| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | ln(y) vs ln(x) linear graph |
| Difficulty | Moderate -0.8 This is a straightforward logarithmic linearization question requiring students to take logs of both sides to get ln(y) = ln(C) - n·ln(x), substitute two data points to form simultaneous equations, and solve for n and C. Part (ii) tests basic understanding that the log form is linear (y = mx + c). The algebra is routine and the concept is a standard textbook exercise with no novel problem-solving required. |
| Spec | 1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State or imply \(n\ln x + \ln y = \ln C\) | B1 | EITHER path |
| Substitute \(x\)- and \(y\)-values and solve for \(n\) | M1 | |
| Obtain \(n = 1.50\) | A1 | |
| Solve for \(C\) | M1 | |
| Obtain \(C = 6.00\) | A1 | [5] |
| Obtain two correct equations by substituting \(x\)- and \(y\)-values in \(x^n y = C\) | B1 | OR path |
| Solve for \(n\) | M1 | |
| Obtain \(n = 1.50\) | A1 | |
| Solve for \(C\) | M1 | |
| Obtain \(C = 6.00\) | A1 | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State that the graph of \(\ln y\) against \(\ln x\) has equation \(n\ln x + \ln y = \ln C\) which is *linear* in \(\ln y\) and \(\ln x\), or has equation of the form \(nX + Y = \ln C\), where \(X = \ln x\) and \(Y = \ln y\), and is thus a straight line | B1 | [1] |
## Question 3:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply $n\ln x + \ln y = \ln C$ | B1 | EITHER path |
| Substitute $x$- and $y$-values and solve for $n$ | M1 | |
| Obtain $n = 1.50$ | A1 | |
| Solve for $C$ | M1 | |
| Obtain $C = 6.00$ | A1 | [5] |
| Obtain two correct equations by substituting $x$- and $y$-values in $x^n y = C$ | B1 | OR path |
| Solve for $n$ | M1 | |
| Obtain $n = 1.50$ | A1 | |
| Solve for $C$ | M1 | |
| Obtain $C = 6.00$ | A1 | [5] |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State that the graph of $\ln y$ against $\ln x$ has equation $n\ln x + \ln y = \ln C$ which is *linear* in $\ln y$ and $\ln x$, or has equation of the form $nX + Y = \ln C$, where $X = \ln x$ and $Y = \ln y$, and is thus a straight line | B1 | [1] |
---3 The variables $x$ and $y$ satisfy the equation $x ^ { n } y = C$, where $n$ and $C$ are constants. When $x = 1.10$, $y = 5.20$, and when $x = 3.20 , y = 1.05$.\\
(i) Find the values of $n$ and $C$.\\
(ii) Explain why the graph of $\ln y$ against $\ln x$ is a straight line.
\hfill \mbox{\textit{CAIE P3 2010 Q3 [6]}}