| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Model y=ax^b: linearise and find constants from graph/data |
| Difficulty | Moderate -0.3 This is a standard logarithmic modeling question requiring students to linearize a power law relationship and extract parameters from data. Part (i) is routine algebraic manipulation, part (ii) involves plotting and reading from a graph (standard C2 skill), and part (iii) is straightforward substitution. While it requires understanding of logarithms and graph work, it follows a well-established textbook pattern with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02z Models in context: use functions in modelling1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form |
| \(x\) | 2 | 3 | 4 | 5 | 6 |
| \(y\) | 4.6 | 5.0 | 5.3 | 5.5 | 5.7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = ax^b \Rightarrow \log y = \log ax^b\) | M1 | |
| \(\Rightarrow \log y = \log a + \log x^b\), i.e. \(\log y = \log a + b\log x\) | A1 | |
| This is of the form \(y = mx + c\), so plotting gives a straight line where intercept is \(\log a\) and gradient is \(b\) | B1 | Explanation [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Correct \(\log x\) values: \(0.3, 0.48, 0.60, 0.70, 0.78\) | B1 | |
| Correct \(\log y\) values: \(0.66, 0.70, 0.72, 0.74, 0.76\) | B1 | |
| Correct plot | B1 | |
| Straight line drawn | B1 | |
| Straight line so model is appropriate | B1 | |
| Gradient \(\approx \frac{1}{5} \Rightarrow b = \frac{1}{5}\) | B1 | |
| Intercept \(\approx 0.6 \Rightarrow a = 4\), i.e. \(y = 4x^{0.2}\) | B1 | Or substitute if origin not on graph [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = 4\times(2.8)^{0.2} \approx 4.91\) | M1, A1 | [2] |
## Question 10:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = ax^b \Rightarrow \log y = \log ax^b$ | M1 | |
| $\Rightarrow \log y = \log a + \log x^b$, i.e. $\log y = \log a + b\log x$ | A1 | |
| This is of the form $y = mx + c$, so plotting gives a straight line where intercept is $\log a$ and gradient is $b$ | B1 | Explanation **[3]** |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct $\log x$ values: $0.3, 0.48, 0.60, 0.70, 0.78$ | B1 | |
| Correct $\log y$ values: $0.66, 0.70, 0.72, 0.74, 0.76$ | B1 | |
| Correct plot | B1 | |
| Straight line drawn | B1 | |
| Straight line so model is appropriate | B1 | |
| Gradient $\approx \frac{1}{5} \Rightarrow b = \frac{1}{5}$ | B1 | |
| Intercept $\approx 0.6 \Rightarrow a = 4$, i.e. $y = 4x^{0.2}$ | B1 | Or substitute if origin not on graph **[7]** |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = 4\times(2.8)^{0.2} \approx 4.91$ | M1, A1 | **[2]** |
---10 A function $y = \mathrm { f } ( x )$ may be modelled by the equation $y = a x ^ { b }$.\\
(i) Show why, if this is so, then plotting $\log y$ against $\log x$ will produce a straight line graph. Explain how $a$ and $b$ may be determined experimentally from the graph.\\
(ii) Values of $x$ and $y$ are given below. By plotting a graph of logy against log $x$, show that the model above is appropriate for this set of data and find values of $a$ and $b$ given that $a$ is an integer and $b$ can be written as a fraction with a denominator less than 10 .
\begin{center}
\begin{tabular}{ | l | l | l | l | l | l | }
\hline
$x$ & 2 & 3 & 4 & 5 & 6 \\
\hline
$y$ & 4.6 & 5.0 & 5.3 & 5.5 & 5.7 \\
\hline
\end{tabular}
\end{center}
(iii) Use your formula from part (ii) to estimate the value of $y$ when $x = 2.8$.
\hfill \mbox{\textit{OCR MEI C2 Q10 [12]}}