document.write( "Question 416652: How do I set this up so I can solve it, it's a word proble.
\n" ); document.write( "An anthropologist studying the bones of a prehistoric person finds there is so little remaining Carbon-14 in the bones that instruments cannot measure it. This means there is less than 0.5% of the amount of Carbon-14 the bones would have contained when the person was alive. How long ago did the person die?
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Algebra.Com's Answer #291874 by stanbon(75887)\"\" \"About 
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An anthropologist studying the bones of a prehistoric person finds there is so little remaining Carbon-14 in the bones that instruments cannot measure it.
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\n" ); document.write( "This means there is less than 0.5% of the amount of Carbon-14 the bones would have contained when the person was alive.
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\n" ); document.write( "How long ago did the person die?
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\n" ); document.write( "Carbon 14 half-life is 5730 years
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\n" ); document.write( "Equation to use:
\n" ); document.write( "A(t) = Ao(1/2)^(t/5730)
\n" ); document.write( "0.0005Ao = Ao(1/2)^(t/5730)
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\n" ); document.write( "(1/2)^(t/5730) = 0.0005
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\n" ); document.write( "Take the ln of both sides:
\n" ); document.write( "(t/5730)ln(1/2) = ln(0.0005)
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\n" ); document.write( "t/5730 = 10.9658
\n" ); document.write( "t > 62,833.94 year old
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.\r
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