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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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.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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you need the half life of c-14 or else the "k-value" without it you cannot solve this.
k = .0001
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 \[ln(.8)=-.0001t\] \[t=\frac{ln(.8)}{-.0001}=2231\]

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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 \[.8=e^{-.0001t}\]
then what?
oh then to get the variable out of the exponent take the natural log.
\[e^y = x\] same as \[y=ln(x)\]
so if \[e^{-.0001t}=.8\] same as \[-.0001t=ln(.8)\]
then divide to -.0001 to get t. done

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