After a shipwreck, 120 rats manage to swim from the wreckage to a deserted island. The population on the island grows exponentially according to the model n(t)-ne", where n(t)is the number of rats at time t in months. After 15 months there are 280 rats on the island a. Find a function that models the population t months after the arrival of the rats. b. What will the population be 3 years after the shipwreck? Round to a whole number. c. When will the population reach 2000 rats?

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Complete Question

The  complete question is shown on the first uploaded image  

Answer:

a

  [tex]n(t) = 120 e^{(0.056t)}[/tex]

b

  [tex]n(36) = 901 \ rats[/tex]

c

 [tex]t = 50 \ months[/tex]

Step-by-step explanation:

From the question we are told that

    The  original number of  rats that swim from the wreckage to a deserted island is

   The population on the island grows exponentially according to the model

          [tex]n(t) = n_o e^{rt}[/tex]

     The  number of rats after t =  15 months is  280

So

     [tex]n(15) = 120 e^{15 * r } = 280[/tex]

=>   [tex]120 e^{15 * r } = 280[/tex]

=>   [tex]e^{15r} = 2.33[/tex]

=>  [tex]15r = 0.8459[/tex]

=>   [tex]r = 0.056[/tex]

Therefore the population t months after the arrival of the rats is mathematically represented as

      [tex]n(t) = 120 e^{(0.056t)}[/tex]

Here t is in months

Considering question b

   For  t =  3 years  =  12 * 3 =  36 months

   Then the number of rats that will be present is mathematically represented as

     [tex]n(36) = 120 e^{0.056 * 36 }[/tex]

      [tex]n(36) = 901 \ rats[/tex]

Considering question c

Now  the number of rats considered is  n(t) =  2000

So  

     [tex]n(t) = 2000 = 120e^{0.056t}[/tex]

=>    [tex]2000 = 120e^{0.056t}[/tex]

=>  [tex]16.67 = e^{0.056t}[/tex]

Taking natural log of both sides

=>   [tex]0.056 t = 2.81[/tex]

=>   [tex]t = 50 \ months[/tex]

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