| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2005 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Quadratic in exponential form |
| Difficulty | Moderate -0.3 This is a standard C3 exponential equation question with routine techniques: part (a) is basic logarithm manipulation, part (b) follows a guided substitution to form a quadratic that factors easily, then requires back-substitution and taking logarithms. The scaffolding makes it slightly easier than average, though it requires multiple steps and careful algebraic manipulation. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(2e^x = 5\); \(e^x = \frac{5}{2}\) | M1 | Exact |
| \(x = \ln\frac{5}{2}\) \((0.916)\) | A1 | A0 if further wrong work; Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(2e^x + 5e^{-x} = 7\); \(2e^{2x}+5 = 7e^x\) | M1 | Dealing with \(e^{-x}\) |
| \(2y^2 - 7y + 5 = 0\) | A1 | AG; Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \((2y-5)(y-1)=0\) | M1 | Attempt to solve; \(y = \frac{5}{2}, 1\) (SC B1) |
| \(x = \ln\frac{5}{2}\) | A1 | \(e^x = \frac{5}{2}\) |
| \(x = 0\) (or \(\ln 1\)) | A1 | \(e^x = 1\); Total: 3 marks |
## Question 5:
### Part (a)
| Working | Marks | Guidance |
|---------|-------|---------|
| $2e^x = 5$; $e^x = \frac{5}{2}$ | M1 | Exact |
| $x = \ln\frac{5}{2}$ $(0.916)$ | A1 | A0 if further wrong work; Total: 2 marks |
### Part (b)(i)
| Working | Marks | Guidance |
|---------|-------|---------|
| $2e^x + 5e^{-x} = 7$; $2e^{2x}+5 = 7e^x$ | M1 | Dealing with $e^{-x}$ |
| $2y^2 - 7y + 5 = 0$ | A1 | AG; Total: 2 marks |
### Part (b)(ii)
| Working | Marks | Guidance |
|---------|-------|---------|
| $(2y-5)(y-1)=0$ | M1 | Attempt to solve; $y = \frac{5}{2}, 1$ (SC B1) |
| $x = \ln\frac{5}{2}$ | A1 | $e^x = \frac{5}{2}$ |
| $x = 0$ (or $\ln 1$) | A1 | $e^x = 1$; Total: 3 marks |
---5
\begin{enumerate}[label=(\alph*)]
\item Solve the equation $2 \mathrm { e } ^ { x } = 5$, giving your answer as an exact natural logarithm.
\item \begin{enumerate}[label=(\roman*)]
\item By substituting $y = \mathrm { e } ^ { x }$, show that the equation $2 \mathrm { e } ^ { x } + 5 \mathrm { e } ^ { - x } = 7$ can be written as
$$2 y ^ { 2 } - 7 y + 5 = 0$$
\item Hence solve the equation $2 \mathrm { e } ^ { x } + 5 \mathrm { e } ^ { - x } = 7$, giving your answers as exact values of $x$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2005 Q5 [7]}}