| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2020 |
| Session | June |
| 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 conversion from log-linear form to exponential form using standard index/log laws (10^(log V) = V), then interpretation and substitution. Part (a) is routine manipulation, part (b) tests understanding of the model's meaning, and part (c) is direct substitution. Slightly easier than average as it follows a standard template with clear scaffolding and no problem-solving insight required. |
| Spec | 1.02z Models in context: use functions in modelling1.06c Logarithm definition: log_a(x) as inverse of a^x1.06f Laws of logarithms: addition, subtraction, power rules1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\log_{10} V = 0.072t + 2.379\) shown equivalent to \(V = ab^t\) | B1 | Either direction; shows intermediate lines |
| Correct equation in \(a\) or correct equation in \(b\) | M1 | E.g. \(\log_{10} a = 2.379\) or \(\log_{10} b = 0.072\) |
| \(a = 239\) or \(b = 1.18\) | A1 | Allow \(a = \text{awrt } 240\) or \(b = \text{awrt } 1.2\) |
| \(V = 239 \times 1.18^t\) or \(a = 239, b = 1.18\) stated | A1 | Not awrt; exact values required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(ab\) is the total number of views of the advert 1 day after it went live | B1 | Condone not seeing "total"; do not allow "number of views at the start" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitute \(t = 20\): \(V = 239 \times 1.18^{20}\) | M1 | Into either equation; uses correct method to find \(V\) |
| Awrt 6500 or 6600 | A1 |
# Question 12:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\log_{10} V = 0.072t + 2.379$ shown equivalent to $V = ab^t$ | B1 | Either direction; shows intermediate lines |
| Correct equation in $a$ or correct equation in $b$ | M1 | E.g. $\log_{10} a = 2.379$ or $\log_{10} b = 0.072$ |
| $a = 239$ or $b = 1.18$ | A1 | Allow $a = \text{awrt } 240$ or $b = \text{awrt } 1.2$ |
| $V = 239 \times 1.18^t$ or $a = 239, b = 1.18$ stated | A1 | Not awrt; exact values required |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $ab$ is the total number of views of the advert 1 day after it went live | B1 | Condone not seeing "total"; do not allow "number of views at the start" |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $t = 20$: $V = 239 \times 1.18^{20}$ | M1 | Into either equation; uses correct method to find $V$ |
| Awrt 6500 or 6600 | A1 | |
---\begin{enumerate}
\item An advertising agency is monitoring the number of views of an online advert.
\end{enumerate}
The equation
$$\log _ { 10 } V = 0.072 t + 2.379 \quad 1 \leqslant t \leqslant 30 , t \in \mathbb { N }$$
is used to model the total number of views of the advert, $V$, in the first $t$ days after the advert went live.\\
(a) Show that $V = a b ^ { t }$ where $a$ and $b$ are constants to be found.
Give the value of $a$ to the nearest whole number and give the value of $b$ to 3 significant figures.\\
(b) Interpret, with reference to the model, the value of $a b$.
Using this model, calculate\\
(c) the total number of views of the advert in the first 20 days after the advert went live. Give your answer to 2 significant figures.
\hfill \mbox{\textit{Edexcel AS Paper 1 2020 Q12 [7]}}