| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Solve by showing reduces to polynomial |
| Difficulty | Standard +0.3 Part (a) is trivial recall of logarithm definition. Part (b) requires applying log laws to form a quadratic equation and solving it, which is a standard C2 exercise with clear scaffolding ('by forming a quadratic equation'). The algebraic manipulation is straightforward once log laws are applied, making this slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.06c Logarithm definition: log_a(x) as inverse of a^x1.06d Natural logarithm: ln(x) function and properties1.06f Laws of logarithms: addition, subtraction, power rules |
| Answer | Marks | Guidance |
|---|---|---|
| \(b = a^c\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(2\log_2(x + 7) - \log_2(x + 5) = 3\) | M1 | A law of logs used correctly on a correct expression. |
| \(\log_2(x + 7)^2 - \log_2(x + 5) = 3\) | M1 | A further correct use of law of logs on a correct expression. |
| \(\log_2 \frac{(x + 7)^2}{x + 5} = 3\) | B1 | 3=3log₂2 or 3 = log₂ 2³ (= log₂ 8) seen or eg log f(x) = 3 ⇒ f(x)=2³ (=8) OE |
| \(\Rightarrow \frac{(x + 7)^2}{x + 5} = 2^3\) | A1 | Correct equation having eliminated logs and fractions |
| \(\Rightarrow x^2 + 14x + 49 = 8x + 40\) | ||
| \(\Rightarrow x^2 + 6x + 9 (= 0)\) | A1 | |
| Since \(6^2 - 4(1)(9) = 0\), (there is only) one value of \(x\) (which satisfies the given equation). | A1 | OE CSO Need conclusion which is also correctly justified |
| Total | 7 |
**8(a)**
$b = a^c$ | B1 | | 1
**8(b)**
$2\log_2(x + 7) - \log_2(x + 5) = 3$ | M1 | A law of logs used correctly on a correct expression.
$\log_2(x + 7)^2 - \log_2(x + 5) = 3$ | M1 | A further correct use of law of logs on a correct expression.
$\log_2 \frac{(x + 7)^2}{x + 5} = 3$ | B1 | 3=3log₂2 or 3 = log₂ 2³ (= log₂ 8) seen or eg log f(x) = 3 ⇒ f(x)=2³ (=8) OE
$\Rightarrow \frac{(x + 7)^2}{x + 5} = 2^3$ | A1 | Correct equation having eliminated logs and fractions
$\Rightarrow x^2 + 14x + 49 = 8x + 40$ | |
$\Rightarrow x^2 + 6x + 9 (= 0)$ | A1 |
Since $6^2 - 4(1)(9) = 0$, (there is only) one value of $x$ (which satisfies the given equation). | A1 | OE CSO Need conclusion which is also correctly justified | 6
Total | | 7
---8
\begin{enumerate}[label=(\alph*)]
\item Given that $\log _ { a } b = c$, express $b$ in terms of $a$ and $c$.
\item By forming a quadratic equation, show that there is only one value of $x$ which satisfies the equation $2 \log _ { 2 } ( x + 7 ) - \log _ { 2 } ( x + 5 ) = 3$.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2013 Q8 [7]}}