| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Two unrelated log parts: one non-log algebraic part |
| Difficulty | Standard +0.3 Part (a) is a standard quadratic-in-disguise requiring substitution u=e^x, factorising, then taking logarithms—routine C3 technique. Part (b) is a proof by contradiction following a well-established template (assume rational, manipulate to show 3=5^k which is impossible), requiring more careful reasoning but still a standard proof type at this level. Combined, slightly above average difficulty due to the proof component. |
| Spec | 1.01d Proof by contradiction1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \((e^x - 3)(e^x - 5) = 0\) | M1 A1 | |
| \(e^x = 3, 5\) | M1 A1 | |
| \(x = \ln 3, \ln 5\) | ||
| (b) assume \(\log_3 2\) is rational | B1 | |
| \(\therefore \log_3 2 = \frac{p}{q}\) where \(p\) and \(q\) are integers and \(q \neq 0\) | M1 | |
| \(\Rightarrow 2^q = 3\) | M1 | |
| \(\Rightarrow 2^p = 3^q\) | A1 | |
| 2 and 3 are co-prime ∴ only solution is \(p = q = 0\) | M1 | |
| but \(q \neq 0\) ∴ contradiction ∴ \(\log_3 2\) is irrational | A1 | (10) |
(a) $(e^x - 3)(e^x - 5) = 0$ | M1 A1 |
$e^x = 3, 5$ | M1 A1 |
$x = \ln 3, \ln 5$ |
(b) assume $\log_3 2$ is rational | B1 |
$\therefore \log_3 2 = \frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$ | M1 |
$\Rightarrow 2^q = 3$ | M1 |
$\Rightarrow 2^p = 3^q$ | A1 |
2 and 3 are co-prime ∴ only solution is $p = q = 0$ | M1 |
but $q \neq 0$ ∴ contradiction ∴ $\log_3 2$ is irrational | A1 | **(10)**\begin{enumerate}[label=(\alph*)]
\item Find, as natural logarithms, the solutions of the equation
$$e^{2x} - 8e^x + 15 = 0.$$ [4]
\item Use proof by contradiction to prove that $\log_5 3$ is irrational. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q4 [10]}}