- #1
KevinL
- 37
- 0
I feel like I'm doing everything correctly but my answer for (b) doesn't make any sense.
Assume that bacteria in aculture grows according to an exponential growth model. If the number of bacteria grows from 50 to 1000 in 12 hours:
a)How many bacteria will be present after 18 hours?
b)How long does i take for the number of bacteria to double?
A) db/dt = kb, b(0) = 50
b(t)=ce^(kt)
b(0)=50=ce^(k*0)
c=50
b(12)=50e^(k*12) = 1000
e^(k*12)=20
k=12*ln(20)
k=35.94 >>>> .359
Now that I have c and k I can find how much bacteria there is at 18 hours. So:
b(18) = 50e^(.359*18) = 32017
B) I am assuming they mean double as in get to 2000 bacteria. So:
2000=50e^(.359*t)
40=e^(.359t)
ln(40)/.359 = t
10.2 = t
How can it be at 2000 at 10 hours when I already know that 2 hours later its only at 1000? I must have screwed something up.
Assume that bacteria in aculture grows according to an exponential growth model. If the number of bacteria grows from 50 to 1000 in 12 hours:
a)How many bacteria will be present after 18 hours?
b)How long does i take for the number of bacteria to double?
A) db/dt = kb, b(0) = 50
b(t)=ce^(kt)
b(0)=50=ce^(k*0)
c=50
b(12)=50e^(k*12) = 1000
e^(k*12)=20
k=12*ln(20)
k=35.94 >>>> .359
Now that I have c and k I can find how much bacteria there is at 18 hours. So:
b(18) = 50e^(.359*18) = 32017
B) I am assuming they mean double as in get to 2000 bacteria. So:
2000=50e^(.359*t)
40=e^(.359t)
ln(40)/.359 = t
10.2 = t
How can it be at 2000 at 10 hours when I already know that 2 hours later its only at 1000? I must have screwed something up.