pre calc question

Question

The doubling period of a bacterial population is  20 minutes. At time t=110 minutes, the bacterial population was 70000.

What was the initial population at time t=0?    

Find the size of the bacterial population after 4 hours. 

Answer 1

2= e^20r

ln2 =20r

r= ln2/20= about .0347

B=Bo(e^.0347t)

70,000=Bo(e^.0347(110))

Bo = 70,000/e^3.8123

= about 1,546.81= initial bacteria at time t=0

B= 1546,81e^.0347t

4 hours = 4(60)= 240 minutes

B= 1546.81e^.0347(240)

= about 6,400,909 after 4 hours