An internet analytics company measured the number of people watching a video posted on a social media platform. The company found 129 people had watched the video and that the number of people who had watched it was increasing by 30% every 3 hours.
(a)
Write an exponential function for the number of people A who had watched the video n hours after the initial observation. (Enter a mathematical expression.)

Answer:

(a) The exponential function representing the number of people who had watched the video [tex]t[/tex] hours after the initial observation is [tex]n(t) = 129\cdot 1.3^{\frac{t}{3} }[/tex].

(b) As [tex]n(120) < 5\times 10^{6}[/tex], we conclude that this video is not going "viral".

Step-by-step explanation:

Statement is incomplete. The complete statement is:

An internet analytics company measured the number of people watching a video posted on a social media platform. The company found 129 people had watched the video and that the number of people who had watched it was increasing by 30% every 3 hours.

(a) Write an exponential function for the number of people A who had watched the video n hours after the initial observation.

(b) A video is said to go "viral" if the number of people who have watched the video exceeds 5 million within 5 days (120 hours). Would this video be considered to have gone viral?

(a) From the statement of the problem we get the following relationship:

[tex]\frac{n_{i+1}}{n_{i}}= r^{[(i+1)-i]}[/tex] (Eq. 1)

Where:

[tex]n_{i}[/tex] - i-th number of people watching a video, dimensionless.

[tex]n_{i+1}[/tex] - (i+1)-th number of people watching a video, dimensionless.

[tex]r[/tex] - Increase ratio, dimensionless.

For Induction Theorem, we get the following relatioship for [tex]i = 0[/tex]:

[tex]\frac{n_{1}}{n_{o}} = r^{1-0}[/tex] (Eq. 2)

For [tex]i = m[/tex], the following relationship is constructed:

[tex]\left(\frac{n_{1}}{n_{o}} \right)\cdot \left(\frac{n_{2}}{n_{1}} \right)\cdot ...\cdot \left(\frac{n_{m+1}}{n_{m}} \right) = r^{m}[/tex]

[tex]\frac{n_{m+1}}{n_{o}} = r^{m}[/tex](Eq. 3)

And for [tex]i = m+1[/tex], we have the following expression:

[tex]\left(\frac{n_{1}}{n_{0}} \right)\cdot \left(\frac{n_{2}}{n_{1}} \right)\cdot ...\cdot \left(\frac{n_{m+2}}{n_{m}} \right) = r^{m+1}[/tex]

[tex]\frac{n_{m+2}}{n_{0}} = r^{m+1}[/tex](Eq. 4)

If we multiply (Eq. 3) by the (m+1)-th ratio based on (Eq. 1):

[tex]\left(\frac{n_{m+1}}{n_{o}}\right)\cdot \left(\frac{n_{m+2}}{n_{m+1}} \right) = r^{m} \cdot r[/tex]

[tex]\frac{n_{m+2}}{n_{o}} = r^{m+1}[/tex]

Which is (Eq. 4) and the exponential function is represented by:

[tex]n_{i+1} = n_{o}\cdot r^{i}[/tex], [tex]i\in \mathbb{N}_{O}[/tex] (Eq. 5)

As the number of people is increased at constant rate every 3 hours, we get that [tex]i[/tex] is:

[tex]i = \frac{t}{3}[/tex], [tex]t \in \mathbb{R}[/tex], [tex]t \geq 0[/tex] (Eq. 6)

Where [tex]t[/tex] is the time, measured in hours.

Then, the exponential function is:

[tex]n(t) = n_{o}\cdot r^{\frac{t}{3} }[/tex] (Eq. 7)

Where [tex]n_{o}[/tex] is the initial number of people watching the video, dimensionless.

If we know that [tex]n_{o} = 129[/tex] and [tex]r = 1.3[/tex], then the exponential function representing the number of people who had watched the video [tex]t[/tex] hours after the initial observation is:

[tex]n(t) = 129\cdot 1.3^{\frac{t}{3} }[/tex]

(b) If we know that [tex]t = 120\,h[/tex], then we evaluate the exponential function:

[tex]n (120) = 129\cdot 1.3^{\frac{120}{3} }[/tex]

[tex]n(120) = 4.659\times 10^{6}[/tex]

As [tex]n(120) < 5\times 10^{6}[/tex], we conclude that this video is not going "viral".

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