October 2020 Paper 1 Q12

EdexcelCurrent spec7 marksLogs & Exponentials Modelling

12. An advertising agency is monitoring the number of views of an online advert.

The equation

\[\log_{10} V = 0.072t + 2.379 \qquad 1 \leqslant t \leqslant 30,\ t \in \mathbb{N}\]

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 \(V = ab^t\) 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.

(4)

(b) Interpret, with reference to the model, the value of \(ab\). (1)

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. (2)

Mark scheme (a)

Scheme	Marks	AO
\(\log_{10} V = 0.072t + 2.379\) \(\Rightarrow V = 10^{0.072t + 2.379}\) \(\Rightarrow V = 10^{0.072t} \times 10^{2.379}\) or \(V = ab^t\) \(\Rightarrow \log_{10} V = \log_{10} a + \log_{10} b^t\) \(\Rightarrow \log_{10} V = \log_{10} a + t\log_{10} b\)	B1	2.1
States either \(a = 10^{2.379}\) or \(b = 10^{0.072}\) or States either \(\log_{10} a = 2.379\) or \(\log_{10} b = 0.072\)	M1	1.1b
\(a = 239\) or \(b = 1.18\)	A1	1.1b
Either \(V = 239 \times 1.18^t\) or imply by \(a = 239, b = 1.18\)	A1	1.1b
	(4)

Notes

Condone \(\log_{10}\) written \(\log\) or \(\lg\) written throughout the question

B1: Scored for showing that \(\log_{10} V = 0.072t + 2.379\) can be written in the form \(V = ab^t\) or vice versa
Either starts with \(\log_{10} V = 0.072t + 2.379\) (may be implied) and shows lines\[V = 10^{0.072t + 2.379} \text{ and } V = 10^{0.072t} \times 10^{2.379}\]Or starts with \(V = ab^t\) (implied) and shows the lines\[\log_{10} V = \log_{10} a + \log_{10} b^t \text{ and } \log_{10} V = \log_{10} a + t\log_{10} b\]

M1: For a correct equation in \(a\) or a correct equation in \(b\)

A1: Finds either constant. Allow \(a =\) awrt 240 or \(b =\) awrt 1.2 following a correct method

A1: Correct solution: Look for \(V = 239 \times 1.18^t\) or \(a = 239, b = 1.18\)
Note that this is NOT awrt

Mark scheme (b)

Scheme	Marks	AO
The value of \(ab\) is the (total) number of views of the advert 1 day after it went live.	B1	3.4
	(1)

Notes

B1: See scheme. Condone not seeing total. Do not allow number of views at the start or similar.

Mark scheme (c)

Scheme	Marks	AO
Substitutes \(t = 20\) in either equation and finds \(V\) Eg \(V = 239 \times 1.18^{20}\)	M1	3.4
Awrt 6500 or 6600	A1	1.1b
	(2)
	(7 marks)

Notes

M1: Substitutes \(t = 20\) in either their \(V = 239 \times 1.18^t\) or \(\log_{10} V = 0.072t + 2.379\) and uses a correct method to find \(V\)

A1: Awrt 6500 or 6600