Respuesta :
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:
[tex] y = 123 e^{bt}[/tex]
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:
[tex] 159.9= 123 e^{3b}[/tex]
We can divide both sides by 123 and we got:
[tex]\frac{159.9}{123}= e^{3b}[/tex]
Now we can apply natural log on both sides and we have:
[tex] Ln(\frac{159.9}{123}) = 3b[/tex]
[tex] b = \frac{Ln(\frac{159.9}{123})}{3} = 0.0874547[/tex]
And our model for this case would be:
[tex] y= 123 e^{0.0874547t}[/tex]
Step-by-step explanation:
For this case we want to construct and exponential model given by this general expression:
[tex] y = ae^{bt}[/tex]
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:
[tex] y = 123 e^{bt}[/tex]
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:
[tex] 159.9= 123 e^{3b}[/tex]
We can divide both sides by 123 and we got:
[tex]\frac{159.9}{123}= e^{3b}[/tex]
Now we can apply natural log on both sides and we have:
[tex] Ln(\frac{159.9}{123}) = 3b[/tex]
[tex] b = \frac{Ln(\frac{159.9}{123})}{3} = 0.0874547[/tex]
And our model for this case would be:
[tex] y= 123 e^{0.0874547t}[/tex]
