Moderate -0.3 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. While it involves multiple steps (finding gradient, finding intercept, exponentiating), these are routine techniques practiced extensively in P2 with no novel problem-solving required, making it slightly easier than average.
5
\includegraphics[max width=\textwidth, alt={}, center]{e814d76c-8757-4cc4-a69c-e3636b4cab16-2_604_887_1667_628}
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 ( \(1.4,0.8\) ) and ( \(2.2,1.2\) ), 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 \(b = 1.65\)
A1
Use gradient and one point correctly to find \(\ln A\)
M1
Obtain \(\ln A = 0.1\)
A1
Obtain \(A = 1.11\)
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 $b = 1.65$ | A1 |
| Use gradient and one point correctly to find $\ln A$ | M1 |
| Obtain $\ln A = 0.1$ | A1 |
| Obtain $A = 1.11$ | A1 | [6] |
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\includegraphics[max width=\textwidth, alt={}, center]{e814d76c-8757-4cc4-a69c-e3636b4cab16-2_604_887_1667_628}
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 ( $1.4,0.8$ ) and ( $2.2,1.2$ ), as shown in the diagram. Find the values of $A$ and $b$, correct to 2 decimal places.
\hfill \mbox{\textit{CAIE P2 2010 Q5 [6]}}