| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Two unrelated log parts: both solve equations |
| Difficulty | Moderate -0.8 Both parts are routine logarithm/exponential equation solving requiring standard techniques: (a) take natural log of both sides, (b) use log laws to combine and solve linear equation. These are textbook exercises testing basic manipulation skills with no problem-solving insight required, making them easier than average A-level questions. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(4x - 3 = \ln 2\) | M1 | |
| \(x = \frac{1}{4}(3 + \ln 2)\) | M1 A1 | |
| (b) \(\ln(2y-1) - \ln(3-y) = \ln\frac{2y-1}{3-y} = 1\) | M1 | |
| \(\frac{2y-1}{3-y} = e\) | A1 | |
| \(2y - 1 = e(3-y)\), \(y(e+2) = 3e+1\) | M1 | |
| \(y = \frac{3e+1}{e+2}\) | A1 | (7 marks) |
**(a)** $4x - 3 = \ln 2$ | M1 |
$x = \frac{1}{4}(3 + \ln 2)$ | M1 A1 |
**(b)** $\ln(2y-1) - \ln(3-y) = \ln\frac{2y-1}{3-y} = 1$ | M1 |
$\frac{2y-1}{3-y} = e$ | A1 |
$2y - 1 = e(3-y)$, $y(e+2) = 3e+1$ | M1 |
$y = \frac{3e+1}{e+2}$ | A1 | (7 marks)Solve each equation, giving your answers in exact form.
\begin{enumerate}[label=(\alph*)]
\item $e^{4x-3} = 2$ [3]
\item $\ln (2y - 1) = 1 + \ln (3 - y)$ [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q2 [7]}}