| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Linear transformation to find constants |
| Difficulty | Moderate -0.3 This is a standard logarithmic modeling question requiring finding a linear equation from two points, then converting back to exponential form. While it involves multiple parts and logarithms, each step follows routine procedures (gradient calculation, y-intercept identification, exponential conversion) with no novel problem-solving required. Slightly easier than average due to clear structure and given points. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| \(\log_{10}V = \frac{1}{25}t + \log_{10}20\) | M1, M1, A1 | Attempts to find the gradient of line \(l\); Uses the equation of a straight line in the form; Correct equation of line \(l\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(A = 20\), \(p = 10^{\frac{1}{25}}\) | M1, M1, A1, A1 | Attempts to rearrange their equation by taking exponentials or takes the log of both sides of the given equation; Completes rearrangement correctly so that both equations are in directly comparable form; States that \(A = 20\); States that \(p = 10^{\frac{1}{25}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(A\) is the initial value of the sculpture; \(p\) is the annual proportional increase in the value of the sculpture | B1, B1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(V = £5024\) | B1 | Substitute 60 into their formula from part b, correct answer |
### Part a:
$\log_{10}V = \frac{1}{25}t + \log_{10}20$ | M1, M1, A1 | Attempts to find the gradient of line $l$; Uses the equation of a straight line in the form; Correct equation of line $l$
### Part b:
$A = 20$, $p = 10^{\frac{1}{25}}$ | M1, M1, A1, A1 | Attempts to rearrange their equation by taking exponentials or takes the log of both sides of the given equation; Completes rearrangement correctly so that both equations are in directly comparable form; States that $A = 20$; States that $p = 10^{\frac{1}{25}}$
### Part c:
$A$ is the initial value of the sculpture; $p$ is the annual proportional increase in the value of the sculpture | B1, B1 | (2 marks)
### Part d:
$V = £5024$ | B1 | Substitute 60 into their formula from part b, correct answer\includegraphics{figure_4}
The value of a sculpture, $£V$, is modelled by the equation $V = Ap^t$, where $A$ and $p$ are constants and $t$ is the number of years since the value of the painting was first recorded on 1st January 1960.
The line $l$ shown in Figure 4 illustrates the linear relationship between $t$ and $\log_{10}V$ for $t \geq 0$.
The line $l$ passes through the point $(0, \log_{10}20)$ and $(50, \log_{10}2000)$.
\begin{enumerate}[label=(\alph*)]
\item Write down the equation of the line $l$. [3]
\item Using your answer to part a or otherwise, find the values of $A$ and $p$. [4]
\item With reference to the model, interpret the values of the constant $A$ and $p$. [2]
\item Use your model, to predict the value of the sculpture, on 1st January 2020, giving your answer to the nearest pounds. [1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 Q13 [10]}}