| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Solve using substitution or auxiliary variable |
| Difficulty | Moderate -0.3 This is a straightforward C2 logarithm question requiring basic log laws (power rule, base change) and solving a linear equation in t. Part (a) is direct application of rules, part (b) involves simple substitution and algebra. Slightly easier than average due to the scaffolded structure guiding students through the substitution method. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| (a) (i) \(= 2 \log_3 x = 2t\) | M1 A1 | |
| (ii) \(= \frac{\log_3 x}{\log_3 9} = \frac{\log_3 x}{2} = \frac{1}{2}t\) | M1 A1 | |
| (b) \(2t - \frac{1}{2}t = 4\) | M1 | |
| \(t = \frac{8}{3}\) | M1 | |
| \(\log_3 x = \frac{8}{3}\), \(x = 3^{\frac{8}{3}} = 18.7\) | M1 A1 | (7) |
(a) (i) $= 2 \log_3 x = 2t$ | M1 A1 |
(ii) $= \frac{\log_3 x}{\log_3 9} = \frac{\log_3 x}{2} = \frac{1}{2}t$ | M1 A1 |
(b) $2t - \frac{1}{2}t = 4$ | M1 |
$t = \frac{8}{3}$ | M1 |
$\log_3 x = \frac{8}{3}$, $x = 3^{\frac{8}{3}} = 18.7$ | M1 A1 | (7)5. (a) Given that $t = \log _ { 3 } x$, find expressions in terms of $t$ for
\begin{enumerate}[label=(\roman*)]
\item $\log _ { 3 } x ^ { 2 }$,
\item $\log _ { 9 } x$.\\
(b) Hence, or otherwise, find to 3 significant figures the value of $x$ such that
$$\log _ { 3 } x ^ { 2 } - \log _ { 9 } x = 4 .$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q5 [7]}}