| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Two unrelated log parts: one non-log algebraic part |
| Difficulty | Moderate -0.3 This is a straightforward application of logarithm laws (combining logs, converting between log and exponential form, and change of base). Part (a) requires basic manipulation to isolate k, while part (b) involves chaining logarithm relationships but follows a standard pattern. Both parts are routine C2-level exercises requiring recall and direct application rather than problem-solving insight, making it slightly easier than average. |
| Spec | 1.06c Logarithm definition: log_a(x) as inverse of a^x1.06f Laws of logarithms: addition, subtraction, power rules |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2\log_k x - \log_k 5 = 1\) | ||
| \(\log_k x^2 - \log_k 5 = 1\) | M1 | Using power law |
| \(\log_k\frac{x^2}{5} = 1\) | M1 | Using subtraction law |
| \(k^1 = \frac{x^2}{5}\) | M1 | Converting to exponential form |
| \(k = \frac{x^2}{5}\) | A1 | Correct answer, no logarithms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\log_a y = \frac{3}{2}\), so \(y = a^{3/2}\) | B1 | Converting first equation |
| \(\log_4 a = b+2\), so \(a = 4^{b+2}\) | M1 | Converting second equation |
| \(y = \left(4^{b+2}\right)^{3/2} = 4^{\frac{3}{2}(b+2)} = 2^{3(b+2)} = 2^{3b+6}\) | A1 | So \(p = 3b+6\) (or equivalent) |
# Question 8:
## Part (a) – Express $k$ in terms of $x$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\log_k x - \log_k 5 = 1$ | | |
| $\log_k x^2 - \log_k 5 = 1$ | M1 | Using power law |
| $\log_k\frac{x^2}{5} = 1$ | M1 | Using subtraction law |
| $k^1 = \frac{x^2}{5}$ | M1 | Converting to exponential form |
| $k = \frac{x^2}{5}$ | A1 | Correct answer, no logarithms |
## Part (b) – Show $y = 2^p$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\log_a y = \frac{3}{2}$, so $y = a^{3/2}$ | B1 | Converting first equation |
| $\log_4 a = b+2$, so $a = 4^{b+2}$ | M1 | Converting second equation |
| $y = \left(4^{b+2}\right)^{3/2} = 4^{\frac{3}{2}(b+2)} = 2^{3(b+2)} = 2^{3b+6}$ | A1 | So $p = 3b+6$ (or equivalent) |
---8
\begin{enumerate}[label=(\alph*)]
\item Given that $2 \log _ { k } x - \log _ { k } 5 = 1$, express $k$ in terms of $x$. Give your answer in a form not involving logarithms.
\item Given that $\log _ { a } y = \frac { 3 } { 2 }$ and that $\log _ { 4 } a = b + 2$, show that $y = 2 ^ { p }$, where $p$ is an expression in terms of $b$.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{1c06ba04-575c-4eb8-b4aa-0a7510838cd2-09_2102_1717_605_150}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2011 Q8 [7]}}