MATH SOLVE

4 months ago

Q:
# 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, what is the size of the colony after 4 days? How long is it until there are 70,000 mosquitoes?

Accepted Solution

A:

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 amounte is the Euler's constantk is the amount of increaset=1 day is the timeA=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.