| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | log(y) vs x: convert and interpret |
| Difficulty | Moderate -0.5 This is a straightforward log-linear conversion question requiring finding a linear equation from two points (basic coordinate geometry) then converting to exponential form using log laws. While it involves multiple steps, each is routine: gradient calculation, y-intercept identification, and applying 10^(mx+c) = k·b^x. Slightly easier than average due to the clear structure and standard technique. |
| Spec | 1.06g Equations with exponentials: solve a^x = b1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\log_{10} y = \frac{5}{16}x + 1.5\) | M1 A1 | M1: Complete attempt to get equation of line condoning \(\log_{10} y \leftrightarrow y \leftrightarrow l\) and incorrect sign on gradient. Allow \((\log_{10} y) = \pm\frac{1.5}{4.8}x + 1.5\) or \(\frac{y-0}{x+4.8} = \pm\frac{1.5}{4.8}\). If via simultaneous equations, scored when candidate reaches \(m = \pm 0.3125\), \(c = 1.5\). A1: Correct equation e.g. \(\log_{10} y = \frac{5}{16}x + 1.5\) or equivalent such as \(16\log_{10} y = 5x + 24\). \(\log_{10}\) must not appear as "ln" but allow "log" or "lg" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\log_{10} y = \frac{5}{16}x + 1.5 \Rightarrow y = 10^{\frac{5}{16}x + 1.5}\) | M1 | "Removes" the logs in their equation |
| \(\Rightarrow y = 10^{\frac{5}{16}x} \times 10^{1.5}\) | M1 | "Correct" strategy to obtain values of \(k\) and \(b\), or proceeding correctly to \(y = 10^{\frac{5}{16}x} \times 10^{1.5}\). Allow \(k = 10^{1.5} (=31.6)\) and \(b = 10^{\frac{5}{16}} (=2.05)\) |
| \(y = 31.6 \times 2.05^x\) | A1 | Correct equation, no errors seen. Condone \(k=31.6, b=2.05\) with equation implied. Values must be 31.6 and 2.05, not values rounding to these or exact values like \(10\sqrt{10}\) |
# Question 3:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\log_{10} y = \frac{5}{16}x + 1.5$ | M1 A1 | M1: Complete attempt to get equation of line condoning $\log_{10} y \leftrightarrow y \leftrightarrow l$ and incorrect sign on gradient. Allow $(\log_{10} y) = \pm\frac{1.5}{4.8}x + 1.5$ or $\frac{y-0}{x+4.8} = \pm\frac{1.5}{4.8}$. If via simultaneous equations, scored when candidate reaches $m = \pm 0.3125$, $c = 1.5$. A1: Correct equation e.g. $\log_{10} y = \frac{5}{16}x + 1.5$ or equivalent such as $16\log_{10} y = 5x + 24$. $\log_{10}$ must not appear as "ln" but allow "log" or "lg" |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\log_{10} y = \frac{5}{16}x + 1.5 \Rightarrow y = 10^{\frac{5}{16}x + 1.5}$ | M1 | "Removes" the logs in their equation |
| $\Rightarrow y = 10^{\frac{5}{16}x} \times 10^{1.5}$ | M1 | "Correct" strategy to obtain values of $k$ and $b$, or proceeding correctly to $y = 10^{\frac{5}{16}x} \times 10^{1.5}$. Allow $k = 10^{1.5} (=31.6)$ and $b = 10^{\frac{5}{16}} (=2.05)$ |
| $y = 31.6 \times 2.05^x$ | A1 | Correct equation, no errors seen. Condone $k=31.6, b=2.05$ with equation implied. Values must be 31.6 and 2.05, not values rounding to these or exact values like $10\sqrt{10}$ |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5abaa077-1da4-4023-b442-194f6972095b-06_648_885_287_591}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The line $l$ in Figure 1 shows a linear relationship between $\log _ { 10 } y$ and $x$.\\
The line passes through the points $( 0,1.5 )$ and $( - 4.8,0 )$ as shown.
\begin{enumerate}[label=(\alph*)]
\item Write down an equation for $l$.
\item Hence, or otherwise, express $y$ in the form $k b ^ { x }$, giving the values of the constants $k$ and $b$ to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2023 Q3 [5]}}