**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

#1**+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**+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