SOLUTION: Having trouble getting started and really appreciate any assistance/direction. An animal skeleton was un-earthed during a certain geological excavation. It has been determined

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Question 115799: Having trouble getting started and really appreciate any assistance/direction.
An animal skeleton was un-earthed during a certain geological excavation. It has been determined that the skeleton lost 84% of its carbon-14 content. How old was the animal at the time of its death?
Thank you.
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
An animal skeleton was un-earthed during a certain geological excavation. It has been determined that the skeleton lost 84% of its carbon-14 content. How old was the animal at the time of its death?
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Using Google you can find the half-life of carbon 14 to be 5700 yrs.
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Then the half-life formula is:
A(t) = Ao (1/2)^(t/5700)
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Your Problem becomes:
0.84Ao = Ao (1/2)^(t/5700)
0.84 = (1/2)^(t/5700)
Take the log of both sides to get:
log(0.84) = (t/5700)*log(1/2)
t/5700 = log(0.84)/log(1/2)
t/5700 = 0.2515388...
t = 1433.77 yrs.
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Cheers,
Stan H.