| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Two unrelated log parts: one non-log algebraic part |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing basic logarithm laws (power rule, subtraction rule) with routine algebraic manipulation. Part (a) requires simple application of the power law, parts (b) and (c) are 'show that' questions with clear targets requiring only direct substitution and standard log rules. No problem-solving insight needed, just mechanical application of learned techniques. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(3\log_x x = \log_x 8 \Rightarrow \log_x x^3 = \log_x 8\) | M1 |
| \(x^3 = 8 \Rightarrow x = 2\) | A1 | |
| Total: 2 | ||
| (b) | \(3\log_6 6 - \log_6 8 = \log_6 6^3 - \log_6 8\) | M1 |
| \(= \log_6\frac{6^3}{8}\) | M1 | Correct use of a different log law |
| \(= \log_6\frac{216}{8} = \log_6 27\) | A1 | CSO AG (be convinced) |
| (c)(i) | \(\{p =\}3\log_{10}3 - \log_{10}8\) | M1 |
| \(p = \log_{10}\frac{3^3}{8} = \log_{10}\frac{27}{8}\) | A1 | AG (be convinced) |
| (ii) | Gradient of \(PQ = \frac{q - p}{6 - 3}\) | M1 |
| \(....... = \frac{\log_{10}27 - \log_{10}\frac{27}{8}}{3}\) | A1 | Any correct exact form |
| \(....... = \frac{1}{3}\log_{10}(27 \div \frac{27}{8})\) | m1 | Correct use of log law |
| Gradient \(= \frac{1}{3}\log_{10}8 = \log_{10}2\) | A1 | AG (be convinced) |
| Total: 11 | ||
| TOTAL: 75 |
(a) | $3\log_x x = \log_x 8 \Rightarrow \log_x x^3 = \log_x 8$ | M1 | OE use of the log law |
| $x^3 = 8 \Rightarrow x = 2$ | A1 | |
| | **Total: 2** | |
(b) | $3\log_6 6 - \log_6 8 = \log_6 6^3 - \log_6 8$ | M1 | Correct use of one log law |
| $= \log_6\frac{6^3}{8}$ | M1 | Correct use of a different log law |
| $= \log_6\frac{216}{8} = \log_6 27$ | A1 | CSO AG (be convinced) |
(c)(i) | $\{p =\}3\log_{10}3 - \log_{10}8$ | M1 | Substitute $x = 3$ |
| $p = \log_{10}\frac{3^3}{8} = \log_{10}\frac{27}{8}$ | A1 | AG (be convinced) |
(ii) | Gradient of $PQ = \frac{q - p}{6 - 3}$ | M1 | used $\frac{\text{difference in y-coords}}{\text{difference in x-coords}}$ |
| | | |
| $....... = \frac{\log_{10}27 - \log_{10}\frac{27}{8}}{3}$ | A1 | Any correct exact form |
| | | |
| $....... = \frac{1}{3}\log_{10}(27 \div \frac{27}{8})$ | m1 | Correct use of log law |
| | | |
| Gradient $= \frac{1}{3}\log_{10}8 = \log_{10}2$ | A1 | AG (be convinced) |
| | **Total: 11** | |
| | **TOTAL: 75** | |9
\begin{enumerate}[label=(\alph*)]
\item Solve the equation $3 \log _ { a } x = \log _ { a } 8$.
\item Show that
$$3 \log _ { a } 6 - \log _ { a } 8 = \log _ { a } 27$$
\item \begin{enumerate}[label=(\roman*)]
\item The point $P ( 3 , p )$ lies on the curve $y = 3 \log _ { 10 } x - \log _ { 10 } 8$.
Show that $p = \log _ { 10 } \left( \frac { 27 } { 8 } \right)$.
\item The point $Q ( 6 , q )$ also lies on the curve $y = 3 \log _ { 10 } x - \log _ { 10 } 8$.
Show that the gradient of the line $P Q$ is $\log _ { 10 } 2$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2007 Q9 [11]}}