Answer:
- initial size: 75
- doubling time: 7.51 minutes
- after 115 minutes: about 3,056,900
- reaches 11,000: 54.03 minutes
Explanation:
For given points (t1, y1), (t2, y2), I like to write the exponential function as ...
y(t) = y1·(y2/y1)^((t-t1)/(t2-t1))
This can be converted to other forms (such as a·b^t or a·e^(kt)) fairly easily, but those tend not to reproduce the given numbers exactly as this form does.
Using (15, 300) and (35, 1900) as our data values, the exponential function can be written as ...
y(t) = 300·(19/3)^((t-15)/20)
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a) The initial size of the culture is the value of y(0).
y(0) = 300·(19/3)^(-15/20) ≈ 75.144
y(0) ≈ 75 . . . initial population
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b) The doubling period will be the value of t that satisfies ...
(19/3)^(t/20) = 2
Taking logarithms, we have ...
(t/20)·log(19/3) = log(2)
t = 20·log(2)/log(19/3) ≈ 7.5104 . . . . minutes
The doubling time is about 7.51 minutes.
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c) Evaluating the formula for t=115, we have ...
y(115) = 300·(19/3)^(100/20) ≈ 3056912.346
The count after 115 minutes will be about 3,056,900.
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d) Solving y(t) = 11,000, we have ...
11000 = 300·(19/3)^((t-15)/20)
11000/300 = (19/3)^((t-15)/20)
log(110/3) = (t-15)/20·log(19/3)
t = 20·log(110/3)/log(19/3) + 15 ≈ 54.027
It will take about 54.03 minutes for the count to reach 11,000.
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I find a graphing calculator to be a nice tool for solving problems like this.