| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2012 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Solve by showing reduces to polynomial |
| Difficulty | Moderate -0.3 This is a straightforward C2 logarithm question requiring basic log laws (power rule, addition rule) and solving a quadratic. Part (a) is pure manipulation, part (b) reduces to a simple quadratic equation. Slightly easier than average due to clear structure and standard techniques, though requires careful algebraic manipulation. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\log_3 3x^2 = \log_3 3 + \log_3 x^2\) or \(\log y - \log x^2 = \log 3\) | B1 | B1: correct use of addition rule (or subtraction rule) |
| \(\log_3 x^2 = 2\log_3 x\) | B1 | B1: replacing \(\log x^2\) by \(2\log x\). Not \(\log 3x^2\) by \(2\log 3x\) — this is B0 |
| Using \(\log_3 3 = 1\) | B1 (3) | B1: replacing \(\log 3\) by 1 (or use of \(3^1=3\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(3x^2 = 28x - 9\) | M1 | M1: removing logs to get equation in \(x\); must be accurate with no errors from part (b) |
| Solves \(3x^2 - 28x + 9 = 0\) to give \(x = \frac{1}{3}\) or \(x = 9\) | M1, A1 (3) | M1: attempting to solve three-term quadratic. A1: two correct answers |
# Question 4:
## Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| $\log_3 3x^2 = \log_3 3 + \log_3 x^2$ or $\log y - \log x^2 = \log 3$ | B1 | B1: correct use of addition rule (or subtraction rule) |
| $\log_3 x^2 = 2\log_3 x$ | B1 | B1: replacing $\log x^2$ by $2\log x$. **Not** $\log 3x^2$ by $2\log 3x$ — this is **B0** |
| Using $\log_3 3 = 1$ | B1 (3) | B1: replacing $\log 3$ by 1 (or use of $3^1=3$) |
## Part (b)
| Working | Marks | Guidance |
|---------|-------|----------|
| $3x^2 = 28x - 9$ | M1 | M1: removing logs to get equation in $x$; must be accurate with no errors from part (b) |
| Solves $3x^2 - 28x + 9 = 0$ to give $x = \frac{1}{3}$ or $x = 9$ | M1, A1 (3) | M1: attempting to solve three-term quadratic. A1: two correct answers |
---4. Given that $y = 3 x ^ { 2 }$,
\begin{enumerate}[label=(\alph*)]
\item show that $\log _ { 3 } y = 1 + 2 \log _ { 3 } x$
\item Hence, or otherwise, solve the equation
$$1 + 2 \log _ { 3 } x = \log _ { 3 } ( 28 x - 9 )$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2012 Q4 [6]}}