Moderate -0.5 This is a standard linear transformation question requiring students to take logarithms of an exponential equation to obtain a straight line, then use two given points to find the gradient and intercept. The algebraic manipulation is straightforward with clear steps: ln y = ln A + px + p gives a linear form, find p from gradient, then find A from the y-intercept. Slightly easier than average as it's a well-practiced technique with no conceptual surprises.
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\includegraphics[max width=\textwidth, alt={}, center]{772c14a1-f79a-4147-a293-0ff34f930e20-04_577_569_260_788}
The variables \(x\) and \(y\) satisfy the equation \(y = A \mathrm { e } ^ { p x + p }\), where \(A\) and \(p\) are constants. The graph of \(\ln y\) against \(x\) is a straight line passing through the points \(( 1,2.835 )\) and \(( 6,6.585 )\), as shown in the diagram. Find the values of \(A\) and \(p\).
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\includegraphics[max width=\textwidth, alt={}, center]{772c14a1-f79a-4147-a293-0ff34f930e20-04_577_569_260_788}
The variables $x$ and $y$ satisfy the equation $y = A \mathrm { e } ^ { p x + p }$, where $A$ and $p$ are constants. The graph of $\ln y$ against $x$ is a straight line passing through the points $( 1,2.835 )$ and $( 6,6.585 )$, as shown in the diagram. Find the values of $A$ and $p$.\\
\hfill \mbox{\textit{CAIE P2 2019 Q3 [5]}}