Under ideal conditions, a population of e. coli bacteria can double every 20 minutes. This behavior can be modeled by the exponential function:
N(t)=N(lower case 0)(2^0.05t)
If the initial number of e. coli bacteria is 5, how many bacteria will be present in 4 hours?
2 answers
I am having a hard time trying to figure this out. I have watched a video on how to do it but unable to understand. If i know what to put where in the equation I would be able to figure it out. I know I have to change 4 hours to 240 minutes.
The equation would be:
n(t) = 5 (2^(t/20) , where t is in minutes
(note that t/20 = (1/20)t = .05t)
So they want the count after 4 hrs or 240 minutes
n(240) = 5 (2^(240/20))
= 5 (2^12)
= 5(4096)
= 20480
you could do this the long way:
now --- 5
after 20 min -- 10
after 40 min -- 20
after 60 min -- 40
after 80 min -- 80
after 100 min -- 160
after 120 min -- 320
after 140 min -- 640
after 160 min -- 1280
after 180 min -- 2560
after 200 min -- 5120
after 220 min -- 10240
after 240 min or after 4 hrs -- 20480
n(t) = 5 (2^(t/20) , where t is in minutes
(note that t/20 = (1/20)t = .05t)
So they want the count after 4 hrs or 240 minutes
n(240) = 5 (2^(240/20))
= 5 (2^12)
= 5(4096)
= 20480
you could do this the long way:
now --- 5
after 20 min -- 10
after 40 min -- 20
after 60 min -- 40
after 80 min -- 80
after 100 min -- 160
after 120 min -- 320
after 140 min -- 640
after 160 min -- 1280
after 180 min -- 2560
after 200 min -- 5120
after 220 min -- 10240
after 240 min or after 4 hrs -- 20480
1 answer