| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | log(y) vs x: convert and interpret |
| Difficulty | Standard +0.3 This is a standard log-linear modelling question requiring students to find the equation of a line from two points, then convert back to exponential form using logarithm properties. The steps are routine: find gradient, write line equation, use log₁₀(N) = log₁₀(a) + t·log₁₀(b) to extract a and b, then solve for T. While it requires multiple techniques, each step follows a well-practiced procedure with no novel insight needed, making it slightly easier than average. |
| 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 | Mark | Guidance |
| Gradient \(= \frac{3.85-3.08}{5-0}\) \(\left(=0.154 \text{ or } \frac{77}{500}\right)\) | M1 | Correct attempt to find gradient between given coordinates; may be implied by \(10^{0.154}\) or awrt 1.43 or \(\frac{77}{500}\) |
| \(\log_{10} N = 3.08 + 0.154t\) oe | A1 | Must be in terms of \(\log_{10} N\) and \(t\); condone log \(N\) or lg \(N\); do not accept ln \(N\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = 10^{3.08}\) = awrt 1200 | B1 | |
| \(\log_{10} b = \text{"0.154"}\) or \(b = 10^{\text{"0.154"}}\) | M1 | May be implied by awrt 1.43; allow log or lg but not ln |
| \(b =\) awrt 1.43 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\log_{10} 500000 = 3.08 + 0.154T\) or \(T = \log_{\text{"1.43"}}\left(\frac{500000}{\text{"1200"}}\right)\) | M1 | Substitutes \(N=500000\); valid method to find \(T\); can be left as a logarithm |
| \((T=)\) awrt 17 | A1 | Cannot be left as a logarithm; mark can only be scored if equation in (a) is correct OR values of \(a\) and \(b\) are correct in (b) |
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient $= \frac{3.85-3.08}{5-0}$ $\left(=0.154 \text{ or } \frac{77}{500}\right)$ | M1 | Correct attempt to find gradient between given coordinates; may be implied by $10^{0.154}$ or awrt 1.43 or $\frac{77}{500}$ |
| $\log_{10} N = 3.08 + 0.154t$ oe | A1 | Must be in terms of $\log_{10} N$ and $t$; condone log $N$ or lg $N$; do not accept ln $N$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = 10^{3.08}$ = awrt 1200 | B1 | |
| $\log_{10} b = \text{"0.154"}$ or $b = 10^{\text{"0.154"}}$ | M1 | May be implied by awrt 1.43; allow log or lg but not ln |
| $b =$ awrt 1.43 | A1 | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\log_{10} 500000 = 3.08 + 0.154T$ or $T = \log_{\text{"1.43"}}\left(\frac{500000}{\text{"1200"}}\right)$ | M1 | Substitutes $N=500000$; valid method to find $T$; can be left as a logarithm |
| $(T=)$ awrt 17 | A1 | Cannot be left as a logarithm; mark can only be scored if equation in (a) is correct OR values of $a$ and $b$ are correct in (b) |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{44035bf8-f54c-472a-b969-b4fa4fa3d203-10_677_839_251_516}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The number of subscribers to an online video streaming service, $N$, is modelled by the equation
$$N = a b ^ { t }$$
where $a$ and $b$ are constants and $t$ is the number of years since monitoring began.\\
The line in Figure 1 shows the linear relationship between $t$ and $\log _ { 10 } N$\\
The line passes through the points $( 0,3.08 )$ and $( 5,3.85 )$
Using this information,
\begin{enumerate}[label=(\alph*)]
\item find an equation for this line.
\item Find the value of $a$ and the value of $b$, giving your answers to 3 significant figures.
When $t = T$ the number of subscribers is 500000
According to the model,
\item find the value of $T$
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2022 Q4 [7]}}