Thread: Did I do this bacteria growth problem correctly?

A certain type of bacteria increases continuously at a rate proportional to the number present. if there are 500 bacteria at a given time and 1000 present 2 hours later, how many will there be 5 hours from the initial time given?

dP/dt = kP
dP=kPdt
∫dP/P = k ∫dt
ln|P| = kt + C
P = Ce^kt
P= P0e^kt
1000 = 500e^2k
2 = e^2k
ln2= 2k
k=ln(2) /2
P = 500e^5(ln2/2)
P=2828.4

Originally Posted by wmw88

A certain type of bacteria increases continuously at a rate proportional to the number present. if there are 500 bacteria at a given time and 1000 present 2 hours later, how many will there be 5 hours from the initial time given?

dP/dt = kP
dP=kPdt
∫dP/P = k ∫dt
ln|P| = kt + C
P = Ce^kt
P= P0e^kt
1000 = 500e^2k
2 = e^2k
ln2= 2k
k=ln(2) /2
P = 500e^5(ln2/2)
P=2828.4

looks fine.