Moderate -0.5 This is a standard logarithmic linearization question requiring students to recognize that ln(y) = ln(K) + m·ln(x) gives a straight line, find the gradient m from two points using (10.2-2.0)/(6-0), then find K from the y-intercept. It's routine application of logarithm laws and straight-line equations with no novel problem-solving required, making it slightly easier than average.
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\includegraphics[max width=\textwidth, alt={}, center]{d90dc270-b304-4b42-8e0e-37641b8a03b8-2_556_1113_680_516}
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 \(( 0,2.0 )\) and \(( 6,10.2 )\), as shown in the diagram. Find the values of \(K\) and \(m\), correct to 2 decimal places.
Equate intercept on axis for \(\ln y\) to \(\ln K\)
M1
Obtain \(7.39\) for \(K\)
A1
Attempt calculation of gradient of line
M1
Obtain \(1.37\) for \(m\)
A1
[5]
State or imply that $\ln y = \ln K + m\ln x$ | B1 |
Equate intercept on axis for $\ln y$ to $\ln K$ | M1 |
Obtain $7.39$ for $K$ | A1 |
Attempt calculation of gradient of line | M1 |
Obtain $1.37$ for $m$ | A1 | [5]
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\includegraphics[max width=\textwidth, alt={}, center]{d90dc270-b304-4b42-8e0e-37641b8a03b8-2_556_1113_680_516}
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 $( 0,2.0 )$ and $( 6,10.2 )$, as shown in the diagram. Find the values of $K$ and $m$, correct to 2 decimal places.
\hfill \mbox{\textit{CAIE P2 2011 Q3 [5]}}