SOLUTION: The doubling period of a bacterial population is 20 minutes. At time t=90 minutes, the bacterial population was 50000. What was the initial population at time t=0 ?

Question 993293: The doubling period of a bacterial population is 20 minutes. At time t=90 minutes, the bacterial population was 50000.
What was the initial population at time t=0 ?
Find the size of the bacterial population after 3 hours
*formula: p(t)=a(b)^t
**Note:b=1+r

Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39623) (Show Source): You can put this solution on YOUR website!
and you have some information for doubling time. Twenty minutes is hour. Try to use just that much and see what you can find.

Try taking a as the initial population when t=0.
,
.

, or b=1.26 approximately.
The model might work as .

The next part of the description is the given point (1.5, 50000), and you want to know a, or value of p when t=0 instead of t=1.5.

and for convenience, recall where the "1.26" came from:

...the rendering does not look properly aligned but that is an equation, formula for a = fifty thousand over (cube root of two) to the three-halves power...

, simple radical form, although denominator is not rationalized.

This would be about .
Answer by MathTherapy(10555) (Show Source): You can put this solution on YOUR website!
The doubling period of a bacterial population is 20 minutes. At time t=90 minutes, the bacterial population was 50000.
What was the initial population at time t=0 ?
Find the size of the bacterial population after 3 hours
*formula: p(t)=a(b)^t
**Note:b=1+r

r, or growth rate = .035264924 ≈ 3.526%, per minute
Initial population at time = 0 minutes: 2,209.708691 ≈
Population at time = 3 hours (180 minutes): 1,131,370.8499 ≈
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