Moderate -0.3 This is a standard logarithmic transformation question requiring students to recognize that ln(y) = ln(A) - x·ln(b) gives a straight line, find the gradient and intercept from two points, then solve for A and b. It's slightly easier than average as it's a routine textbook exercise with clear steps and no conceptual surprises, though it does require careful algebraic manipulation.
5
\includegraphics[max width=\textwidth, alt={}, center]{0355f624-3a35-4b9e-8520-af011a0fb6db-3_512_732_251_705}
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 \(( 1,2.9 )\) and \(( 3.5,1.4 )\), as shown in the diagram. Find the values of \(A\) and \(b\), correct to 2 decimal places.
Form a numerical expression for the gradient of the line
M1
Obtain \(b = 1.82\)
A1
Use gradient and one point correctly to find \(\ln A\)
M1
Obtain \(\ln A = 3.5\)
A1
Obtain \(A = 33.12\)
A1
[6]
State or imply $\ln y = \ln A - x \ln b$ | B1 |
Form a numerical expression for the gradient of the line | M1 |
Obtain $b = 1.82$ | A1 |
Use gradient and one point correctly to find $\ln A$ | M1 |
Obtain $\ln A = 3.5$ | A1 |
Obtain $A = 33.12$ | A1 |
| | [6] |
5\\
\includegraphics[max width=\textwidth, alt={}, center]{0355f624-3a35-4b9e-8520-af011a0fb6db-3_512_732_251_705}
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 $( 1,2.9 )$ and $( 3.5,1.4 )$, as shown in the diagram. Find the values of $A$ and $b$, correct to 2 decimal places.
\hfill \mbox{\textit{CAIE P2 2012 Q5 [6]}}