| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2009 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Quadratic in exponential form |
| Difficulty | Moderate -0.3 Part (a) is a straightforward logarithm application. Part (b) involves a standard substitution technique to convert to a quadratic, then factorising and applying logarithms. This is a routine C3 exercise testing familiar methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.06c Logarithm definition: log_a(x) as inverse of a^x1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(e^x = \frac{4}{3}\) | M1 | |
| \(x = \ln\frac{4}{3}\) | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(3y+\frac{20}{y}=19\) or \(3e^{2x}+20=19e^x\) | ||
| \(3y^2-19y+20=0\) | B1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((3y-4)(y-5)=0\) | ||
| \(y=\frac{4}{3}, 5\) | B1 | |
| \(\therefore x=\ln\frac{4}{3}, \ln 5\) | M1 A1 | 3 |
# Question 7:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $e^x = \frac{4}{3}$ | M1 | |
| $x = \ln\frac{4}{3}$ | A1 | 2 | |
## Part (b)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $3y+\frac{20}{y}=19$ or $3e^{2x}+20=19e^x$ | | |
| $3y^2-19y+20=0$ | B1 | 1 | AG |
## Part (b)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $(3y-4)(y-5)=0$ | | |
| $y=\frac{4}{3}, 5$ | B1 | |
| $\therefore x=\ln\frac{4}{3}, \ln 5$ | M1 A1 | 3 | $\ln(\text{their}+\text{ve } y\text{'s})$ |
---7
\begin{enumerate}[label=(\alph*)]
\item Given that $3 \mathrm { e } ^ { x } = 4$, find the exact value of $x$.
\item \begin{enumerate}[label=(\roman*)]
\item By substituting $y = \mathrm { e } ^ { x }$, show that the equation $3 \mathrm { e } ^ { x } + 20 \mathrm { e } ^ { - x } = 19$ can be written as $3 y ^ { 2 } - 19 y + 20 = 0$.
\item Hence solve the equation $3 \mathrm { e } ^ { x } + 20 \mathrm { e } ^ { - x } = 19$, giving your answers as exact values. (3 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2009 Q7 [6]}}