Question 107571: A certain bacterial infection has 110,000 organisms at 12:00 p.m. on Wednesday at 12:30 on the same day the population was 140,000. If this growth rate has been steady, when did the infection start?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A certain bacterial infection has 110,000 organisms at 12:00 p.m. on Wednesday at 12:30 on the same day the population was 140,000. If this growth rate has been steady, when did the infection start?
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Form: A(t)= ab^t where t is # of minutes after 12:00
From 12:00 to 12:30 is 30 minutes.
You have point (0,110000), so
110000 = ab^0
Therefore a = 110000
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You also have point (30,140,000)
Also: 140,000 = 110000b^30
1.2727.. = b^30
b = 1.272727...^(1/30)
b = 1.008071133...
EQUATION:
A(t) = 110000*1.008071133..^t
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When it started A(t)=1
1 = 110000*1.008071133..^t
1.008071133..^t = 1/110000
t = log(1/110000)/log(1.008071133)
t = -1444.037 minutes
The infection started 1444.037 minutes before 12:00
or 24.067 hours = 24 hrs 4 minutes before 12:00pm on Wednesday.
That would be 11:56 am on Tuesday.
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Cheers,
Stan H.
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