| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Finding x from given y value |
| Difficulty | Moderate -0.3 This is a straightforward multi-part exponential modelling question requiring routine substitution (part a), logarithmic rearrangement (part b), algebraic manipulation to show a given result (part c), and simple iteration (part d). All techniques are standard C3 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.06g Equations with exponentials: solve a^x = b1.06i Exponential growth/decay: in modelling context1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks |
|---|---|
| \(P = 30 + 50e^{0.002 \times 30} = 83.1\) | M1 |
| ∴ population = 83 100 (3sf) | A1 |
| Answer | Marks |
|---|---|
| \(30 + 50e^{0.002t} > 84\) | M1 |
| \(e^{0.002t} > \frac{54}{50}\) | A1 |
| \(t > \frac{1}{0.002}\ln\frac{54}{50}, t > 38.5\) ∴ 2018 | M1 A1 |
| Answer | Marks |
|---|---|
| \(30 + 50e^{0.003t} = 26 + 50e^{0.002t}, \frac{50e^{0.003t}}{50e^{0.002t}} - 50e^{0.002t} = 4\) | M1 |
| \(e^{0.001t} - 1 = 0.08e^{-0.002}\) | |
| \(0.001t = \ln(1 + 0.08e^{-0.002})\) | M1 |
| \(t = 1000\ln(1 + 0.08e^{-0.002})\) | A1 |
| Answer | Marks |
|---|---|
| \(t_1 = 69.887, t_2 = 67.251, t_3 = 67.595\) | M1 A2 |
| ∴ 2047 | A1 |
| (13) |
**Part (a)**
$P = 30 + 50e^{0.002 \times 30} = 83.1$ | M1 |
∴ population = 83 100 (3sf) | A1 |
**Part (b)**
$30 + 50e^{0.002t} > 84$ | M1 |
$e^{0.002t} > \frac{54}{50}$ | A1 |
$t > \frac{1}{0.002}\ln\frac{54}{50}, t > 38.5$ ∴ 2018 | M1 A1 |
**Part (c)**
$30 + 50e^{0.003t} = 26 + 50e^{0.002t}, \frac{50e^{0.003t}}{50e^{0.002t}} - 50e^{0.002t} = 4$ | M1 |
$e^{0.001t} - 1 = 0.08e^{-0.002}$ |
$0.001t = \ln(1 + 0.08e^{-0.002})$ | M1 |
$t = 1000\ln(1 + 0.08e^{-0.002})$ | A1 |
**Part (d)**
$t_1 = 69.887, t_2 = 67.251, t_3 = 67.595$ | M1 A2 |
∴ 2047 | A1 |
| (13) |
---\begin{enumerate}
\item The population in thousands, $P$, of a town at time $t$ years after $1 ^ { \text {st } }$ January 1980 is modelled by the formula
\end{enumerate}
$$P = 30 + 50 \mathrm { e } ^ { 0.002 t }$$
Use this model to estimate\\
(a) the population of the town on $1 { } ^ { \text {st } }$ January 2010,\\
(b) the year in which the population first exceeds 84000 .
The population in thousands, $Q$, of another town is modelled by the formula
$$Q = 26 + 50 \mathrm { e } ^ { 0.003 t }$$
(c) Show that the value of $t$ when $P = Q$ is a solution of the equation
$$t = 1000 \ln \left( 1 + 0.08 \mathrm { e } ^ { - 0.002 t } \right) .$$
(d) Use the iteration formula
$$t _ { n + 1 } = 1000 \ln \left( 1 + 0.08 \mathrm { e } ^ { - 0.002 t _ { n } } \right)$$
with $t _ { 0 } = 50$ to find $t _ { 1 } , t _ { 2 }$ and $t _ { 3 }$ and hence, the year in which the populations of these two towns will be equal according to these models.\\
\hfill \mbox{\textit{Edexcel C3 Q6 [13]}}