| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Solve by showing reduces to polynomial |
| Difficulty | Standard +0.3 Part (a) requires routine manipulation of logarithm laws and change of base (converting log₂₇ to log₃), which is standard C2 content. Part (b) involves combining logarithms to form a quadratic equation and checking validity of solutions—a typical multi-step logarithm problem. Both parts are straightforward applications of techniques with no novel insight required, making this 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 |
|---|---|---|
| (a) | \(c = 3^m, \quad d = 27^n\) | M1 |
| (a) | \(d^2 = 3^{6n}\) seen or used | A1 |
| (a) | \(\sqrt{c} = 3^{0.5m}\) seen or used | A1 |
| (a) | \(\frac{\sqrt{c}}{d^2} = 3^{\frac{m}{2}-6n}\) | A1 |
| Altn | \(\frac{1}{2}\log_c c - 2\log_3 d = y\log_3 3\) | (M1) |
| \(\log_3 d = \frac{\log_{27} d}{\log_{27} 3}\) | (A1) | \(\log_3 d = \frac{\log_{27} d}{\log_{27} 3}\) or \(\log_{27} c = \frac{\log_3 c}{\log_3 27}\) seen or used |
| \(\log_3 3 = \frac{1}{3}\) | (A1) | \(\log_{27} 3 = \frac{1}{3}\) or \(\log_3 27 = 3\) seen or used |
| \(y = \frac{1}{2}m - 6n\) | (A1) | (4) |
| (b) | \(1 = \log_4 4\) | B1 |
| (b) | \(\log_4(2x + 3)(2x + 15) = 1 + \log_4(14x + 5)\) | M1 |
| (b) | \(\log_4(2x + 3)(2x + 15) = \log_4 4(14x + 5)\) | A1 |
| (b) | \((2x + 3)(2x + 15) = 4(14x + 5)\) | |
| (b) | \(4x^2 + 36x + 45 = 56x + 20\) | |
| (b) | \(4x^2 - 20x + 25 = 0; \quad (2x - 5)^2 = 0\) | A1 |
| (b) | Only one solution 2.5 | |
| Total | 8 |
(a) | $c = 3^m, \quad d = 27^n$ | M1 | Either $c = 3^m$ or $d = 27^n$ seen or used
(a) | $d^2 = 3^{6n}$ seen or used | A1 | Either $d = 3^{3n}$ or $d^2 = 3^{6n}$ seen or used
(a) | $\sqrt{c} = 3^{0.5m}$ seen or used | A1 |
(a) | $\frac{\sqrt{c}}{d^2} = 3^{\frac{m}{2}-6n}$ | A1 | 4 marks | OE expression for y in terms of m and n.
| **Altn** | $\frac{1}{2}\log_c c - 2\log_3 d = y\log_3 3$ | (M1) | A correct expression in y in terms of logs to base 3 or base 27 where no further log laws are required
| | $\log_3 d = \frac{\log_{27} d}{\log_{27} 3}$ | (A1) | $\log_3 d = \frac{\log_{27} d}{\log_{27} 3}$ or $\log_{27} c = \frac{\log_3 c}{\log_3 27}$ seen or used
| | $\log_3 3 = \frac{1}{3}$ | (A1) | $\log_{27} 3 = \frac{1}{3}$ or $\log_3 27 = 3$ seen or used
| | $y = \frac{1}{2}m - 6n$ | (A1) | (4) | Correct expression for y in terms of m and n OE eg $\frac{\sqrt{c}}{d^2} = 3^{\frac{m}{2}-6n}$
(b) | $1 = \log_4 4$ | B1 | $1 = \log_4 4$ seen or used at any stage.
(b) | $\log_4(2x + 3)(2x + 15) = 1 + \log_4(14x + 5)$ | M1 | Applying a log law correctly to two correct log terms. [Condone missing base]. NB: Lots of other possibilities after correct rearrangements!
(b) | $\log_4(2x + 3)(2x + 15) = \log_4 4(14x + 5)$ | A1 | PI by $(2x + 3)(2x + 15) = 4(14x + 5)$ OE with no errors seen
(b) | $(2x + 3)(2x + 15) = 4(14x + 5)$ | |
(b) | $4x^2 + 36x + 45 = 56x + 20$ | |
(b) | $4x^2 - 20x + 25 = 0; \quad (2x - 5)^2 = 0$ | A1 | 4 marks | Must include statement and correct value
(b) | Only one solution 2.5 | |
| **Total** | **8** |
**Note on (b):** ......... $4x^2 - 20x + 25 = 0, \quad b^2 - 4ac = 400 - 400 = 0$, only one soln (which is $-\frac{b}{2a}$) **2.5**9
\begin{enumerate}[label=(\alph*)]
\item Given that $\log _ { 3 } c = m$ and $\log _ { 27 } d = n$, express $\frac { \sqrt { c } } { d ^ { 2 } }$ in the form $3 ^ { y }$, where $y$ is an expression in terms of $m$ and $n$.
\item Show that the equation
$$\log _ { 4 } ( 2 x + 3 ) + \log _ { 4 } ( 2 x + 15 ) = 1 + \log _ { 4 } ( 14 x + 5 )$$
has only one solution and state its value.\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2016 Q9 [8]}}