| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2019 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Simplify or prove logarithmic identity |
| Difficulty | Standard +0.3 This question requires applying the change of base formula and laws of logarithms systematically, but follows a standard pattern. Part (a) is straightforward manipulation (ln 9 = 2ln 3, ln 27 = 3ln 3, giving 6(ln 3)³), and part (b) requires only the simple numerical fact that ln 3 > 1. While it involves multiple steps, it's a routine application of logarithm laws with no novel insight required, making it slightly easier than average. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.06f Laws of logarithms: addition, subtraction, power rules |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| DR: \(\ln 3^2\), \(\ln 3^3\) seen | B1 | |
| \(\ln 3 \times 2\ln 3 \times 3\ln 3 = 6(\ln 3)^3\) | B1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| DR: \(3 > e\) so \(\ln 3 > 1\) | M1 | Must mention \(e\) |
| \((\ln 3)^3 > 1\) so \(6(\ln 3)^3 > 6\) | E1 | Convincing completion (answer given) |
| [2] |
## Question 7:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| DR: $\ln 3^2$, $\ln 3^3$ seen | B1 | |
| $\ln 3 \times 2\ln 3 \times 3\ln 3 = 6(\ln 3)^3$ | B1 | |
| **[2]** | | |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| DR: $3 > e$ so $\ln 3 > 1$ | M1 | Must mention $e$ |
| $(\ln 3)^3 > 1$ so $6(\ln 3)^3 > 6$ | E1 | Convincing completion (answer given) |
| **[2]** | | |7 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item Express $\ln 3 \times \ln 9 \times \ln 27$ in terms of $\ln 3$.
\item Hence show that $\ln 3 \times \ln 9 \times \ln 27 > 6$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2019 Q7 [4]}}