Moderate -0.5 This is a standard logarithmic transformation question requiring students to recognize that ln y = ln K + m ln x gives a straight line, calculate the gradient m from two points using the standard formula, then find K by substitution. It's slightly easier than average as it's a direct application of a well-practiced technique with no conceptual surprises, though it does require careful arithmetic with logarithms.
3
\includegraphics[max width=\textwidth, alt={}, center]{7e100be2-9768-4fcd-b516-c714e53b0665-2_456_725_1082_712}
The variables \(x\) and \(y\) satisfy the equation \(y = K x ^ { m }\), where \(K\) and \(m\) are constants. The graph of \(\ln y\) against \(\ln x\) is a straight line passing through the points ( \(0.22,3.96\) ) and ( \(1.32,2.43\) ), as shown in the diagram. Find the values of \(K\) and \(m\) correct to 2 significant figures.
Use their gradient value and one point correctly to obtain intercept
M1
Obtain value for \(\ln K\) between \(4.26\) and \(4.28\)
A1
Obtain \(K = 71\) or \(K = 72\) or value rounding to either with no error noted
A1
[6]
State or imply that $\ln y = \ln K + m \ln x$ | B1 |
Form a numerical expression for gradient of line | M1 |
Obtain $-1.39$ or $-1.4$ | A1 |
Use their gradient value and one point correctly to obtain intercept | M1 |
Obtain value for $\ln K$ between $4.26$ and $4.28$ | A1 |
Obtain $K = 71$ or $K = 72$ or value rounding to either with no error noted | A1 | [6]
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3\\
\includegraphics[max width=\textwidth, alt={}, center]{7e100be2-9768-4fcd-b516-c714e53b0665-2_456_725_1082_712}
The variables $x$ and $y$ satisfy the equation $y = K x ^ { m }$, where $K$ and $m$ are constants. The graph of $\ln y$ against $\ln x$ is a straight line passing through the points ( $0.22,3.96$ ) and ( $1.32,2.43$ ), as shown in the diagram. Find the values of $K$ and $m$ correct to 2 significant figures.
\hfill \mbox{\textit{CAIE P2 2015 Q3 [6]}}