| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Simultaneous equations with logarithms |
| Difficulty | Standard +0.3 This is a straightforward application of logarithm laws (power rule, addition/subtraction rules) and change of base, followed by routine algebraic substitution. Part (a) is guided ('show that'), and part (b) requires only standard manipulation of the two equations to find x and y. Slightly above average due to the two-equation system and logarithm base conversions, but still a standard textbook exercise with no novel insight required. |
| Spec | 1.06c Logarithm definition: log_a(x) as inverse of a^x1.06f Laws of logarithms: addition, subtraction, power rules |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(2\log_2 y = 5 - \log_2 x \Rightarrow \log_2 y^2 = 5 - \log_2 x\) | M1 | Uses one correct log law, e.g. index law to write \(2\log_2 y = \log_2 y^2\), or writes 5 as \(\log_2 32\) |
| \(\log_2 y^2 = \log_2 32 - \log_2 x \Rightarrow \log_2 y^2 = \log_2\left(\frac{32}{x}\right)\) | M1 | Uses two correct log laws; award for \(\log_2 y^2 = \log_2(32) - \log_2 x\) or \(\log_2 x + \log_2 y^2 = 5 \Rightarrow \log_2 xy^2 = 5\) |
| \(y^2 = \frac{32}{x}\) | A1 | Proceeds correctly to \(y^2 = \frac{32}{x}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\log_x y = -3 \Rightarrow y = x^{-3}\) | M1 | Undoes the log in the second equation; may appear later in solution |
| Sub \(y = x^{-3}\) into \(y^2 = \frac{32}{x} \Rightarrow x^{-6} = \frac{32}{x} \Rightarrow x^5 = \frac{1}{32} \Rightarrow x = \frac{1}{2}\) | M1, A1 | Combines both equations to form single equation in one variable; \(x = \frac{1}{2}\) or \(y = 8\). Condone \(y = \pm 8\) for this mark |
| Sub \(x = \frac{1}{2}\) into either equation \(\Rightarrow y = 8\) | M1, A1 | Substitutes \(x = \frac{1}{2}\) into equation to find \(y\); \(x = \frac{1}{2}\) and \(y = 8\) only. Note \(x = \frac{1}{2}\) and \(y = \pm 8\) is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Sub \(y^2 = \frac{32}{x}\) into \(\log_x y = -3 \Rightarrow \log_x \sqrt{\frac{32}{x}} = -3\) | 2nd M1 | |
| \(\Rightarrow \sqrt{\frac{32}{x}} = x^{-3}\) | 1st M1 | |
| \(\Rightarrow x^5 = \frac{1}{32} \Rightarrow x = \frac{1}{2}\) | A1 |
## Question 13:
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $2\log_2 y = 5 - \log_2 x \Rightarrow \log_2 y^2 = 5 - \log_2 x$ | M1 | Uses one correct log law, e.g. index law to write $2\log_2 y = \log_2 y^2$, or writes 5 as $\log_2 32$ |
| $\log_2 y^2 = \log_2 32 - \log_2 x \Rightarrow \log_2 y^2 = \log_2\left(\frac{32}{x}\right)$ | M1 | Uses two correct log laws; award for $\log_2 y^2 = \log_2(32) - \log_2 x$ or $\log_2 x + \log_2 y^2 = 5 \Rightarrow \log_2 xy^2 = 5$ |
| $y^2 = \frac{32}{x}$ | A1 | Proceeds correctly to $y^2 = \frac{32}{x}$ |
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $\log_x y = -3 \Rightarrow y = x^{-3}$ | M1 | Undoes the log in the second equation; may appear later in solution |
| Sub $y = x^{-3}$ into $y^2 = \frac{32}{x} \Rightarrow x^{-6} = \frac{32}{x} \Rightarrow x^5 = \frac{1}{32} \Rightarrow x = \frac{1}{2}$ | M1, A1 | Combines both equations to form single equation in one variable; $x = \frac{1}{2}$ or $y = 8$. Condone $y = \pm 8$ for this mark |
| Sub $x = \frac{1}{2}$ into either equation $\Rightarrow y = 8$ | M1, A1 | Substitutes $x = \frac{1}{2}$ into equation to find $y$; $x = \frac{1}{2}$ and $y = 8$ only. Note $x = \frac{1}{2}$ and $y = \pm 8$ is A0 |
**Alt (b):**
| Working | Mark | Guidance |
|---------|------|----------|
| Sub $y^2 = \frac{32}{x}$ into $\log_x y = -3 \Rightarrow \log_x \sqrt{\frac{32}{x}} = -3$ | 2nd M1 | |
| $\Rightarrow \sqrt{\frac{32}{x}} = x^{-3}$ | 1st M1 | |
| $\Rightarrow x^5 = \frac{1}{32} \Rightarrow x = \frac{1}{2}$ | A1 | |
---\begin{enumerate}
\item (a) Show that the equation
\end{enumerate}
$$2 \log _ { 2 } y = 5 - \log _ { 2 } x \quad x > 0 , y > 0$$
may be written in the form $y ^ { 2 } = \frac { k } { x }$ where $k$ is a constant to be found.\\
(b) Hence, or otherwise, solve the simultaneous equations
$$\begin{gathered}
2 \log _ { 2 } y = 5 - \log _ { 2 } x \\
\log _ { x } y = - 3
\end{gathered}$$
for $x > 0 , y > 0$\\
\begin{center}
\end{center}
\hfill \mbox{\textit{Edexcel C12 2016 Q13 [8]}}