| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | ln(y) vs ln(x) linear graph |
| Difficulty | Moderate -0.3 This is a straightforward application of logarithm laws and exponential modelling. Part (a) requires simple substitution and calculator work, part (b) is a standard manipulation using log laws (antilog both sides), and part (c) tests understanding of the model's meaning. While it involves multiple parts, each step follows routine procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02z Models in context: use functions in modelling1.06c Logarithm definition: log_a(x) as inverse of a^x1.06d Natural logarithm: ln(x) function and properties1.06f Laws of logarithms: addition, subtraction, power rules1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\log_{10} M = 1.93\log_{10} 45 + 0.684 \Rightarrow \log_{10} M = 3.8747\) | M1 | Substitutes \(r=45\), proceeds to numerical value; implied by awrt 7500 for \(r=45\) |
| \((M =)\) awrt \(7500\) (kg) | A1 | Condone \(7.5 \times 10^3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct proof that \(\log_{10} M = 1.93\log_{10} r + 0.684 \Leftrightarrow M = pr^q\); e.g. \(\log_{10} M = \log_{10} r^{1.93} + 0.684 \Rightarrow M = 10^{0.684} \times r^{1.93}\) | B1 | Addition/index law of logs explicitly used; condone log/lg for \(\log_{10}\); natural logs is B0 |
| Either \(p = 10^{0.684}\) or \(q = 1.93\) | M1 | May be implied by \(M = pr^q\) with correct \(p\) or \(q\) |
| Both \(p =\) awrt \(4.83\) and \(q = 1.93\) | A1 | May be implied by \(M = 4.83r^{1.93}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| "\(p\)" is the mass (in kg) of a tree with radius 1 cm | B1 | — |
## Question 7:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\log_{10} M = 1.93\log_{10} 45 + 0.684 \Rightarrow \log_{10} M = 3.8747$ | M1 | Substitutes $r=45$, proceeds to numerical value; implied by awrt 7500 for $r=45$ |
| $(M =)$ awrt $7500$ (kg) | A1 | Condone $7.5 \times 10^3$ |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct proof that $\log_{10} M = 1.93\log_{10} r + 0.684 \Leftrightarrow M = pr^q$; e.g. $\log_{10} M = \log_{10} r^{1.93} + 0.684 \Rightarrow M = 10^{0.684} \times r^{1.93}$ | B1 | Addition/index law of logs explicitly used; condone log/lg for $\log_{10}$; natural logs is B0 |
| Either $p = 10^{0.684}$ or $q = 1.93$ | M1 | May be implied by $M = pr^q$ with correct $p$ or $q$ |
| Both $p =$ awrt $4.83$ and $q = 1.93$ | A1 | May be implied by $M = 4.83r^{1.93}$ |
**Part (c):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| "$p$" is the mass (in kg) of a tree with radius 1 cm | B1 | — |
---7 The mass, $M \mathrm {~kg}$, of a species of tree can be modelled by the equation
$$\log _ { 10 } M = 1.93 \log _ { 10 } r + 0.684$$
where $r \mathrm {~cm}$ is the base radius of the tree.\\
The base radius of a particular tree of this species is 45 cm .\\
According to the model,
\begin{enumerate}[label=(\alph*)]
\item find the mass of this tree, giving your answer to 2 significant figures.
\item Show that the equation of the model can be written in the form
$$M = p r ^ { q }$$
giving the values of the constants $p$ and $q$ to 3 significant figures.
\item With reference to the model, interpret the value of the constant $p$.
Q
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2021 Q7 [6]}}