Standard +0.3 This is a standard logarithmic transformation question requiring students to convert y = A × B^(ln x) to linear form ln y = ln A + (ln B)(ln x), identify 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 given points, requiring only substitution and basic algebra.
2
\includegraphics[max width=\textwidth, alt={}, center]{873a104f-e2e2-49bb-b943-583769728fbb-04_554_493_260_826}
The variables \(x\) and \(y\) satisfy the equation \(y = A \times B ^ { \ln x }\), where \(A\) and \(B\) are constants. The graph of \(\ln y\) against \(\ln x\) is a straight line passing through the points (2.2, 4.908) and (5.9, 11.008), as shown in the diagram. Find the values of \(A\) and \(B\) correct to 2 significant figures.
State or imply equation \(\ln y = \ln A + \ln B \ln x\)
B1
Equate gradient of line to \(\ln B\)
M1
Obtain \(\ln B = 1.6486...\) and hence \(B = 5.2\)
A1
Substitute appropriate values to find \(\ln A\)
M1
Obtain \(\ln A = 1.2809...\) and hence \(A = 3.6\)
A1
Or method (simultaneous equations via logs):
Answer
Marks
Guidance
Answer
Mark
Guidance
State or imply equation \(\ln y = \ln A + \ln B \ln x\)
B1
Use given coordinates to obtain a correct equation
B1
Equations are \(4.908 = \ln A + 2.2\ln B\) and \(11.008 = \ln A + 5.9\ln B\)
Use given coordinates to obtain a second correct equation and attempt to solve both simultaneously to obtain at least one of the unknowns \(\ln A\) or \(\ln B\)
M1
Obtain \(\ln B = 1.6486...\) and hence \(B = 5.2\)
A1
Obtain \(\ln A = 1.2809...\) and hence \(A = 3.6\)
A1
Or method (without logs):
Answer
Marks
Guidance
Answer
Mark
Guidance
Use given coordinates to obtain a correct equation
B1
Equations are \(e^{4.908} = AB^{2.2}\) and \(e^{11.008} = AB^{5.9}\)
Use given coordinates to obtain a second correct equation
B1
Solve to obtain \(B\)
M1
M mark dependent on both previous B marks
\(B = 5.2\)
A1
\(A = 3.6\)
A1
Total
5
## Question 2:
**Either method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply equation $\ln y = \ln A + \ln B \ln x$ | B1 | |
| Equate gradient of line to $\ln B$ | M1 | |
| Obtain $\ln B = 1.6486...$ and hence $B = 5.2$ | A1 | |
| Substitute appropriate values to find $\ln A$ | M1 | |
| Obtain $\ln A = 1.2809...$ and hence $A = 3.6$ | A1 | |
**Or method (simultaneous equations via logs):**
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply equation $\ln y = \ln A + \ln B \ln x$ | B1 | |
| Use given coordinates to obtain a correct equation | B1 | Equations are $4.908 = \ln A + 2.2\ln B$ and $11.008 = \ln A + 5.9\ln B$ |
| Use given coordinates to obtain a second correct equation and attempt to solve both simultaneously to obtain at least one of the unknowns $\ln A$ or $\ln B$ | M1 | |
| Obtain $\ln B = 1.6486...$ and hence $B = 5.2$ | A1 | |
| Obtain $\ln A = 1.2809...$ and hence $A = 3.6$ | A1 | |
**Or method (without logs):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Use given coordinates to obtain a correct equation | B1 | Equations are $e^{4.908} = AB^{2.2}$ and $e^{11.008} = AB^{5.9}$ |
| Use given coordinates to obtain a second correct equation | B1 | |
| Solve to obtain $B$ | M1 | M mark dependent on both previous B marks |
| $B = 5.2$ | A1 | |
| $A = 3.6$ | A1 | |
| **Total** | **5** | |
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2\\
\includegraphics[max width=\textwidth, alt={}, center]{873a104f-e2e2-49bb-b943-583769728fbb-04_554_493_260_826}
The variables $x$ and $y$ satisfy the equation $y = A \times B ^ { \ln x }$, where $A$ and $B$ are constants. The graph of $\ln y$ against $\ln x$ is a straight line passing through the points (2.2, 4.908) and (5.9, 11.008), as shown in the diagram. Find the values of $A$ and $B$ correct to 2 significant figures.\\
\hfill \mbox{\textit{CAIE P2 2018 Q2 [5]}}