Moderate -0.3 This is a standard quadratic-in-disguise problem requiring substitution u = e^x, solving the resulting quadratic, then taking logarithms. While it involves multiple steps, the technique is routine and commonly practiced in P2/C3 courses, making it slightly easier than average.
\(e^x = \frac{1}{3}\), \(e^x = 27\) may be implied if \(u = e^x\) is stated
Use correct process at least once for solving \(e^x = c\) where \(c > 0\)
M1
Obtain \(-\ln 3\) from a correct solution
A1
Condone use of \(x = e^x\)
Obtain \(3\ln 3\) from a correct solution
A1
Total
5
## Question 1:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt to solve quadratic equation in $e^x$ | **M1** | Either directly or using substitution $u = e^x$ |
| Obtain $e^x = \frac{1}{3}$, $e^x = 27$ | **A1** | $e^x = \frac{1}{3}$, $e^x = 27$ may be implied if $u = e^x$ is stated |
| Use correct process at least once for solving $e^x = c$ where $c > 0$ | **M1** | |
| Obtain $-\ln 3$ from a correct solution | **A1** | Condone use of $x = e^x$ |
| Obtain $3\ln 3$ from a correct solution | **A1** | |
| **Total** | **5** | |
1 Solve the equation $3 \mathrm { e } ^ { 2 x } - 82 \mathrm { e } ^ { x } + 27 = 0$, giving your answers in the form $k \ln 3$.\\
\hfill \mbox{\textit{CAIE P2 2018 Q1 [5]}}