| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Exponential relation to line equation |
| Difficulty | Moderate -0.8 This is a straightforward logarithmic transformation question requiring only routine application of log laws (log of both sides, bringing down powers, rearranging to y = mx + c form). The algebra is simple with no tricky manipulation, and finding the y-intercept requires just substituting x=0. This is easier than average A-level content—purely procedural with no problem-solving or insight required. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State or imply \((y + 1) \log 5 = 3x \log 2\) | M1 | |
| State that this is of the form \(\phi y = hx + c\) and thus a straight line, or equivalent | A1 | [2] |
| (ii) State gradient is \(\frac{3 \ln 2}{\ln 5}\), or equivalent, e.g. \(3\log_5 2\) | B1 | |
| State \((0, -1)\) | B1 | [2] |
**(i)** State or imply $(y + 1) \log 5 = 3x \log 2$ | M1 |
State that this is of the form $\phi y = hx + c$ and thus a straight line, or equivalent | A1 | [2]
**(ii)** State gradient is $\frac{3 \ln 2}{\ln 5}$, or equivalent, e.g. $3\log_5 2$ | B1 |
State $(0, -1)$ | B1 | [2]4 The variables $x$ and $y$ satisfy the equation $5 ^ { y + 1 } = 2 ^ { 3 x }$.\\
(i) By taking logarithms, show that the graph of $y$ against $x$ is a straight line.\\
(ii) Find the exact value of the gradient of this line and state the coordinates of the point at which the line cuts the $y$-axis.
\hfill \mbox{\textit{CAIE P2 2013 Q4 [4]}}