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Math Help - Logirithms-- Doubling problem

  1. #1
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    Logirithms-- Doubling problem

    If some one could answer this it would be great.

    A population of bacteria is quadrupling in a glass of milk on the counter every half hour.

    A)Write a function for this growth P(t), where "t" is in hours
    B)Write a function for this growth P(t), where "t" is in minutes
    c)If the current bacteria population is 250000, how many bacteria were there 58 minutes ago?

    Thanks!
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  2. #2
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    Re: Logirithms-- Doubling problem

    Quote Originally Posted by Gurp925 View Post
    If some one could answer this it would be great.
    A population of bacteria is quadrupling in a glass of milk on the counter every half hour.

    A)Write a function for this growth P(t), where "t" is in hours
    B)Write a function for this growth P(t), where "t" is in minutes
    c)If the current bacteria population is 250000, how many bacteria were there 58 minutes ago?
    Surely, you are not asking us to do your work for you. Are you?

    So that means you have tried something?
    Post it so we can see where you are.
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  3. #3
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    Re: Logirithms-- Doubling problem

    hello
    I'm just trying..don't know whether right or not..
    (a) P(t)=4^(2t)
    (b) P(t)=4^(t/30)
    (c) 4^(t/30)=250000
    4^(t/30 -58/30)=250000(4^(-58/30))
    =17137.89
    Thanks from Gurp925
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  4. #4
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    Re: Logirithms-- Doubling problem

    Quote Originally Posted by Trefoil2727 View Post
    hello
    I'm just trying..don't know whether right or not..
    (a) P(t)=4^(2t)
    (b) P(t)=4^(t/30)
    A good start but P(0)= 4^0= 1 and you were not told that. Put in a constant to represent the initial number of bacteria.

    (c) 4^(t/30)=250000
    No, the problem said "the current bacteria is 250000". That is not true for any time, t. There are two ways to do this:
    1) Take t= 0 as "current", find the constant so that P(0)= 250000, and then find P(-58).
    2) Take t= 58 as "current", find the constant so that P(58)= 250000, and then find P(0).

    4^(t/30 -58/30)=250000(4^(-58/30))
    =17137.89
    Thanks from Gurp925
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  5. #5
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    Re: Logirithms-- Doubling problem

    This is what I got

    A)P(t)=A(4^t.5)
    B)P(t)=A(4^t/30)
    Last edited by Gurp925; May 11th 2013 at 01:59 PM.
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  6. #6
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    Re: Logirithms-- Doubling problem

    So for C I solve For P(0) for the constant, then when i have the constant I can solve for P(-58)? But when solving for P(0)= 250000 I get P(0)=250000=A(240^0/30) It turns to the same answer of course so what am I doing wrong?

    This is what I set up P(-58)=250000(4^-58/30)=17137.89 Seems as though it can be right seeing as it is less then 250000 lol.
    Last edited by Gurp925; May 11th 2013 at 02:02 PM.
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  7. #7
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    Re: Logirithms-- Doubling problem

    Quote Originally Posted by Plato View Post
    Surely, you are not asking us to do your work for you. Are you?


    So that means you have tried something?
    Post it so we can see where you are.

    Of course not my mistake for not posting my work I apologise, I have posted my work above.
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