| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Natural logarithm equation solving |
| Difficulty | Moderate -0.3 Part (a) is a straightforward one-step natural logarithm equation requiring only exponentiation and simple algebra. Part (b) is more substantial, requiring the substitution u = e^y to form a quadratic equation, then solving and taking logarithms, but this is a standard C3 technique. Overall slightly easier than average due to the routine nature of both parts, though (b) provides some modest challenge. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(2x - 3 = e\) | M1 | |
| \(x = \frac{1}{2}(e+3)\) | M1 A1 | |
| (b) \(3e^{2y} - 16e^y + 5 = 0\) | M1 | |
| \((3e^y - 1)(e^y - 5) = 0\) | M1 | |
| \(e^y = \frac{1}{3}, 5\) | A1 | |
| \(y = \ln \frac{1}{3}, \ln 5\) | M1 A1 | (8 marks) |
**(a)** $2x - 3 = e$ | M1 |
$x = \frac{1}{2}(e+3)$ | M1 A1 |
**(b)** $3e^{2y} - 16e^y + 5 = 0$ | M1 |
$(3e^y - 1)(e^y - 5) = 0$ | M1 |
$e^y = \frac{1}{3}, 5$ | A1 |
$y = \ln \frac{1}{3}, \ln 5$ | M1 A1 | (8 marks)
---3. Solve each equation, giving your answers in exact form.
\begin{enumerate}[label=(\alph*)]
\item $\quad \ln ( 2 x - 3 ) = 1$
\item $3 \mathrm { e } ^ { y } + 5 \mathrm { e } ^ { - y } = 16$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q3 [8]}}