Question 81408
A skeleton contains one hundredth of its original amount of its original amount of carbon-14. If carbon-14 has a half-life of 5750 years, to the nearest 1000 years, how old is the skeleton?
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A(t) = A(o)(1/2)^(t/5750)
0.01A(o) = A(o)(1/2)^(t/5750)
0.01 = 0.5^(t/5750)
Take the log of both sides to get:
log 0.01 = (t/5750)*log0.5
t/5750 = log0.01/log0.5 =6.64385619...
t = 38202 yrs
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Cheers,
Stan H.