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Math Help - Help with log problem

  1. #1
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    Help with log problem

    I am not sure what to use for this problem, I think it is a log is there any formula I could use to solve this ? Any help appreciated, I am studying problems from old exams to practicing for my exam !!


    A ship is leaking oil, it doubles the area it covers every 7 days. If the initial leakage area is 15 square metres how long would it take to cover 25000 metres squared.
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  2. #2
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    I had typed out a nice explanation but firefox had an error so here it goes again. I have learned to copy and temporarily save as I type now.

    It seems to be an exponential function since it doubles every week, x. So 25000 = 15\cdot 2^x

    Dividing by 15, we get \frac{5000}{3} = 2^x

    Taking the log base two of each side, we get \log_2{\frac{5000}{3}} = \log_2{2^x} = x\cdot \log_2{2} = x

    So \boxed{x \approx 10.71~ weeks} or if needed in days, multiply by 7, of course.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by gregorio View Post
    I am not sure what to use for this problem, I think it is a log is there any formula I could use to solve this ? Any help appreciated, I am studying problems from old exams to practicing for my exam !!


    A ship is leaking oil, it doubles the area it covers every 7 days. If the initial leakage area is 15 square metres how long would it take to cover 25000 metres squared.
    After 7 days the covered area is 2 x 15 m^2
    After 14 days the covered area is 4 x 15 m^2
    After 21 days the covered area is 2^3 x 15 m^2

    After 7N days the covered area is 2^N x 15 m^2

    If after 7N days the area covered is 25000 m^2, we have:

    25000 = 2^N x 15.

    Now take logs (any base it does not matter):

    log(25000) = N log(2) + log(15),

    so:

    N = [log(25000) - log(15)]/log(2) = log(5000/3)/log(2) ~=10.7

    So the oil covers the required area after about 7 x 10.7 ~= 75 days

    RonL
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