Moderate -0.5 This is a straightforward logarithmic transformation question requiring students to recognize that ln(ky) = ln k + cx gives a linear relationship, then use two points to find the gradient (c) and y-intercept (ln k). The algebra is routine and the method is a standard textbook exercise, making it slightly easier than average.
\includegraphics{figure_4}
The variables \(x\) and \(y\) satisfy the equation \(ky = e^{cx}\), where \(k\) and \(c\) are constants. The graph of \(\ln y\) against \(x\) is a straight line passing through the points \((2.80, 0.372)\) and \((5.10, 2.21)\), as shown in the diagram.
Find the values of \(k\) and \(c\). Give each value correct to 2 significant figures. [4]
State or imply that lnk + lny = cx or lny = cx + ln etc.
Answer
Marks
Guidance
k
B1
Allow lnk + lny = cxlne
Carry out a completely correct method for finding lnk or c
M1
Equations must have been formulated correctly.
Obtain value c = 0.80
A1
AWRT
Allow 0.8 for 0.80.
Not a fraction.
Accept in the equation ky = ecx.
Answer
Marks
Guidance
Obtain value k = 6.5
A1
AWRT
Not a fraction.
Accept in the equation ky = ecx.
Answer
Marks
Guidance
Question
Answer
Marks
Question 4:
4 | 1
State or imply that lnk + lny = cx or lny = cx + ln etc.
k | B1 | Allow lnk + lny = cxlne
Carry out a completely correct method for finding lnk or c | M1 | Equations must have been formulated correctly.
Obtain value c = 0.80 | A1 | AWRT
Allow 0.8 for 0.80.
Not a fraction.
Accept in the equation ky = ecx.
Obtain value k = 6.5 | A1 | AWRT
Not a fraction.
Accept in the equation ky = ecx.
Question | Answer | Marks | Guidance
\includegraphics{figure_4}
The variables $x$ and $y$ satisfy the equation $ky = e^{cx}$, where $k$ and $c$ are constants. The graph of $\ln y$ against $x$ is a straight line passing through the points $(2.80, 0.372)$ and $(5.10, 2.21)$, as shown in the diagram.
Find the values of $k$ and $c$. Give each value correct to 2 significant figures. [4]
\hfill \mbox{\textit{CAIE P3 2024 Q4 [4]}}