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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. |
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Subject:
Re: MATH PROBLEM
Answered By: maniac-ga on 21 Dec 2004 19:02 PST Rated: |
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:
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thanks for your help. would you two more? |
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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 |
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