| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Two unrelated log parts: one non-log algebraic part |
| Difficulty | Moderate -0.3 Part (a) is straightforward log evaluation using basic laws (log_3 27 = 3 is direct, then subtract log_3 4). Part (b) requires recognizing 4^x = (2^2)^x = 2^{2x} and substituting y = 2^x to get a quadratic, which is a standard C2 technique but requires more problem-solving than pure recall. Overall slightly easier than average due to being a routine two-part question with well-practiced methods. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks |
|---|---|
| (a) \(= 3 - \log_8 8^3\) | B1 M1 A1 |
| \(= 3 - \frac{3}{8} = \frac{7}{8}\) | A1 |
| (b) \((2^x)^2 - 3(2 \times 2^x) = 0\) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(2^x(2^x - 6) = 0\) | M1 | |
| \(2^x = 0\) (no solutions) or \(6\) | A1 | |
| \(x = \frac{\lg 6}{\lg 2} = 2.58\) (3sf) | M1 A1 |
**(a)** $= 3 - \log_8 8^3$ | B1 M1 A1 |
$= 3 - \frac{3}{8} = \frac{7}{8}$ | A1 |
**(b)** $(2^x)^2 - 3(2 \times 2^x) = 0$ | M1 |
$(2^x)^2 - 6(2^x) = 0$
$2^x(2^x - 6) = 0$ | M1 |
$2^x = 0$ (no solutions) or $6$ | A1 |
$x = \frac{\lg 6}{\lg 2} = 2.58$ (3sf) | M1 A1 | | (9 marks)\begin{enumerate}[label=(\alph*)]
\item Evaluate
$$\log_3 27 - \log_3 4.$$ [4]
\item Solve the equation
$$4^x - 3(2^{x+1}) = 0.$$ [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q5 [9]}}