Question

Suppose the population of a certain bacteria in a laboratory sample is 100.if it doubles in the population every 5hrs,what is the qrow rate?how many bacteria will there be in 3 days?

Answer 1

Let `t` be the time in hours, `P_o` the initial population so `P=P_o e^{kt}` is the population at `t`. We're given `2=e^{5k}` so `k=1/5 ln 2`. We're asked for `P` at `t=3(24)=72` which is

#P= 100 e^{(1/5 ln 2)(72)} = 1638400 cdot 2^(2/5) approx 2161881#

Answer 2

Growth rate is `0.138629` and bacteria population after
`72` hours will be `2161882`

Explanation:

`B_0= 100 , B_5=200 , B_72= ? , k = ?`

`B_t = B_0*e^(kt); k , t` are rate of growth and time in hours.

`B_5 = B_0*e^(kt) or 200 =100 *e^(k*5)` or

`e^(5 k)= 200/100 or e^(5 k)= 2` . Taking natural log on both

sides we get , `5*k*ln (e)= ln (2) or k = ln(2)/5 ; [ln (e)=1]`

`:.k~~0.138629`. Growth rate is `0.138629` and growth

equation is `B_t = B_0*e^(0.138629*t)`

Bacteria population after `72` hours will be

`B_72 = 100*e^(0.138629*72)=2161882` [Ans]