Moderate -0.8 This is a straightforward logarithmic transformation question requiring students to recognize that ln(y) = ln(A) + n·ln(x) gives a linear relationship, then use two points to find the gradient (n) and y-intercept (ln A). The calculation is routine: n = (1.7-2.3)/(4.0-0) = -0.15, and A = e^2.3 ≈ 10. This is easier than average as it's a standard textbook exercise with clearly marked points and no conceptual obstacles.
3 Two variable quantities \(x\) and \(y\) are related by the equation
$$y = A x ^ { n }$$
where \(A\) and \(n\) are constants.
\includegraphics[max width=\textwidth, alt={}, center]{9b103197-7ba0-427a-b983-34edb51b6cca-2_422_697_977_740}
When a graph is plotted showing values of \(\ln y\) on the vertical axis and values of \(\ln x\) on the horizontal axis, the points lie on a straight line. This line crosses the vertical axis at the point ( \(0,2.3\) ) and also passes through the point (4.0,1.7), as shown in the diagram. Find the values of \(A\) and \(n\).
State or imply the relation \(\ln y = \ln A + n \ln x\)
B1
State or imply \(\ln A = 2.3\)
B1
Obtain answer \(A \approx 9.97\)
B1
Calculate gradient of the given line
M1
Obtain answer \(n = -0.15\)
A1
5
State or imply the relation $\ln y = \ln A + n \ln x$ | B1 |
State or imply $\ln A = 2.3$ | B1 |
Obtain answer $A \approx 9.97$ | B1 |
Calculate gradient of the given line | M1 |
Obtain answer $n = -0.15$ | A1 | 5
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3 Two variable quantities $x$ and $y$ are related by the equation
$$y = A x ^ { n }$$
where $A$ and $n$ are constants.\\
\includegraphics[max width=\textwidth, alt={}, center]{9b103197-7ba0-427a-b983-34edb51b6cca-2_422_697_977_740}
When a graph is plotted showing values of $\ln y$ on the vertical axis and values of $\ln x$ on the horizontal axis, the points lie on a straight line. This line crosses the vertical axis at the point ( $0,2.3$ ) and also passes through the point (4.0,1.7), as shown in the diagram. Find the values of $A$ and $n$.
\hfill \mbox{\textit{CAIE P2 2002 Q3 [5]}}