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Thread: A couple of Exponential Functions Questions

  1. #1
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    A couple of Exponential Functions Questions

    8). The oldest rock knows to data found in Minnesota contains 2g of carbon 14(C^14). Using geological methods toestimate the original size of the rock, scentists estimate it to originally have contained 4.096kg of C^14 .The Half-life of c^14 is 5760 years.What is the approximate age of rock?

    9). Land developers had to delay a housein development because the land had been previously used as a nucleur waste pit for Radium-D, an isotope has decayed to 1/√2 of its original. When could they build ifRadium-D has a half life of 22 years.

    both use the y=ka^x, i just cant seem to get the answer right when i plugin the values its prol cuz i subed it in wrong,if some1 cud kindly help me it wud be apreciatted.
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    Quote Originally Posted by usm_67 View Post
    8). The oldest rock knows to data found in Minnesota contains 2g of carbon 14(C^14). Using geological methods toestimate the original size of the rock, scentists estimate it to originally have contained 4.096kg of C^14 .The Half-life of c^14 is 5760 years.What is the approximate age of rock?

    9). Land developers had to delay a housein development because the land had been previously used as a nucleur waste pit for Radium-D, an isotope has decayed to 1/√2 of its original. When could they build ifRadium-D has a half life of 22 years.

    both use the y=ka^x, i just cant seem to get the answer right when i plugin the values its prol cuz i subed it in wrong,if some1 cud kindly help me it wud be apreciatted.
    The decay equation is indeed
    $\displaystyle y = ke^{bt}$

    How do we use this?

    Well we know the half-life in each case, so what does that mean? It means that the when the time is equal to the half-life only half of the sample is left.

    When t = 0
    $\displaystyle y = ke^{(0)} = k$

    So k is the amount of sample initially, before it started to decay. So when t is equal to the half-life $\displaystyle \tau$ we only have (1/2)k of the substance:
    $\displaystyle \frac{1}{2}k = ke^{b \tau}$

    $\displaystyle \frac{1}{2} = e^{b \tau}$

    $\displaystyle ln \left ( \frac{1}{2} \right ) = b \tau$

    $\displaystyle b =\frac{ln \left ( \frac{1}{2} \right )}{\tau} = - \frac{ln(2)}{\tau}$

    So the decay equation is:
    $\displaystyle y = ke^{-(ln(2) \cdot t)/\tau} = k2^{-t/\tau}$

    Try using this form for your questions. If you still have problems go ahead and let us know.

    -Dan
    Last edited by topsquark; Jul 31st 2007 at 04:34 AM.
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    well bro u totally lost me with the equation because he din teach us anythin like that, can u please solve it so the answer for first one is 63360 years and second one is 11 years
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    Quote Originally Posted by usm_67 View Post
    8). The oldest rock knows to data found in Minnesota contains 2g of carbon 14(C^14). Using geological methods toestimate the original size of the rock, scentists estimate it to originally have contained 4.096kg of C^14 .The Half-life of c^14 is 5760 years.What is the approximate age of rock?
    $\displaystyle y = k2^{-t/\tau}$

    $\displaystyle \tau = 5760~yr$
    $\displaystyle k = 4096~g$
    $\displaystyle y = 2~g$

    Thus
    $\displaystyle 2 = 4096 \cdot 2^{-t/5760}$

    Solve for t.

    -Dan
    Last edited by topsquark; Jul 31st 2007 at 04:35 AM.
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    what du u mean when u say t/5760 how wud i go by solvin that i get the rest of it
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    Quote Originally Posted by usm_67 View Post
    what du u mean when u say t/5760 how wud i go by solvin that i get the rest of it
    You might want to find an English forum also.
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    lol my bad i was in a hurry, got the letters mixed up
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    someone please helpp.!!!!!!!!!!!!!!!!!!!i dont understand it
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark View Post
    $\displaystyle 2 = 4096 \cdot 2^{-t/5760}$
    (I apparently left of the "-" in the exponent. Sorry about that!)

    $\displaystyle \frac{2}{4096} = 2^{-t/5760}$

    $\displaystyle 2^{-t/5760} = \frac{1}{2048}$

    $\displaystyle -\frac{t}{5760} = log_2 \left ( \frac{1}{2048} \right )$

    $\displaystyle t = -5760 \cdot log_2 \left ( \frac{1}{2048} \right )$

    Now, the nice thing about this is that:
    $\displaystyle t = -5760 \cdot log_2 \left ( \frac{1}{2^{11}} \right )$

    $\displaystyle t = -5760 \cdot -11 = 63360$
    So the answer is 63360 years.

    -Dan
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