Question 978184: The doubling period of a baterial population is 10 minutes. At time t = 110 minutes, the baterial population was 80000. Round your answers to at least 1 decimal place.
What was the initial population at time t = 0 ?
Find the size of the baterial population after 3 hours
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! A=Ao(1+r)^t, where t=number of 10 minute intervals
A/Ao=2=(1+r)^t
ln of both sides
0.693=t(ln(1+r)
t=1
0.6931=ln(1+r)
2=(1+r); r=1
80000=Ao(2)^11
ln 80000=ln Ao+11 ln(2)
11.2898=ln Ao + 7.623
3.6667=ln Ao
Take e of both sides.
e=39.12 or 39, initially
39[(2)^11]=79872, and with rounding, that is 80000
after 3 hours, t=18
39*2^18=
10,223,616. Note, must round down to 39 first, since can't have a fractional bacterium.
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