Question

A given bacteria culture initially contains 2500 bacteria and doubles every half hour. The number of bacteria p at a given time t is given by the formula p(t)=2500ekt for some constant k. (You will need to find k to answer the following.) (a) Find the size of the bacterial population after 50 minutes. (b) Find the size of the bacterial population after 8 hours.

Answer 1

After half hour=30 min you have 2×2500=5000 bacteria so you can use this into your equation to write:
$5000=2500 e^{30k}$
and:
$e^{30t}= \frac{5000}{2500}=2$
Apply the natural logarithm ln to both sides and get:
$30t=ln(2)$
so $k=0.0231$ in units of $\frac{1}{min}$
so your equation is now:
$p(t)=2500 e^{0.0231t}$
1] it t= 50 min substituting you get:
$p(50)=2500 e^{0.0231*50}=7935$ bacteria
2] if t= 8 h = 480 min you get:
$p(480)=2500 e^{0.0231*480}=1.6* 10^{8}$ bacteria