SOLUTION: 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

Algebra ->  Graphs -> SOLUTION: 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      Log On


   

Question 212974This 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 100,000,000 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 Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Starting with 25,000 bacteria present at time zero, you can write:
N%28t%29+=+25000%282%29%5E%28t%2F20%29 Note: The exponent is really = t%2F20
At what time,t, will the bacteria count reach 100,000,000? Set N(t) = 100,000,000 and solve for t.
100000000+=+25000%282%29%5E%28t%2F20%29 Divide both sides by 25000.
4000+=+%282%29%5E%28t%2F20%29 Take the logarithm of both sides.
log%28%284000%29%29+=+log%28%282%29%29%5E%28t%2F20%29 Apply the power rule for logarithms.
log%28%284000%29%29+=+%28t%2F20%29log%28%282%29%29 Divide both sides by log%28%282%29%29
t%2F20+=+log%28%284000%29%29%2Flog%28%282%29%29 Multiply both sides by 20.
t+=+20%2Alog%28%284000%29%29%2Flog%28%282%29%29 Evaluate.
t+=+239.3156Minutes. Round up to:
t+=+240Minutes. Convert to hours. 240%2F60+=+4
t+=+4hours. Add this to 11:00am to get 15:00 hours. Subtract 12 hours.
Time = 3:00pm
Infection can occur at 3:00pm