Question

An internet analytics company measured the number of people watching a video posted on a social media platform. The company found 123 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.) A = 125(1.3)^n

Answer 1

Answer:

For this case we know from the initial conditions that the value for a = 123 since is the starting value, so our model is given by:

$y = 123 e^{bt}$

Now we can use the other condition given "the number of people who had watched it was increasing by 30% every 3 hours", so for example after the first t =3 hours we will have a value for y = 1.3*123= 159.9, and using this condition we have this:

$159.9= 123 e^{3b}$

We can divide both sides by 123 and we got:

$\frac{159.9}{123}= e^{3b}$

Now we can apply natural log on both sides and we have:

$Ln(\frac{159.9}{123}) = 3b$

$b = \frac{Ln(\frac{159.9}{123})}{3} = 0.0874547$

And our model for this case would be:

$y= 123 e^{0.0874547t}$

Step-by-step explanation:

For this case we want to construct and exponential model given by this general expression:

$y = ae^{bt}$

Where a is the initial amount and b the growth/decay constant.

For this case we know from the initial conditions that the value for a = 123 since is the starting value, so our model is given by:

$y = 123 e^{bt}$

Now we can use the other condition given "the number of people who had watched it was increasing by 30% every 3 hours", so for example after the first t =3 hours we will have a value for y = 1.3*123= 159.9, and using this condition we have this:

$159.9= 123 e^{3b}$

We can divide both sides by 123 and we got:

$\frac{159.9}{123}= e^{3b}$

Now we can apply natural log on both sides and we have:

$Ln(\frac{159.9}{123}) = 3b$

$b = \frac{Ln(\frac{159.9}{123})}{3} = 0.0874547$

And our model for this case would be:

$y= 123 e^{0.0874547t}$