SOLUTION: The number of bacteria present in a culture after t minutes is given as B=10e^kt. If they are 593 bacteria present after nine minutes, find k

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Question 1081277: The number of bacteria present in a culture after t minutes is given as B=10e^kt. If they are 593 bacteria present after nine minutes, find k
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
593=10%2Ae%5E%284%2Ak%29

Take logarithms of both sides,... and solve for k. Use Natural Logarithm.

ln%28593%29=ln%2810%29%2Bln%28e%5E%284k%29%29
ln%28593%29=ln%2810%29%2B4k%2Aln%28e%29
4k%2A1%2Bln%2810%29=ln%28593%29
You finish the steps.
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Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
The number of bacteria present in a culture after t minutes is given as B=10e^kt. If they are 593 bacteria present after nine minutes, find k
B+=+10e%5E%28kt%29

593+=+10e%5E%289k%29 ------ Substituting 593 for B, and 9 for t

59.3+=+e%5E%289k%29 ------- Dividing each side by 10

9k+=+ln+%2859.3%29 ------ Converting to LOGARITHMIC (natural) form



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