| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2009 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Combine logs into single logarithm |
| Difficulty | Moderate -0.8 This is a straightforward C2 question testing basic logarithm laws (addition, subtraction, multiplication) with routine algebraic manipulation. Part (a) requires direct application of standard rules, part (b) is a standard 'solve using logs' question, and part (c) involves simple base conversion and combining expressions. All parts are textbook exercises requiring recall and mechanical application rather than problem-solving or insight. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| \(\log_k 40\) | B1 | Accept '\(k = 40\)'. 1 mark total |
| Answer | Marks | Guidance |
|---|---|---|
| \(\log_k 8\) | B1 | Accept '\(k = 8\)'. 1 mark total |
| Answer | Marks | Guidance |
|---|---|---|
| \(\log_k 125\) | B1 | Accept '\(k = 125\)' but not '\(k = 5^3\)'. 1 mark total |
| Answer | Marks | Guidance |
|---|---|---|
| \(\log_{10}[(1.5)^{3x}] = \log_{10} 7.5\) | M1 | Correct statement having taken logs of both sides of \((1.5)^{3x} = 7.5\) OE PI or \(3x = \log_{1.5} 7.5\) seen |
| \(3x \log_{10} 1.5 = \log_{10} 7.5\) | m1 | \(\log 1.5^{3x} = 3x \log 1.5\) OE |
| \(x = \frac{\lg 7.5}{3\lg 1.5} = 1.65645\ldots = 1.656 \text{ to 3dp}\) | A1 | Both method marks must have been awarded with clear use of logarithms seen. 3 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(\log_2 p = m \Rightarrow p = 2^m\); \(\log_8 q = n \Rightarrow q = 8^n\) | M1 | Either \(p = 2^m\) or \(q = 8^n\) seen or used |
| \(p = 2^m\) and \(q = 2^{3n}\) | m1 | Writing \(8^n = 2^{3n}\) and having \(p = 2^m\) |
| \(pq = 2^m \times (2^3)^n = 2^m \times 2^{3n}\) so \(pq = 2^{m+3n}\) | A1 | Accept \(y = m + 3n\). 3 marks total |
**6(a)(i)**
$\log_k 40$ | B1 | Accept '$k = 40$'. 1 mark total
**6(a)(ii)**
$\log_k 8$ | B1 | Accept '$k = 8$'. 1 mark total
**6(a)(iii)**
$\log_k 125$ | B1 | Accept '$k = 125$' but not '$k = 5^3$'. 1 mark total
**6(b)**
$\log_{10}[(1.5)^{3x}] = \log_{10} 7.5$ | M1 | Correct statement having taken logs of both sides of $(1.5)^{3x} = 7.5$ OE PI or $3x = \log_{1.5} 7.5$ seen
$3x \log_{10} 1.5 = \log_{10} 7.5$ | m1 | $\log 1.5^{3x} = 3x \log 1.5$ OE
$x = \frac{\lg 7.5}{3\lg 1.5} = 1.65645\ldots = 1.656 \text{ to 3dp}$ | A1 | Both method marks must have been awarded with clear use of logarithms seen. 3 marks total
**6(c)**
$\log_2 p = m \Rightarrow p = 2^m$; $\log_8 q = n \Rightarrow q = 8^n$ | M1 | Either $p = 2^m$ or $q = 8^n$ seen or used
$p = 2^m$ and $q = 2^{3n}$ | m1 | Writing $8^n = 2^{3n}$ and having $p = 2^m$
$pq = 2^m \times (2^3)^n = 2^m \times 2^{3n}$ so $pq = 2^{m+3n}$ | A1 | Accept $y = m + 3n$. 3 marks total
**Total for Q6: 9 marks**
---6
\begin{enumerate}[label=(\alph*)]
\item Write each of the following in the form $\log _ { a } k$, where $k$ is an integer:
\begin{enumerate}[label=(\roman*)]
\item $\log _ { a } 4 + \log _ { a } 10$;
\item $\log _ { a } 16 - \log _ { a } 2$;
\item $3 \log _ { a } 5$.
\end{enumerate}\item Use logarithms to solve the equation $( 1.5 ) ^ { 3 x } = 7.5$, giving your value of $x$ to three decimal places.
\item Given that $\log _ { 2 } p = m$ and $\log _ { 8 } q = n$, express $p q$ in the form $2 ^ { y }$, where $y$ is an expression in $m$ and $n$.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2009 Q6 [9]}}