Question 212782This question is from textbook Elementary and Intermediate Algebra
: Growth of bacteria. The bacteria Escherichia coli (E. coli) are commonly found in the human bladder. Suppose that 3000 of the bacteria are present at time t=0. Then t minutes later, the number of bacteria present is N(t) = 3000(2)^t/20. if 100,000 bacteria accumulate, a bladder infection can occur. If, at 11:00am, a patient's bladder contains 25,000 E. coli bacteria, at what time can infection occur?
This question is from textbook Elementary and Intermediate Algebra
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Growth of bacteria. The bacteria Escherichia coli (E. coli) are commonly found in the human bladder. Suppose that 3000 of the bacteria are present at time t=0. Then t minutes later, the number of bacteria present is N(t) = 3000(2)^t/20. if 100,000 bacteria accumulate, a bladder infection can occur. If, at 11:00am, a patient's bladder contains 25,000 E. coli bacteria, at what time can infection occur?
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100,000 = 3000*2^(t/20)
2^(t/20) = 100/3
(t/20)ln2 = ln(100/3)
t = 20[ln(100/3)]/[ln(2)]
t = 101.18 minutes (time after time = 0 to reach 100,000)
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25000 = 3000*2^(t/2)
2^(t/2) = 25/3
t = 2[ln(25/3)/ln(2)]
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t = 6.18 minutes (time after time = 0 to reach 25000)
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Time from 25000 to 100000: 101.18-6.18 = 95 minutes
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95 minutes after 11:00 the time will be 12:35 PM
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Cheers,
Stan H.
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