difficulty	Easy -1.2 This is a straightforward application of basic logarithm laws (power rule and subtraction rule) with no problem-solving required. Part (a) is routine algebraic manipulation, and part (b) tests only direct recall of fundamental logarithm properties. Easier than average for A-level.
Spec	1.06c Logarithm definition: log_a(x) as inverse of a^x 1.06f Laws of logarithms: addition, subtraction, power rules

Given that $$\log _ { a } x = 3 \log _ { a } 6 - \log _ { a } 8$$ where $a$ is a positive constant, show that $x = 27$.
Write down the value of:
1. $\quad \log _ { 4 } 1$;
2. $\log _ { 4 } 4$;
3. $\log _ { 4 } 2$;
4. $\quad \log _ { 4 } 8$.

Show mark scheme Show mark scheme source

5(a)

Answer	Marks	Guidance
$\log_a x = \log_a 6^3 - \log_a 8$; $\log_a x = \log_a(6^3 \div 8)$; $x = 6^3 \div 8 = 27$	M1, M1, A1	A law of logs used correctly; A different law of logs used correctly; CSO AG (be convinced); ALT: $\log_a x = 3\log_a 6 - 3\log_a 2$ (M1); $\frac{1}{3}\log_a x = \log_a\frac{6}{2}$ (M1); $x^{\frac{1}{3}} = 3 \Rightarrow x = 27$ (A1 CSO)
Total for 5(a): 3 marks

5(b)(i)

Answer	Marks
$\log_4 1 = 0$	B1

5(b)(ii)

Answer	Marks
$\log_4 4 = 1$	B1

5(b)(iii)

Answer	Marks
$\log_4 2 = 0.5$	B1

5(b)(iv)

Answer	Marks	Guidance
$\log_4 8 = 1.5$	B1	SC in (b): For all four answers ¼, 1, ½; 2 give 0/4; otherwise mark each independently.
Total for 5(b): 4 marks
Total for Question 5: 7 marks

**5(a)**
| $\log_a x = \log_a 6^3 - \log_a 8$; $\log_a x = \log_a(6^3 \div 8)$; $x = 6^3 \div 8 = 27$ | M1, M1, A1 | A law of logs used correctly; A different law of logs used correctly; CSO AG (be convinced); **ALT**: $\log_a x = 3\log_a 6 - 3\log_a 2$ (M1); $\frac{1}{3}\log_a x = \log_a\frac{6}{2}$ (M1); $x^{\frac{1}{3}} = 3 \Rightarrow x = 27$ (A1 CSO) |
| Total for 5(a): 3 marks |

**5(b)(i)**
| $\log_4 1 = 0$ | B1 | |

**5(b)(ii)**
| $\log_4 4 = 1$ | B1 | |

**5(b)(iii)**
| $\log_4 2 = 0.5$ | B1 | |

**5(b)(iv)**
| $\log_4 8 = 1.5$ | B1 | SC in (b): For all four answers ¼, 1, ½; 2 give 0/4; otherwise mark each independently. |
| Total for 5(b): 4 marks |
| **Total for Question 5: 7 marks** |

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Show LaTeX source

5
\begin{enumerate}[label=(\alph*)]
\item Given that

$$\log _ { a } x = 3 \log _ { a } 6 - \log _ { a } 8$$

where $a$ is a positive constant, show that $x = 27$.
\item Write down the value of:
\begin{enumerate}[label=(\roman*)]
\item $\quad \log _ { 4 } 1$;
\item $\log _ { 4 } 4$;
\item $\log _ { 4 } 2$;
\item $\quad \log _ { 4 } 8$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA C2 2005 Q5 [7]}}

This paper (8 questions)

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Q1 8 Q2 10 Q3 6 Q4 9 Q5 7 Q6 10 Q7 11 Q8 12

Answer	Marks	Guidance
\(\log_a x = \log_a 6^3 - \log_a 8\); \(\log_a x = \log_a(6^3 \div 8)\); \(x = 6^3 \div 8 = 27\)	M1, M1, A1	A law of logs used correctly; A different law of logs used correctly; CSO AG (be convinced); ALT: \(\log_a x = 3\log_a 6 - 3\log_a 2\) (M1); \(\frac{1}{3}\log_a x = \log_a\frac{6}{2}\) (M1); \(x^{\frac{1}{3}} = 3 \Rightarrow x = 27\) (A1 CSO)
Total for 5(a): 3 marks

Answer	Marks	Guidance
\(\log_4 8 = 1.5\)	B1	SC in (b): For all four answers ¼, 1, ½; 2 give 0/4; otherwise mark each independently.
Total for 5(b): 4 marks
Total for Question 5: 7 marks