| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | ln(y) vs ln(x) linear graph |
| Difficulty | Moderate -0.8 This is a standard textbook exercise on logarithmic linearisation of power laws. Part (a) requires knowing to plot ln(Q) vs ln(P), and part (b) asks for the gradient interpretation (n = gradient). It's routine bookwork with no problem-solving or novel application required, making it easier than average. |
| Spec | 1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form |
| Answer | Marks | Guidance |
|---|---|---|
| Plot \(\log P\) against \(\log Q\) | B1 | Either way round |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\log Q = \log a + n\log P\) | (Draw a line of best fit) | |
| So plotting \(\log P\) against \(\log Q\) gives a straight line \(Y = mX + c\) with gradient \(n\) | M1 | Comparison with \(Y = mX + c\) |
| \(Y = \log Q\) and \(X = \log P\) | A1 | Clarify which way round |
| [2] |
# Question 7:
## Part (a):
| Plot $\log P$ against $\log Q$ | B1 | Either way round |
|---|---|---|
| | [1] | |
## Part (b):
| $\log Q = \log a + n\log P$ | | (Draw a line of best fit) |
|---|---|---|
| So plotting $\log P$ against $\log Q$ gives a straight line $Y = mX + c$ with gradient $n$ | M1 | Comparison with $Y = mX + c$ |
| $Y = \log Q$ and $X = \log P$ | A1 | Clarify which way round |
| | [2] | |
---7 The relationship between the variables $P$ and $Q$ is modelled by the formula\\
$Q = a P ^ { n }$\\
where $a$ and $n$ are constants.\\
Some values of $P$ and $Q$ are obtained from an experiment.
\begin{enumerate}[label=(\alph*)]
\item State appropriate quantities to plot so that the resulting points lie approximately in a straight line.
\item Explain how to use such a graph to estimate the value of $n$.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q7 [3]}}