Moderate -0.8 This is a standard logarithmic transformation question requiring students to recognize that ln y = ln K + p ln x gives a straight line, then use two points to find the gradient p and y-intercept ln K. It's routine bookwork with straightforward arithmetic, requiring only recall of logarithm laws and basic coordinate geometry—easier than average A-level questions.
\includegraphics{figure_2}
The variables \(x\) and \(y\) satisfy the equation \(y = Kx^p\), where \(K\) and \(p\) are constants. The graph of \(\ln y\) against \(\ln x\) is a straight line passing through the points \((1.28, 3.69)\) and \((2.11, 4.81)\), as shown in the diagram. Find the values of \(K\) and \(p\) correct to 2 decimal places. [5]
Question 2:
2 | State or imply lny=lnK + plnx
Calculate gradient of line
Obtain p=1.35
Substitute to find K
Obtain K =7.11 or K =7.12 | B1
M1
A1
M1
A1 | [5]
\includegraphics{figure_2}
The variables $x$ and $y$ satisfy the equation $y = Kx^p$, where $K$ and $p$ are constants. The graph of $\ln y$ against $\ln x$ is a straight line passing through the points $(1.28, 3.69)$ and $(2.11, 4.81)$, as shown in the diagram. Find the values of $K$ and $p$ correct to 2 decimal places. [5]
\hfill \mbox{\textit{CAIE P2 2016 Q2 [5]}}