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
\includegraphics[max width=\textwidth, alt={}, center]{beb8df77-e091-4248-812b-20e885c42e37-2_453_771_386_685}
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 \(( 0,2.14 )\) and \(( 5,4.49 )\), as shown in the diagram. Find the values of \(A\) and \(b\), correct to 1 decimal place.
Attempt gradient of line or equivalent (or use of correct substitution)
M1
Obtain \(0.47 = \ln b\) or equivalent and hence \(b = 1.6\)
A1
[5]
State or imply that $\ln y = \ln A + x \ln b$ | B1 |
Equate intercept on y-axis to $\ln A$ | M1 |
Obtain $\ln A = 2.14$ and hence $A = 8.5$ | A1 |
Attempt gradient of line or equivalent (or use of correct substitution) | M1 |
Obtain $0.47 = \ln b$ or equivalent and hence $b = 1.6$ | A1 | [5]
2\\
\includegraphics[max width=\textwidth, alt={}, center]{beb8df77-e091-4248-812b-20e885c42e37-2_453_771_386_685}
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 $( 0,2.14 )$ and $( 5,4.49 )$, as shown in the diagram. Find the values of $A$ and $b$, correct to 1 decimal place.
\hfill \mbox{\textit{CAIE P2 2012 Q2 [5]}}