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The population of a colony of a certain insect obeys the law of uninhibited growth. If there are 500 insects initially and there are 800 after one day, what is the size of the colony after 3 days?

Guest Mar 4, 2015

Best Answer 

 #1
avatar+93333 
+5

 

$$\\P=500e^{kt}\\\\
800=500e^{k*1}\\\\
1.6=e^k\\\\
ln(1.6)=k\\\\
so\\\\
P=500e^{ln(1.6)*t}\\\\
when\;\; t=3\\\\
P=500e^{ln(1.6)*3}\\\\
P=500e^{3ln(1.6)}\\\\$$

 

$${\mathtt{500}}{\mathtt{\,\times\,}}{{\mathtt{e}}}^{\left({\mathtt{3}}{\mathtt{\,\times\,}}{ln}{\left({\mathtt{1.6}}\right)}\right)} = {\mathtt{2\,047.999\: \!999\: \!999\: \!999\: \!5}}$$

 

There will  be 2048 insects.   

Melody  Mar 4, 2015
 #1
avatar+93333 
+5
Best Answer

 

$$\\P=500e^{kt}\\\\
800=500e^{k*1}\\\\
1.6=e^k\\\\
ln(1.6)=k\\\\
so\\\\
P=500e^{ln(1.6)*t}\\\\
when\;\; t=3\\\\
P=500e^{ln(1.6)*3}\\\\
P=500e^{3ln(1.6)}\\\\$$

 

$${\mathtt{500}}{\mathtt{\,\times\,}}{{\mathtt{e}}}^{\left({\mathtt{3}}{\mathtt{\,\times\,}}{ln}{\left({\mathtt{1.6}}\right)}\right)} = {\mathtt{2\,047.999\: \!999\: \!999\: \!999\: \!5}}$$

 

There will  be 2048 insects.   

Melody  Mar 4, 2015
 #2
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A population of insects obeys the law of uninhibited growth. If there are 650 insects to begin with, and there are 800 after 2 days, how many will theure be after 5 days?

Guest Nov 15, 2016

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