| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2015 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Solve by showing reduces to polynomial |
| Difficulty | Standard +0.3 This is a standard logarithm manipulation question requiring systematic application of log laws (power rule, product rule, converting log to exponential form) to reduce to a quadratic equation, then solving with domain checking. While multi-step, each technique is routine for C1/C2 level with no novel insight required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.06f Laws of logarithms: addition, subtraction, power rules |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use or state \(2\log_4(2x+3) = \log_4(2x+3)^2\) | M1 | Uses power law for logs |
| Use or state \(\log_4 4 = 1\) or \(4^1 = 4\) | M1 | Connects 1 with 4 correctly |
| Use or state \(\log_4 x + \log_4(2x-1) = \log_4 x(2x-1)\) or \(\log_4(2x+3)^2 - \log_4 x = \log_4 \frac{(2x+3)^2}{x}\) | M1 | Uses addition or subtraction law correctly |
| \((2x+3)^2 = 4x(2x-1)\) or equivalent including correct rational equations | A1 | Correct equation (unsimplified) after correct work, e.g. \(\frac{(2x+3)^2}{x(2x-1)} = 4\) |
| \(4x^2 - 16x - 9 = 0\) | A1* | Obtains printed answer correctly (given answer so needs previous A mark and correct expansion) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((2x+1)(2x-9) = 0\) | M1 | Uses solution of their quadratic or of printed quadratic |
| \(x = -\frac{1}{2}\) or \(x = \frac{9}{2}\) | A1 | \(x = 4.5\) and discards \(x = -0.5\); giving \(x = -\frac{1}{2}, \frac{9}{2}\) is A0 |
# Question 6:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use or state $2\log_4(2x+3) = \log_4(2x+3)^2$ | M1 | Uses power law for logs |
| Use or state $\log_4 4 = 1$ or $4^1 = 4$ | M1 | Connects 1 with 4 correctly |
| Use or state $\log_4 x + \log_4(2x-1) = \log_4 x(2x-1)$ or $\log_4(2x+3)^2 - \log_4 x = \log_4 \frac{(2x+3)^2}{x}$ | M1 | Uses addition or subtraction law correctly |
| $(2x+3)^2 = 4x(2x-1)$ or equivalent including correct rational equations | A1 | Correct equation (unsimplified) after correct work, e.g. $\frac{(2x+3)^2}{x(2x-1)} = 4$ |
| $4x^2 - 16x - 9 = 0$ | A1* | Obtains printed answer correctly (given answer so needs previous A mark and correct expansion) |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(2x+1)(2x-9) = 0$ | M1 | Uses solution of their quadratic or of printed quadratic |
| $x = -\frac{1}{2}$ or $x = \frac{9}{2}$ | A1 | $x = 4.5$ and discards $x = -0.5$; giving $x = -\frac{1}{2}, \frac{9}{2}$ is A0 |
---6. Given that
$$2 \log _ { 4 } ( 2 x + 3 ) = 1 + \log _ { 4 } x + \log _ { 4 } ( 2 x - 1 ) , \quad x > \frac { 1 } { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item show that
$$4 x ^ { 2 } - 16 x - 9 = 0$$
\item Hence solve the equation
$$2 \log _ { 4 } ( 2 x + 3 ) = 1 + \log _ { 4 } x + \log _ { 4 } ( 2 x - 1 ) , \quad x > \frac { 1 } { 2 }$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2015 Q6 [7]}}