Edexcel P3 2020 January — Question 3 5 marks

Exam BoardEdexcel
ModuleP3 (Pure Mathematics 3)
Year2020
SessionJanuary
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLaws of Logarithms
TypeLinear relationship between log variables
DifficultyModerate -0.8 This is a straightforward two-part question requiring only basic skills: finding the equation of a line from two points (standard GCSE/AS skill), then converting between logarithmic and exponential forms using log laws. No problem-solving or insight needed—purely mechanical application of well-practiced techniques.
Spec1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form

3. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{1c700103-ecab-4a08-b411-3f445ed88885-08_599_883_299_536} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows a linear relationship between \(\log _ { 10 } y\) and \(\log _ { 10 } x\) The line passes through the points \(( 0,4 )\) and \(( 6,0 )\) as shown.
  1. Find an equation linking \(\log _ { 10 } y\) with \(\log _ { 10 } x\)
  2. Hence, or otherwise, express \(y\) in the form \(p x ^ { q }\), where \(p\) and \(q\) are constants to be found.

AnswerMarks Guidance
(a) Implies equation of line is of the form \(\log_{10} y = \pm\frac{2}{3}\log_{10} x + 4\)M1
States \(\log_{10} y = -\frac{2}{3}\log_{10} x + 4\) o.e.A1 (2)
(b) Applies one correct log law E.g. \(\log_{10} y = -\frac{2}{3}\log_{10} x + 4 \rightarrow \log_{10} y = \log_{10} x^{-\frac{2}{3}} + 4\)M1
Full attempt to undo the logs \(\log_{10} y = \log_{10} x^{-\frac{2}{3}} + \log_{10} 10^4 \rightarrow y = x^{-\frac{2}{3}} \times 10^4\)dM1
\(\rightarrow y = 10\,000x^{-\frac{2}{3}}\) o.e.A1 (3)
Total: 5 marks
**(a)** Implies equation of line is of the form $\log_{10} y = \pm\frac{2}{3}\log_{10} x + 4$ | M1 |
States $\log_{10} y = -\frac{2}{3}\log_{10} x + 4$ o.e. | A1 | (2)

**(b)** Applies one correct log law E.g. $\log_{10} y = -\frac{2}{3}\log_{10} x + 4 \rightarrow \log_{10} y = \log_{10} x^{-\frac{2}{3}} + 4$ | M1 |
Full attempt to undo the logs $\log_{10} y = \log_{10} x^{-\frac{2}{3}} + \log_{10} 10^4 \rightarrow y = x^{-\frac{2}{3}} \times 10^4$ | dM1 |
$\rightarrow y = 10\,000x^{-\frac{2}{3}}$ o.e. | A1 | (3)

**Total: 5 marks**

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3.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{1c700103-ecab-4a08-b411-3f445ed88885-08_599_883_299_536}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

Figure 1 shows a linear relationship between $\log _ { 10 } y$ and $\log _ { 10 } x$\\
The line passes through the points $( 0,4 )$ and $( 6,0 )$ as shown.
\begin{enumerate}[label=(\alph*)]
\item Find an equation linking $\log _ { 10 } y$ with $\log _ { 10 } x$
\item Hence, or otherwise, express $y$ in the form $p x ^ { q }$, where $p$ and $q$ are constants to be found.

\begin{center}

\end{center}
\end{enumerate}

\hfill \mbox{\textit{Edexcel P3 2020 Q3 [5]}}