Respuesta :
Answer:
The size of the colony after 4 ​days is 8351.15.
8 days long there are 70,000 mosquitoes.
Step-by-step explanation:
Given : The population of a colony of mosquitoes obeys the law of uninhibited growth. If there are 1000 mosquitoes initially and there are 1700 after 1​ day.
To find : What is the size of the colony after 4 ​days and How long is it until there are 70,000 ​mosquitoes?
Solution :
Let the uninhibited growth is defined by a function,
[tex]A=A_0e^{kt}[/tex]
Where, [tex]A_0=1000[/tex] is the initial amount
e is the Euler's constant
k is the amount of increase
t=1 day is the time
A=1700 is the amount
Substitute all the values in the formula,
[tex]1700=1000e^{k(1)}[/tex]
[tex]\frac{1700}{1000}=e^{k(1)}[/tex]
[tex]1.7=e^{k}[/tex]
Taking natural log both side,
[tex]\ln(1.7)=\ln(e^{k})[/tex]
[tex]0.5306=k[/tex]
Now, The size of the colony after 4 ​days is
[tex]A=1000e^{(0.5306)(4)}[/tex]
[tex]A=1000e^{2.1224}[/tex]
[tex]A=1000\times 8.35115[/tex]
[tex]A=8351.15[/tex]
Therefore, The size of the colony after 4 ​days is 8351.15.
When there are 70,000 ​mosquitoes the time is
[tex]70000=1000e^{(0.5306)(t)}[/tex]
[tex]\frac{70000}{1000}=e^{(0.5306)(t)}[/tex]
[tex]70=e^{(0.5306)(t)}[/tex]
Taking ln both side,
[tex]\ln(70)=\ln(e^{0.5306t})[/tex]
[tex]4.248=0.5306t[/tex]
[tex]t=\frac{4.248}{0.5306}[/tex]
[tex]t=8[/tex]
Therefore, 8 days long there are 70,000 mosquitoes.
