| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | log(y) vs x: convert and interpret |
| Difficulty | Moderate -0.3 This is a straightforward log-linear to exponential conversion question requiring standard algebraic manipulation. Students substitute t=2 into the log equation, convert using 10^(log S) = S, and identify p and q by comparing forms. The interpretation in part (c) is routine. Slightly easier than average due to direct application of logarithm laws without complex problem-solving. |
| Spec | 1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\log_{10} S = 4.5 - 0.006 \times 2 \Rightarrow S = 10^{4.5-0.006\times2} = 30800\text{ km}^2\) | M1A1 | M1: Substitute \(t=2\) and find a value for \(S\) via \(10^{\square}\); A1: awrt \(30800\text{ km}^2\) including units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\log_{10} S = 4.5 - 0.006t \Rightarrow S = 10^{4.5-0.006t}\) (or \(p=10^{4.5}\) or \(q=10^{-0.006}\)) | M1 | Writes correct form or awards for either value; must raise to base 10 |
| \(S = 10^{4.5-0.006t} = 10^{4.5} \times (10^{-0.006})^t\) (or \(p=10^{4.5}\) and \(q=10^{-0.006}\)) | dM1 | Must take out factor correctly; accept \(\left(\frac{1}{10^{0.006}}\right)\) or \(\left(\frac{1}{1.01}\right)\) for \(q\) |
| \(S = 31600 \times (0.986)^t\) | A1 | Must be awrt 3 s.f.; equation including \(S\), not just \(p\) and \(q\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| E.g. The proportion of area covered by coral reefs retained from year to year | B1 | Must reference: sea floor area/coral reefs/each year/percentage or proportional decrease or remaining. Do not accept "rate of change" or "decrease in area per year" or "\(q\) is the gradient" |
# Question 6:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\log_{10} S = 4.5 - 0.006 \times 2 \Rightarrow S = 10^{4.5-0.006\times2} = 30800\text{ km}^2$ | M1A1 | M1: Substitute $t=2$ and find a value for $S$ via $10^{\square}$; A1: awrt $30800\text{ km}^2$ including units |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\log_{10} S = 4.5 - 0.006t \Rightarrow S = 10^{4.5-0.006t}$ (or $p=10^{4.5}$ or $q=10^{-0.006}$) | M1 | Writes correct form or awards for either value; must raise to base 10 |
| $S = 10^{4.5-0.006t} = 10^{4.5} \times (10^{-0.006})^t$ (or $p=10^{4.5}$ and $q=10^{-0.006}$) | dM1 | Must take out factor correctly; accept $\left(\frac{1}{10^{0.006}}\right)$ or $\left(\frac{1}{1.01}\right)$ for $q$ |
| $S = 31600 \times (0.986)^t$ | A1 | Must be awrt 3 s.f.; equation including $S$, not just $p$ and $q$ |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| E.g. The proportion of area covered by coral reefs retained from year to year | B1 | Must reference: sea floor area/coral reefs/each year/percentage or proportional decrease or remaining. Do not accept "rate of change" or "decrease in area per year" or "$q$ is the gradient" |
---\begin{enumerate}
\item An area of sea floor is being monitored.
\end{enumerate}
The area of the sea floor, $S \mathrm {~km} ^ { 2 }$, covered by coral reefs is modelled by the equation
$$S = p q ^ { t }$$
where $p$ and $q$ are constants and $t$ is the number of years after monitoring began.\\
Given that
$$\log _ { 10 } S = 4.5 - 0.006 t$$
(a) find, according to the model, the area of sea floor covered by coral reefs when $t = 2$\\
(b) find a complete equation for the model in the form
$$S = p q ^ { t }$$
giving the value of $p$ and the value of $q$ each to 3 significant figures.\\
(c) With reference to the model, interpret the value of the constant $q$
\hfill \mbox{\textit{Edexcel P3 2023 Q6 [6]}}