Moderate -0.3 This is a standard logarithmic transformation question requiring students to recognize that ln y = ln k - x ln a gives a linear relationship, then read off the y-intercept (ln k) and gradient (-ln a) from the graph. While it requires understanding of logarithms and exponential relationships, it's a routine textbook exercise with clear steps and no novel problem-solving required. Slightly easier than average due to being a direct application of a standard technique.
\includegraphics{figure_2}
Two variable quantities \(x\) and \(y\) are related by the equation
$$y = k(a^{-x}),$$
where \(a\) and \(k\) are constants. Four pairs of values of \(x\) and \(y\) are measured experimentally. The result of plotting \(\ln y\) against \(x\) is shown in the diagram. Use the diagram to estimate the values of \(a\) and \(k\). [5]
State or imply at any stage \(\ln y = \ln k - x \ln a\)
B1
Equate estimate of \(\ln y\)-intercept to \(\ln k\)
M1
Obtain value for \(k\) in the range \(9.97 \pm 0.51\)
A1
Calculate gradient of the line of data points
M1
Obtain value for \(a\) in the range \(2.12 \pm 0.11\)
A1
Total: [5]
State or imply at any stage $\ln y = \ln k - x \ln a$ | B1 |
Equate estimate of $\ln y$-intercept to $\ln k$ | M1 |
Obtain value for $k$ in the range $9.97 \pm 0.51$ | A1 |
Calculate gradient of the line of data points | M1 |
Obtain value for $a$ in the range $2.12 \pm 0.11$ | A1 |
**Total: [5]**
\includegraphics{figure_2}
Two variable quantities $x$ and $y$ are related by the equation
$$y = k(a^{-x}),$$
where $a$ and $k$ are constants. Four pairs of values of $x$ and $y$ are measured experimentally. The result of plotting $\ln y$ against $x$ is shown in the diagram. Use the diagram to estimate the values of $a$ and $k$. [5]
\hfill \mbox{\textit{CAIE P2 2003 Q2 [5]}}