SOLUTION: A bacteria culture grows by the exponential model y = 200ekt. How many bacteria
are there initially? If the number of bacteria triples in 2 hours, find the number of
bacteria af
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-> SOLUTION: A bacteria culture grows by the exponential model y = 200ekt. How many bacteria
are there initially? If the number of bacteria triples in 2 hours, find the number of
bacteria af
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Question 252075: A bacteria culture grows by the exponential model y = 200ekt. How many bacteria
are there initially? If the number of bacteria triples in 2 hours, find the number of
bacteria after 5 hours. Found 2 solutions by drk, MRperkins:Answer by drk(1908) (Show Source):
You can put this solution on YOUR website! All exponential models are of the form where A is "initial value", k is "growth rate", t is time units, and Y is "ending value".
So, initially there were 200 bacteria. Since we triple, Y = 3*200 = 600. Now for time; t = 2. We don't know the growth rate, k. This is step 1 in these kind of problems.
Our formula now becomes first then and taking a natural log (LN) of both sides we get ln(3) = 2k, and finally solving for k, we get k = ln(3)/2.
Knowing k, we can now find how many there are after 5 hours.
Y = 200*e^(5*ln(3)/2) becomes Y = 3118 rounded to the nearest bacteria.
You can put this solution on YOUR website! Ok, I have worked this problem for you, scanned it and just need a place to send it. You find k using the tripling at t=2 and then once you have k, you can find y when t=5. I get k=(ln 3)/2.
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Anyone wishing to get a copy of this .pdf file, email me with algebra in the subject line and ask for pdf file Dec 220002. I will send the file without obligation.