| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | log(y) vs x: convert and interpret |
| Difficulty | Moderate -0.3 This is a straightforward exponential modelling question requiring standard logarithm manipulation (converting log-linear to exponential form), solving a logarithmic equation, and evaluating the model at a given time. All techniques are routine P3/C3 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.06g Equations with exponentials: solve a^x = b1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\log_{10} D = 1.04 + 0.38t \Rightarrow D = 10^{1.04 + 0.38t}\), or \(a = 10^{1.04}\) or \(b = 10^{0.38}\) | M1 | Correct application of log laws; may be implied by \(a \approx 11\) or \(b \approx 2.4\) |
| \(a = \text{awrt } 10.96\) or \(b = \text{awrt } 2.399\) | A1 | May be seen embedded in equation |
| \(D = 10.96 \times 2.399^t\) | A1 | Correct equation with correct awrt values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(45000 = \text{"10.96"} \times \text{"2.399"}^T \Rightarrow T = \ldots\) or \(\log_{10} 45000 = 1.04 + 0.38T \Rightarrow T = \ldots\) | M1 | Proceeds to \(T = \ldots\); must use correct form |
| \(\text{awrt } 9.51\) | A1 | Units not required but if given must be correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Substituting \(t = 12\) into \(D = \text{"10.96"} \times \text{"2.399"}^{12}\) or \(\log_{10} D = 1.04 + 0.38 \times 12\), or substituting \(D = 350000\) and solving for \(t\) | M1 | Any one valid substitution proceeding to a value |
| \(D = \text{awrt } \pounds400\,000 \Rightarrow\) yes, or \(t = \text{awrt } 11.9 \Rightarrow\) yes | A1 | Must include minimal conclusion e.g. "yes" |
# Question 3:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\log_{10} D = 1.04 + 0.38t \Rightarrow D = 10^{1.04 + 0.38t}$, or $a = 10^{1.04}$ or $b = 10^{0.38}$ | M1 | Correct application of log laws; may be implied by $a \approx 11$ or $b \approx 2.4$ |
| $a = \text{awrt } 10.96$ **or** $b = \text{awrt } 2.399$ | A1 | May be seen embedded in equation |
| $D = 10.96 \times 2.399^t$ | A1 | Correct equation with correct awrt values |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $45000 = \text{"10.96"} \times \text{"2.399"}^T \Rightarrow T = \ldots$ or $\log_{10} 45000 = 1.04 + 0.38T \Rightarrow T = \ldots$ | M1 | Proceeds to $T = \ldots$; must use correct form |
| $\text{awrt } 9.51$ | A1 | Units not required but if given must be correct |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| Substituting $t = 12$ into $D = \text{"10.96"} \times \text{"2.399"}^{12}$ or $\log_{10} D = 1.04 + 0.38 \times 12$, **or** substituting $D = 350000$ and solving for $t$ | M1 | Any one valid substitution proceeding to a value |
| $D = \text{awrt } \pounds400\,000 \Rightarrow$ yes, **or** $t = \text{awrt } 11.9 \Rightarrow$ yes | A1 | Must include minimal conclusion e.g. "yes" |\begin{enumerate}
\item The amount of money raised for a charity is being monitored.
\end{enumerate}
The total amount raised in the $t$ months after monitoring began, $\pounds D$, is modelled by the equation
$$\log _ { 10 } D = 1.04 + 0.38 t$$
(a) Write this equation in the form
$$D = a b ^ { t }$$
where $a$ and $b$ are constants to be found. Give each value to 4 significant figures.
When $t = T$, the total amount of money raised is $\pounds 45000$\\
According to the model,\\
(b) find the value of $T$, giving your answer to 3 significant figures.
The charity aims to raise a total of $\pounds 350000$ within the first 12 months of monitoring.\\
According to the model,\\
(c) determine whether or not the charity will achieve its aim.
\hfill \mbox{\textit{Edexcel P3 2024 Q3 [7]}}