Question 1100500
p(t) = 281 * e^(.09 * t)


t is the time in days.


the population will reach 1120 insects when p(t) = 1120.


formula becomes 1120 = 281 * e^(.09 * t)


divide both sides by 281 to get:
1120/281 = e^(.09 * t)
take the natural log of both sides of this equaiton to get:
ln(1120/281) = ln(e^(.09 * t))
this becomes:
ln(1120/281) = .09 * t * ln(e) which becomes:
ln(1120/281) = .09 * t
solve for t to get:
t = ln(1120/281) / .09 = 15.36365883.


the population will reach 1120 in 15.36365883 days.


281 * e^(.09 * 15.36365883) = 1120, confirming the solution is correct.


2 basic principles involved.


ln(e^x) = x*ln(e).
ln(e) = 1


to find out when the insect population will double, do the following.


p(t) = 281 * e^(.09 * t)


let p(t) = 2 * 281 = 562


formula becomes 562 = 281 * e^(.09 * t)


do the following:


divide both sides of the equation by 281 to get:
562/281 = e^(.09 * t)
simplify to get:
2 = e^(.09 * t)
take the natural log of both sides of the equation to get:
ln(2) = ln(e^(.09 * t))
this becomes:
ln(2) = .09 * t * ln(e) which becomes:
ln(2) = .09 * t
solve for t to get:
t = ln(2) / .09 = 7.70163534


the insect population will double in 7.70163534 days.


281 * e^(.09 * 7.70163534) = 562, confirming the solution is correct, because 562 is double 281.


rounding your solutions to the nearest integer results in.


population will reach 1120 in 15 years.
population will double in 8 years.