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?

2 Answers
Jun 26, 2018

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#

Jun 26, 2018

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]