| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | June |
| Marks | 6 |
| 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 application of logarithm laws (ln a + ln b = ln(ab)) requiring one step. Part (b) requires multiplying by e^x to form a quadratic in e^x, then solving - a standard C3 technique but slightly more involved than typical routine questions. Overall slightly easier than average due to being a direct application of well-practiced methods with no novel problem-solving required. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\ln 3x = \ln 6\) or \(\ln x = \ln\left(\frac{6}{3}\right)\) [implied by 0.69...] or \(\ln\left(\frac{3x}{6}\right) = 0\) | M1 | |
| \(x = 2\) (only this answer) | A1 (cso) | (2 marks) |
| (b) \((e^x)^2 - 4e^x + 3 = 0\) (any 3 term form) | M1 | |
| \((e^x - 3)(e^x - 1) = 0\) | M1 dep | |
| \(e^x = 3\) or \(e^x = 1\) (Solving quadratic) | M1 | |
| \(x = \ln 3\), \(x = 0\) (or \(\ln 1\)) | M1 A1 | (4 marks) |
**(a)** $\ln 3x = \ln 6$ or $\ln x = \ln\left(\frac{6}{3}\right)$ [implied by 0.69...] or $\ln\left(\frac{3x}{6}\right) = 0$ | M1 | |
$x = 2$ (only this answer) | A1 (cso) | (2 marks) |
**(b)** $(e^x)^2 - 4e^x + 3 = 0$ (any 3 term form) | M1 | |
$(e^x - 3)(e^x - 1) = 0$ | M1 dep | |
$e^x = 3$ or $e^x = 1$ (Solving quadratic) | M1 | |
$x = \ln 3$, $x = 0$ (or $\ln 1$) | M1 A1 | (4 marks) |
**Notes:**
- Answer $x = 2$ with no working or incorrect working: M1A1
- Beware $x = 2$ from $\ln x = \frac{\ln 6}{\ln 3} = \ln 2$: M0A0
- $\ln x = \ln 6 - \ln 3 \Rightarrow x = e^{(\ln 6 - \ln 3)}$ allow M1, $x = 2$ (no wrong working) A1
- 1st M1 for attempting to multiply through by $e^x$: Allow $y$, $X$, even $x$, for $e^x$. Be generous for M1 (e.g. $e^{2x} + 3 = 4$, $e^x + 3 = 4e^x$)
- 2nd M1 for solving quadratic (may be by formula or completing the square) as far as getting two values for $e^x$ or $y$ or $X$ etc
- 3rd M1 for converting their answer(s) of the form $e^x = k$ to $x = \ln k$ (must be exact)
- A1 is for $\ln 3$ **and** $\ln 1$ or $0$ (Both required and no further solutions)
---Find the exact solutions to the equations
\begin{enumerate}[label=(\alph*)]
\item $\ln x + \ln 3 = \ln 6$,
\item $\mathrm { e } ^ { x } + 3 \mathrm { e } ^ { - x } = 4$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2007 Q1 [6]}}