SOLUTION: if a bacteria population doubles every eight minutes, what is the starting population if there are 450 in one hour.

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Question 1174889: if a bacteria population doubles every eight minutes, what is the starting population if there are 450 in one hour.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
p=I%282%29%5E%28x%2F8%29, x is time in minutes.
-
450=I%282%29%5E%2860%2F8%29
450=I%2A2%5E%2815%2F2%29
I=450%2F2%5E%2815%2F2%29
I=450/(2^(15/2))

Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
.

60 minutes comprise  60%2F8 = 15%2F2 = 7.5 doubling periods.


THEREFORE,


    450 = x%2A2%5E7.5,


where x is the starting number of bacteria.


From this equation,


    x = 450%2F2%5E7.5 = use your calculator = 2.49 bacteria.


You may round it as you want.  


I prefer to round to the closest greater integer number, which is 3.

Do not accept any other answer/solution,

and ignore the post by @josgarithmetic, which is INCORRECT and may confuse you.


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For bacteria growth problems, see the lesson
    - Bacteria growth problems
in this site.

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Logarithms".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.