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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

User Lgriffiths
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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:


(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)

User Anthony Harley
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