Question 202641: A certain bacterial infection has 110,000 organisms at 12:00pm. At 12:30 pm 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 12:00pm. At 12:30 pm the population was 140,000. If this growth rate has been steady, when did the infection start?
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A(t) = Ao*(r)^t
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r is the rate of growth
t is the time in minutes
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140,000 = 110,000*r^(30)
14/11 = r^(30)
Take the log:
30*log(r) = log(14/11)
log(r) = [log(14/11)]/30
log(r) = 0.0034911784
r = 10^0.0034911784
r = 1.008071133 per minute
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1 = 110,000(1.008071133)^t
1.008071133)^t = 110,000^-1
t*log(1.998071133) = -log(110,000)
t = -16.7705 minutes
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12:00 - 16.77 minutes is approximately 11:43 am
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Cheers,
Stan H.
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