Google Answers Logo
View Question
 
Q: MATH PROBLEM ( Answered 4 out of 5 stars,   5 Comments )
Question  
Subject: MATH PROBLEM
Category: Science
Asked by: bedhog-ga
List Price: $2.00
Posted: 21 Dec 2004 11:52 PST
Expires: 20 Jan 2005 11:52 PST
Question ID: 445643
The number of bacteria present in a culture after  T  minutes is given
as B=100e^kt (to the kt). If there are 52, bacteria present after 13
minutes, find  k.
Answer  
Subject: Re: MATH PROBLEM
Answered By: maniac-ga on 21 Dec 2004 19:02 PST
Rated:4 out of 5 stars
 
Hello Bedhog,

Let us walk through the formula and explain how to get to the correct answer.

[0] What you know...
  B = 100 * e^(k*t)
  B = 52 when t = 13

You must get the unknown value (k) on one side of the equation without
any other factors to "solve" the equation.

[1] Divide both sides by 100.
  B/100 = e^(k*t)

[2] Take the logarithm of both sides to eliminate the exponent.
  ln (B/100) = k*t

[3] Divide both sides by t
  ln (B/100)/t = k
or
  k = ln (B/100)/t

Now we have the unknown value (k) on one side and the other side has
only known values.

[4] Plug in the known values
  k = ln (52/100)/13
  52/100 = 0.52
  ln (0.52) = -0.653926467
  k = -0.653926467/13 = -0.050302036
or
  k = -0.050302036

[5] Cross check with original formula
  B = 100 * e^(k*t)

Does 52 = 100 * e^(-0.050302036*13)?

The result of the right hand side on the calculator I used was 52, so
this value of k satisifes the formula provided (to the precision of my
calculator). You may get slightly different answers with your
calculator - I suggest you repeat the steps to ensure you get an
answer similar to this one (it may have more / less precision).

Other examples like this one (and other methods to solve them) can be
found with search phrases such as
  solve equation exp
or a site like
  http://library.thinkquest.org/20991/alg2/log.html

If this answer is unclear or you need a more complete explanation,
please make a clarification request.

  --Maniac
bedhog-ga rated this answer:4 out of 5 stars and gave an additional tip of: $2.00
thanks for your help. would you two more?

Comments  
Subject: Re: MATH PROBLEM
From: research_help-ga on 21 Dec 2004 12:30 PST
 
B=100e^kt 
B=52
t=13

52=100e^13k
52/100=e^13k
.52=e^13k
ln .52= ln e^13k
ln .52=13k
(ln .52)/13=k
ln .52 = 1.68
1.68/13=k
k = .1292

Am I right? It's been a long time since I did that kind of math and
wanted to see if I could get it.
Subject: Re: MATH PROBLEM
From: balaqd-ga on 02 Jan 2005 07:21 PST
 
who said ln .52 = 1.68....are u joking?
Subject: Re: MATH PROBLEM
From: naesb81-ga on 03 Jan 2005 08:32 PST
 
ln .52=-.06539

As I recall, the ln of any number 0< n <1 will always be a negative number.
Subject: Re: MATH PROBLEM
From: semantiks-ga on 06 Jan 2005 19:55 PST
 
I'm not sure if this is correct, but it seems logical...

B=100e^kt
where B=52 and t=13

1) 52=100e^13k
2) 52/100=e^13k
3) ln(52/100)=13k
4) k=ln(52/100)/13
5) k=-0.05

Remember this is with time in minutes. If time is to be measured in
seconds, step 4 would have to be divided by 780 instead of 13.
Subject: Re: MATH PROBLEM
From: jacobolus-ga on 08 Jan 2005 22:42 PST
 
Here's the easiest way to get the answer to the original question

://www.google.com/search?hl=en&lr=&safe=off&q=ln%2852%2F100%29%2F13&btnG=Search

Important Disclaimer: Answers and comments provided on Google Answers are general information, and are not intended to substitute for informed professional medical, psychiatric, psychological, tax, legal, investment, accounting, or other professional advice. Google does not endorse, and expressly disclaims liability for any product, manufacturer, distributor, service or service provider mentioned or any opinion expressed in answers or comments. Please read carefully the Google Answers Terms of Service.

If you feel that you have found inappropriate content, please let us know by emailing us at answers-support@google.com with the question ID listed above. Thank you.
Search Google Answers for
Google Answers  


Google Home - Answers FAQ - Terms of Service - Privacy Policy