Posted by Rock on Friday, December 9, 2011 at 8:09pm.
Find the half-life of C14. Call it n years
2^(-t/n) < .005
-t/n < ln(.005)/ln(2) = -7.64
t > 7.64n
check: 2^-7.64 = .005013, so after 7.64 half-lives, we are down to .5% of the original.
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