| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Two unrelated log parts: both solve equations |
| Difficulty | Moderate -0.3 Part (a) is a straightforward logarithm equation requiring basic log laws (bringing terms to one side and using log subtraction = log division), then solving a^3 = 27/8. Part (b) is a standard exponential equation solved by taking logs of both sides and rearranging. Both are routine C2 techniques with no conceptual challenges, though part (b) requires careful algebraic manipulation. Slightly easier than average due to being textbook-standard exercises. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\log_a 27 - \log_a 8 = 3\) | ||
| \(\log_a \frac{27}{8} = 3\) | M1 | |
| \(a^3 = \frac{27}{8}\), \(a = \sqrt[3]{\frac{27}{8}} = \frac{3}{2}\) | M1 A1 | |
| (b) \((x + 3) \lg 2 = (x - 1) \lg 6\) | M1 | |
| \(x(\lg 6 - \lg 2) = 3 \lg 2 + \lg 6\) | M1 | |
| \(x = \frac{3\lg 2 + \lg 6}{\lg 6 - \lg 2} = 3.52\) | M1 A1 | (7) |
**(a)** $\log_a 27 - \log_a 8 = 3$ | |
$\log_a \frac{27}{8} = 3$ | M1 |
$a^3 = \frac{27}{8}$, $a = \sqrt[3]{\frac{27}{8}} = \frac{3}{2}$ | M1 A1 |
**(b)** $(x + 3) \lg 2 = (x - 1) \lg 6$ | M1 |
$x(\lg 6 - \lg 2) = 3 \lg 2 + \lg 6$ | M1 |
$x = \frac{3\lg 2 + \lg 6}{\lg 6 - \lg 2} = 3.52$ | M1 A1 | (7)\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$ such that
$$\log_a 27 = 3 + \log_a 8.$$ [3]
\item Solve the equation
$$2^{x+3} = 6^{x-1},$$
giving your answer to 3 significant figures. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q5 [7]}}