Moderate -0.5 This is a standard logarithmic transformation question requiring students to recognize that ln(y) = ln(a) + x·ln(b) gives a linear relationship, then use two points to find the gradient and intercept. It involves routine algebraic manipulation and exponential/logarithm laws with no novel problem-solving, making it slightly easier than average.
2
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The variables \(x\) and \(y\) satisfy the equation \(y = a \left( b ^ { x } \right)\), where \(a\) and \(b\) are constants. The graph of \(\ln y\) against \(x\) is a straight line passing through the points ( \(0.75,1.70\) ) and ( \(1.53,2.18\) ), as shown in the diagram. Find the values of \(a\) and \(b\) correct to 2 decimal places.
State or imply $\ln y = \ln a + x \ln b$ | B1 |
Equate $\ln b$ to numerical gradient of line | M1 |
Obtain $b = 1.85$ | A1 |
Substitute to find value of $a$ | M1 |
Obtain $a = 3.45$ | A1 | [5]
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\includegraphics[max width=\textwidth, alt={}, center]{293e1e27-77e9-4b19-a152-96d71b75346e-2_654_693_532_724}
The variables $x$ and $y$ satisfy the equation $y = a \left( b ^ { x } \right)$, where $a$ and $b$ are constants. The graph of $\ln y$ against $x$ is a straight line passing through the points ( $0.75,1.70$ ) and ( $1.53,2.18$ ), as shown in the diagram. Find the values of $a$ and $b$ correct to 2 decimal places.
\hfill \mbox{\textit{CAIE P2 2014 Q2 [5]}}