Standard +0.3 This is a standard logarithmic linearization problem requiring students to take ln of both sides to get ln(y) = ln(K) + px·ln(2), identify the gradient as p·ln(2) and y-intercept as ln(K), then use two given points to find these values. It's a routine application of a well-practiced technique with straightforward arithmetic, making it slightly easier than average.
5
\includegraphics[max width=\textwidth, alt={}, center]{de8af872-9f77-4787-8e66-ed199405ca25-2_583_597_1457_772}
The variables \(x\) and \(y\) satisfy the equation \(y = K \left( 2 ^ { p x } \right)\), where \(K\) and \(p\) are constants. The graph of \(\ln y\) against \(x\) is a straight line passing through the points ( \(1.35,1.87\) ) and ( \(3.35,3.81\) ), as shown in the diagram. Find the values of \(K\) and \(p\) correct to 2 decimal places. [0pt]
[6]
\(1.87 = \ln K + 1.35p\ln 2\), \(3.81 = \ln K + 3.35p\ln 2\), \(p\ln 2 = \frac{3.81 - 1.87}{3.35 - 1.35}\) or equivalents
B1
Solve equation(s) to find one constant, dependent on previous B1
M1
Obtain \(p = 1.40\)
A1
Substitute to attempt value of \(K\)
DM1
Obtain \(\ln K = 0.5605\) and hence \(K = 1.75\)
A1 [6]
State or imply $\ln y = \ln K + px\ln 2$ | B1 |
Obtain at least one of:
$1.87 = \ln K + 1.35p\ln 2$, $3.81 = \ln K + 3.35p\ln 2$, $p\ln 2 = \frac{3.81 - 1.87}{3.35 - 1.35}$ or equivalents | B1 |
Solve equation(s) to find one constant, dependent on previous B1 | M1 |
Obtain $p = 1.40$ | A1 |
Substitute to attempt value of $K$ | DM1 |
Obtain $\ln K = 0.5605$ and hence $K = 1.75$ | A1 [6] |
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\includegraphics[max width=\textwidth, alt={}, center]{de8af872-9f77-4787-8e66-ed199405ca25-2_583_597_1457_772}
The variables $x$ and $y$ satisfy the equation $y = K \left( 2 ^ { p x } \right)$, where $K$ and $p$ are constants. The graph of $\ln y$ against $x$ is a straight line passing through the points ( $1.35,1.87$ ) and ( $3.35,3.81$ ), as shown in the diagram. Find the values of $K$ and $p$ correct to 2 decimal places.\\[0pt]
[6]
\hfill \mbox{\textit{CAIE P2 2014 Q5 [6]}}