| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Express log in terms of given variables |
| Difficulty | Moderate -0.3 This is a straightforward application of standard logarithm laws (product, quotient, power rules) with minimal problem-solving required. Parts (a) and (b) are direct substitution exercises, while part (c) combines the results in a simple equation. The question is slightly easier than average because it's highly structured and uses basic log laws that students practice extensively, though it requires more steps than the most trivial questions. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Way 1: \(\log_3(9x)=\log_3 9+\log_3 x = 2+a\) | M1 | Use of \(\log(ab)=\log a+\log b\) |
| \(=2+a\) | A1 | Must be \(a+2\) or \(2+a\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Way 1: \(\log_3\!\left(\frac{x^5}{81}\right)=\log_3 x^5-\log_3 81\) | M1 | Use of \(\log(a/b)=\log a - \log b\) |
| \(\log x^5=5\log x\) or \(\log 81=4\log 3\) or \(\log 81=4\) | M1 | Use of \(n\log a=\log a^n\) |
| \(=5a-4\) | A1 cso | No errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\log_3(9x)+\log_3\!\left(\frac{x^5}{81}\right)=3\) | — | — |
| \(\Rightarrow 2+a+5a-4=3 \Rightarrow a=\frac{5}{6}\) | M1, A1 | M1: uses (a) and (b) to form equation in \(a\); A1: \(a=\) awrt \(0.833\) |
| \(\Rightarrow x=3^{5/6}\) or \(\log_{10}x=a\log_{10}3\) | M1 | Finds \(x\) by use of 3 to a power or change of base performed correctly |
| \(x=2.498\) or awrt | A1 | Accept answer rounding to 2.498 from \(2.498049533\ldots\); lose mark if \(x=-2.498\) also given |
## Question 6:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Way 1: $\log_3(9x)=\log_3 9+\log_3 x = 2+a$ | M1 | Use of $\log(ab)=\log a+\log b$ |
| $=2+a$ | A1 | Must be $a+2$ or $2+a$ |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Way 1: $\log_3\!\left(\frac{x^5}{81}\right)=\log_3 x^5-\log_3 81$ | M1 | Use of $\log(a/b)=\log a - \log b$ |
| $\log x^5=5\log x$ or $\log 81=4\log 3$ or $\log 81=4$ | M1 | Use of $n\log a=\log a^n$ |
| $=5a-4$ | A1 cso | No errors seen |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\log_3(9x)+\log_3\!\left(\frac{x^5}{81}\right)=3$ | — | — |
| $\Rightarrow 2+a+5a-4=3 \Rightarrow a=\frac{5}{6}$ | M1, A1 | M1: uses (a) and (b) to form equation in $a$; A1: $a=$ awrt $0.833$ |
| $\Rightarrow x=3^{5/6}$ or $\log_{10}x=a\log_{10}3$ | M1 | Finds $x$ by use of 3 to a power or change of base performed correctly |
| $x=2.498$ or awrt | A1 | Accept answer rounding to 2.498 from $2.498049533\ldots$; lose mark if $x=-2.498$ also given |6. Given that $\log _ { 3 } x = a$, find in terms of $a$,
\begin{enumerate}[label=(\alph*)]
\item $\log _ { 3 } ( 9 x )$
\item $\log _ { 3 } \left( \frac { x ^ { 5 } } { 81 } \right)$\\
giving each answer in its simplest form.
\item Solve, for $x$,
$$\log _ { 3 } ( 9 x ) + \log _ { 3 } \left( \frac { x ^ { 5 } } { 81 } \right) = 3$$
giving your answer to 4 significant figures.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4f4eac7b-8908-480f-bb39-049944203fff-10_775_1605_221_159}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The line with equation $y = 10$ cuts the curve with equation $y = x ^ { 2 } + 2 x + 2$ at the points $A$ and $B$ as shown in Figure 1. The figure is not drawn to scale.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2013 Q6 [9]}}