| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2017 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Solve using substitution or auxiliary variable |
| Difficulty | Moderate -0.8 This is a straightforward application of logarithm laws with clear scaffolding. Part (a) involves routine manipulation using log(a/b) = log(a) - log(b) and log(√x) = ½log(x), while part (b) is a simple linear equation after substitution. The question explicitly guides students through the method with 'hence', making it easier than average for A-level. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\log_3\left(\frac{x}{9}\right) = \log_3 x - \log_3 9 = y - 2\) | M1A1 | M1: \(\log_3\left(\frac{x}{9}\right) = \log_3 x - \log_3 9\) or \(= \log_3 x + \log_3\frac{1}{9}\). Correct use of subtraction/addition rule. Ignore presence/absence of base and any spurious "\(=0\)"; A1: \(y-2\). Answer left as \(\log_3 3^{y-2}\) scores M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\log_3 \sqrt{x} = \log_3 x^{\frac{1}{2}} = \frac{1}{2}\log_3 x = \frac{1}{2}y\) | B1 | \(\frac{1}{2}y\) or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2\log_3\left(\frac{x}{9}\right) - \log_3\sqrt{x} = 2 \Rightarrow 2(y-2) - \frac{1}{2}y = 2\) | M1 | Uses answers from part (a) to create a linear equation in \(y\). Condone poor brackets e.g. \(2(y-2) = 2y-2\), and the slip \((y-2) - \frac{1}{2}y = 2\) |
| \(y = 4\) | A1 | Correct value for \(y\) |
| \(\log_3 x = 4 \Rightarrow x = 3^4\) | dM1 | Correct method for undoing log. Dependent on first M |
| \(x = 81\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2\log_3\left(\frac{x}{9}\right) - \log_3\sqrt{x} = \log_3\left(\frac{(x/9)^2}{\sqrt{x}}\right)\) or \(= \log_3\frac{x^2}{\sqrt{x}} + \ldots\) Combines two log terms in \(x\) correctly to obtain single log term | M1 | |
| \(\log_3\left(\frac{(x/9)^2}{\sqrt{x}}\right) = 2\) or \(\log_3\left(\frac{x^2}{\sqrt{x}}\right) = 6\) | A1 | Correct equation |
| \(\frac{(x/9)^2}{\sqrt{x}} = 3^2\) or \(\frac{x^2}{\sqrt{x}} = 3^6\) | dM1 | Correct method for undoing log. Dependent on first M |
| \(x = 81\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2\log_3\left(\frac{x}{9}\right) - \log_3\sqrt{x} = 2\log_3\left(\frac{3^y}{9}\right) - \log_3 3^{\frac{y}{2}} = \log_3\left(\frac{3^{\frac{3y}{2}}}{81}\right)\) | M1 | Combines logs correctly |
| \(\log_3\left(\frac{3^{\frac{3y}{2}}}{81}\right) = 2 \Rightarrow y = 4\) | A1 | Correct value for \(y\) |
| \(\log_3 x = 4 \Rightarrow x = 3^4\) | dM1 | Correct method for undoing log. Dependent on the first M |
| \(\Rightarrow x = 81\) | A1 | cao |
## Question 5:
### Part (a)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\log_3\left(\frac{x}{9}\right) = \log_3 x - \log_3 9 = y - 2$ | M1A1 | M1: $\log_3\left(\frac{x}{9}\right) = \log_3 x - \log_3 9$ or $= \log_3 x + \log_3\frac{1}{9}$. Correct use of subtraction/addition rule. Ignore presence/absence of base and any spurious "$=0$"; A1: $y-2$. Answer left as $\log_3 3^{y-2}$ scores M1A0 |
---
### Part (a)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\log_3 \sqrt{x} = \log_3 x^{\frac{1}{2}} = \frac{1}{2}\log_3 x = \frac{1}{2}y$ | B1 | $\frac{1}{2}y$ or equivalent |
---
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\log_3\left(\frac{x}{9}\right) - \log_3\sqrt{x} = 2 \Rightarrow 2(y-2) - \frac{1}{2}y = 2$ | M1 | Uses answers from part (a) to create a **linear** equation in $y$. Condone poor brackets e.g. $2(y-2) = 2y-2$, and the slip $(y-2) - \frac{1}{2}y = 2$ |
| $y = 4$ | A1 | Correct value for $y$ |
| $\log_3 x = 4 \Rightarrow x = 3^4$ | dM1 | Correct method for undoing log. Dependent on first M |
| $x = 81$ | A1 | cao |
**Alternative method (Alt 1b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\log_3\left(\frac{x}{9}\right) - \log_3\sqrt{x} = \log_3\left(\frac{(x/9)^2}{\sqrt{x}}\right)$ or $= \log_3\frac{x^2}{\sqrt{x}} + \ldots$ Combines two log terms in $x$ correctly to obtain single log term | M1 | |
| $\log_3\left(\frac{(x/9)^2}{\sqrt{x}}\right) = 2$ or $\log_3\left(\frac{x^2}{\sqrt{x}}\right) = 6$ | A1 | Correct equation |
| $\frac{(x/9)^2}{\sqrt{x}} = 3^2$ or $\frac{x^2}{\sqrt{x}} = 3^6$ | dM1 | Correct method for undoing log. Dependent on first M |
| $x = 81$ | A1 | cao |
# Question Alt 2(b): (Uses $x = 3^y$)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\log_3\left(\frac{x}{9}\right) - \log_3\sqrt{x} = 2\log_3\left(\frac{3^y}{9}\right) - \log_3 3^{\frac{y}{2}} = \log_3\left(\frac{3^{\frac{3y}{2}}}{81}\right)$ | M1 | Combines logs correctly |
| $\log_3\left(\frac{3^{\frac{3y}{2}}}{81}\right) = 2 \Rightarrow y = 4$ | A1 | Correct value for $y$ |
| $\log_3 x = 4 \Rightarrow x = 3^4$ | dM1 | Correct method for undoing log. **Dependent on the first M** |
| $\Rightarrow x = 81$ | A1 | cao |
---5. (a) Given that
$$y = \log _ { 3 } x$$
find expressions in terms of $y$ for
\begin{enumerate}[label=(\roman*)]
\item $\log _ { 3 } \left( \frac { x } { 9 } \right)$
\item $\log _ { 3 } \sqrt { x }$
Write each answer in its simplest form.\\
(b) Hence or otherwise solve
$$2 \log _ { 3 } \left( \frac { x } { 9 } \right) - \log _ { 3 } \sqrt { x } = 2$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2017 Q5 [7]}}