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anonymous

  • 5 years ago

Situation: A geologist in South America discovers a bird skeleton that contains 80% of it's original amount of c-14. Find the age of the bird skeleton to the nearest year.

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  1. anonymous
    • 5 years ago
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    you need the half life of c-14 or else the "k-value" without it you cannot solve this.

  2. anonymous
    • 5 years ago
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    k = .0001

  3. anonymous
    • 5 years ago
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    ok now we can do it. first of all no matter what you start with you will have 80% when you are done. 80% = .8 so solve \[.8=e^{-.0001t}\]for t \[ln(.8)=-.0001t\] \[t=\frac{ln(.8)}{-.0001}=2231\]

  4. anonymous
    • 5 years ago
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    btw i hope it is clear why it doesn't matter what you start with. if you start with 100 grams then 80% of that is 80 so you would put \[80=100e^{-.0001t}\] and the first step would be to divide by 100 to get \[.8=e^{-.0001t}\]

  5. anonymous
    • 5 years ago
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    then what?

  6. anonymous
    • 5 years ago
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    oh then to get the variable out of the exponent take the natural log.

  7. anonymous
    • 5 years ago
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    \[e^y = x\] same as \[y=ln(x)\]

  8. anonymous
    • 5 years ago
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    so if \[e^{-.0001t}=.8\] same as \[-.0001t=ln(.8)\]

  9. anonymous
    • 5 years ago
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    then divide to -.0001 to get t. done

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