| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| Topic | Exponential Equations & Modelling |
| Type | ln(y) vs x: find constants from two points |
| Difficulty | Moderate -0.8 This is a straightforward application of logarithms to linearize an exponential model. Part (i) requires taking ln of both sides (routine manipulation), and part (ii) involves reading two points from a graph and substituting into ln(P) = ln(a) + bt. This is a standard textbook exercise with no problem-solving insight required, making it easier than average. |
| Spec | 1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form |
**(i)**
State or imply $\ln P = \ln a + bt$ [B1]
State intercept $= \ln a$ [B1]
State gradient $= b$ [B1]
**Subtotal: [3]**
**(ii)**
Obtain $b = 2.5$ [B1]
Attempt to solve $\ln a = 2$ only [M1]
Obtain $a = e^2$ or $7.39$ [A1]
**Subtotal: [3]**
**Total: [6]**4 A population, $P$, is modelled by the equation $P = a \mathrm { e } ^ { b t }$ where $t$ is time in years, and $a$ and $b$ are constants.\\
(i) By considering logarithms, show that a graph of $\ln P$ against $t$ is a straight line. State the intercept on the vertical axis and the gradient.\\
(ii) Use the graph below to obtain values for $a$ and $b$.\\
\includegraphics[max width=\textwidth, alt={}, center]{816a16df-e3a5-48ae-84c6-7f6f5bbba9ca-2_657_750_1530_740}
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2015 Q4 [6]}}