SOLUTION: Can you help please. The question reads: The number of bacteria present in a culture after t minutes is given as B=10e^kt. If there are 2167 bacteria present after 15 minutes, kin

Algebra ->  Customizable Word Problem Solvers  -> Numbers -> SOLUTION: Can you help please. The question reads: The number of bacteria present in a culture after t minutes is given as B=10e^kt. If there are 2167 bacteria present after 15 minutes, kin      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 31649: Can you help please.
The question reads: The number of bacteria present in a culture after t minutes is given as B=10e^kt. If there are 2167 bacteria present after 15 minutes, kind k.
(A) 0.384
(B) 80.678
(C) 0.359
(D) 5.379
Thanks for your help!
Found 2 solutions by mukhopadhyay, Earlsdon:
Answer by mukhopadhyay(490) About Me  (Show Source):
You can put this solution on YOUR website!
Based on the question,
2167 = 10e^15k
=> ln(2167) = ln(10) + 15k
=> 15k = ln(2167) - ln(10) = ln(2167/10) = ln(216.7)
=> k = (ln216.7)/15
=> k = .359
Correct answer: C

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for k:
B+=+10e%5E%28kt%29
Substitute B = 2167 and t = 15
2167+=+10e%5E15k%29 Divide both sides by 10.
216.7+=+e%5E%2815k%29 Take the natural log of both sides.
ln%28216.7%29+=+15kln%28e%29 Substitute ln%28e%29+=+1
ln%28216.7%29+=+15k Divide both sides by 15.
k+=+%28ln%28216.7%29%29%2F15
k+=+0.35856 Round to nearest thousandth.
k = 0.359