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Math Help - Basic logorithmic equation

  1. #1
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    Basic logorithmic equation

    The half-life of iodine-123 is about 13 hours. You begin with 50 grams of iodine-123.
    Write an equation that gives the amount of iodine-123 remaining after t-hours.


    It says that the equation ends up being I(t) = 50(.948)^t. In the middle of finding the final equation there is a part that goes like this: when t = 0 I(t) = 25 so 25 = 50B^13 then it goes into the final equation
    I(t) = 50(.948)^t.

    My question is how do you get .948 from B?
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  2. #2
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    Quote Originally Posted by Drozman03 View Post
    25 = 50B^13



    My question is how do you get .948 from B?
    25 = 50B^{13}

    0.5 = B^{13}

    I would use a spreadsheet from here to trial values for B
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  3. #3
    Member integral's Avatar
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    25 = 50b^{13}

    \frac{1}{2}=b^{13}


    <br /> <br />
b^x=c \therefore log_bc=x
    so
    <br />
log_b\frac{1}{2}=13

    \frac{ln(\frac{1}{2})}{ln(b)}=13
    (log base change formula. Log of base b a equals log a divided by log b or ln a divided by ln b (I like ln))

    \frac{ln(\frac{1}{2})}{13}=ln(b)


    ln\frac{1}{2}=-ln{2}
    because 1/2 is really 2^-1, powers always go before the log.
    b=e\frac{-ln(2)}{13}=.9481
    Last edited by integral; March 24th 2010 at 01:35 PM. Reason: explination/
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