Standard +0.3 This is a standard logarithmic linearization problem requiring students to take logarithms of both sides, identify the gradient and intercept from two points, then solve for the constants. It's slightly easier than average as it's a routine textbook exercise with clear steps: ln both sides, use two-point formula for gradient, substitute back to find constants.
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The variables \(x\) and \(y\) satisfy the equation \(\mathrm { a } ^ { \mathrm { y } } = \mathrm { bx }\), where \(a\) and \(b\) are constants. The graph of \(y\) against \(\ln x\) is a straight line passing through the points ( \(0.336,1.00\) ) and ( \(1.31,1.50\) ), as shown in the diagram.
Find the values of \(a\) and \(b\). Give each value correct to the nearest integer.
Carry out a completely correct method for finding \(\ln a\) or \(\ln b\)
M1
E.g., from \(\ln a = \ln b + 0.336\), \(1.5\ln a = \ln b + 1.31\)
Obtain value \(a = 7\)
A1
Obtain value \(b = 5\)
A1
## Question 3:
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply that $y\ln a = \ln b + \ln x$ | B1 | |
| Carry out a completely correct method for finding $\ln a$ or $\ln b$ | M1 | E.g., from $\ln a = \ln b + 0.336$, $1.5\ln a = \ln b + 1.31$ |
| Obtain value $a = 7$ | A1 | |
| Obtain value $b = 5$ | A1 | |
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The variables $x$ and $y$ satisfy the equation $\mathrm { a } ^ { \mathrm { y } } = \mathrm { bx }$, where $a$ and $b$ are constants. The graph of $y$ against $\ln x$ is a straight line passing through the points ( $0.336,1.00$ ) and ( $1.31,1.50$ ), as shown in the diagram.
Find the values of $a$ and $b$. Give each value correct to the nearest integer.\\
\hfill \mbox{\textit{CAIE P3 2024 Q3 [4]}}