Question

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 1

Answer:

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

(b) As $n(120) < 5\times 10^{6}$, 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:

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

Where:

$n_{i}$ - i-th number of people watching a video, dimensionless.

$n_{i+1}$ - (i+1)-th number of people watching a video, dimensionless.

$r$ - Increase ratio, dimensionless.

For Induction Theorem, we get the following relatioship for $i = 0$:

$\frac{n_{1}}{n_{o}} = r^{1-0}$ (Eq. 2)

For $i = m$, the following relationship is constructed:

$\left(\frac{n_{1}}{n_{o}} \right)\cdot \left(\frac{n_{2}}{n_{1}} \right)\cdot ...\cdot \left(\frac{n_{m+1}}{n_{m}} \right) = r^{m}$

$\frac{n_{m+1}}{n_{o}} = r^{m}$(Eq. 3)

And for $i = m+1$, we have the following expression:

$\left(\frac{n_{1}}{n_{0}} \right)\cdot \left(\frac{n_{2}}{n_{1}} \right)\cdot ...\cdot \left(\frac{n_{m+2}}{n_{m}} \right) = r^{m+1}$

$\frac{n_{m+2}}{n_{0}} = r^{m+1}$(Eq. 4)

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

$\left(\frac{n_{m+1}}{n_{o}}\right)\cdot \left(\frac{n_{m+2}}{n_{m+1}} \right) = r^{m} \cdot r$

$\frac{n_{m+2}}{n_{o}} = r^{m+1}$

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

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

As the number of people is increased at constant rate every 3 hours, we get that $i$ is:

$i = \frac{t}{3}$, $t \in \mathbb{R}$, $t \geq 0$ (Eq. 6)

Where $t$ is the time, measured in hours.

Then, the exponential function is:

$n(t) = n_{o}\cdot r^{\frac{t}{3} }$ (Eq. 7)

Where $n_{o}$ is the initial number of people watching the video, dimensionless.

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

$n(t) = 129\cdot 1.3^{\frac{t}{3} }$

(b) If we know that $t = 120\,h$, then we evaluate the exponential function:

$n (120) = 129\cdot 1.3^{\frac{120}{3} }$

$n(120) = 4.659\times 10^{6}$

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