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

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:

`(159.9)/(123)= e^(3b)`

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

`Ln((159.9)/(123)) = 3b`

`b = (Ln((159.9)/(123)))/(3) = 0.0874547`

And our model for this case would be:

`y= 123 e^(0.0874547t)`

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:

`(159.9)/(123)= e^(3b)`

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

`Ln((159.9)/(123)) = 3b`

`b = (Ln((159.9)/(123)))/(3) = 0.0874547`

And our model for this case would be:

`y= 123 e^(0.0874547t)`