Respuesta :
The question is incomplete. Here is the complete question.
Uninhibited growth can be modeled by exponential functions other than [tex]A(t)=A_{0}e^{kt}[/tex]. for example, if an initial population P₀ requires n units of time to triple, then the function [tex]P(t)=P_{0}(3)^{\frac{t}{n} }[/tex] models the size of the population at time t. An insect population grows exponentially. Complete the parts a through d below.
a) If the population triples in 30 days, and 50 insects are present initially, write an exponential function of the form [tex]P(t)=P_{0}(3)^{\frac{t}{n} }[/tex] that models the population.
b) What will the population be in 47 days?
c) When wil the population reach 750?
d) Express the model from part (a) in the form [tex]A(t)=A_{0}e^{kt}[/tex].
Answer: a) [tex]P(t)=50(3)^{\frac{t}{30} }[/tex]
b) P(t) = 280 insects
c) t = 74 days
d) [tex]A(t)=50e^{0.037t}[/tex]
Step-by-step explanation:
a) n is time necessary to triple the population of insects, i.e., n = 30 and P₀ = 50. So, Exponential equation for growth is
[tex]P(t)=50(3)^{\frac{t}{30} }[/tex]
b) In t = 47 days:
[tex]P(t)=50(3)^{\frac{t}{30} }[/tex]
[tex]P(47)=50(3)^{\frac{47}{30} }[/tex]
[tex]P(47)=50(3)^{1.567}[/tex]
P(47) = 280
In 47 days, population of insects will be 280
c) P(t) = 750
[tex]750=50(3)^{\frac{t}{30} }[/tex]
[tex]\frac{750}{50}=(3)^{\frac{t}{30} }[/tex]
[tex](3)^{\frac{t}{n} }=15[/tex]
Using the property Power Rule of logarithm:
[tex]log(3)^{\frac{t}{30} }=log15[/tex]
[tex]\frac{t}{30}log(3)=log15[/tex]
[tex]t=\frac{log15}{log3} .30[/tex]
t = 74
To reach a population of 750 insects, it will take 74 days
d) To express the population growth into the described form, determine the constant k, using the following:
A(t) = 3A₀ and t = 30
[tex]A(t)=A_{0}e^{kt}[/tex]
[tex]3A_{0}=A_{0}e^{30k}[/tex]
[tex]3=e^{30k}[/tex]
Use Power Rule again:
[tex]ln3=ln(e^{30k})[/tex]
[tex]ln3=30k[/tex]
[tex]k=\frac{ln3}{30}[/tex]
k = 0.037
Equation for exponential growth will be:
[tex]A(t)=50e^{0.037t}[/tex]
