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Math Help - Exponential decay

  1. #1
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    Mar 2008
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    Exponential decay

    i have the table
    t Caffine
    0 125
    1 111
    2 98
    3 87
    4 77
    5 69
    6 61
    7 54
    8 47
    9 42
    10 38
    11 34
    12 30
    13 26
    14 23
    15 21
    16 18
    17 16
    18 14
    19 13
    20 11
    21 10
    22 9
    23 8
    24 7

    The time is in hours and the caffine is the decrease in milligrams. Doing this in excel, i got the exponential decay formula of y = 124.96e-0.1202t which seems to be correct because if i do it to like 2 hours i get
    124.96e^((-0.1202)*2)=98.26, which corresponds to our table showing 98.
    Now i wanted to see how long it would take for the caffine to reach 0 milligrams. I am trying to use the example

    Problem 2: According to our model, when will the population reach 300 thousand?

    To solve this problem we set 100 e0.08t equal to 300 and solve for t.

    100 e0.08t = 300

    e0.08t = 3 Take the natural logarithm of both sides.

    ln e0.08t = ln 3

    0.08t = ln 3

    t = (ln 3)/0.08 = 13.73, approximately.

    Therefore, the population is expected to reach 300 thousand about three fourths of the way through the year 1993.


    Obviously this doesnt relate to my problem, but it shows a way to do it. So if i set my formula to 124.96e^((-0.1202)t)=0 this means i would first have to divide both sides by 124.96 which doesnt really work on the side which is 0, cos when i come to take the natural logarithm, i get ln0 which returns an error on my calculator. Any advice on how i can find out when the caffine miligrams will be 0?
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  2. #2
    Member
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    Feb 2008
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    Westwood, Los Angeles, CA
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    In exponential decay, the y-value (your amount of caffeine) never reaches zero. A percentage of it is always decaying (disappearing) but a percentage always remains. It's like if you start with 1 and keep dividing it by 2. No matter how many times you divide it by 2, you still have some fraction remaining - it's never zero.
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