document.write( "Question 303996: Soil from beneath the Kish Church in Azerbaijan was found to have lost 12% of its carbon-14. How old was the soil? \n" ); document.write( "

Algebra.Com's Answer #217783 by ankor@dixie-net.com(22740) $\"\"$ $\"About$

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If soil was found to have lost 12% of it's carbon-14, then how old was the soil?
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\n" ); document.write( "They should give you the half-life of Carbon 14. Most sources say it's 5730 yrs
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\n" ); document.write( "The decay formula: A = Ao*2^(-t/h)
\n" ); document.write( "where
\n" ); document.write( "A = resulting amt after t yrs
\n" ); document.write( "Ao = initial amt (t=0)
\n" ); document.write( "t = time in yrs
\n" ); document.write( "h = half-life of substance (5730 yrs in this case)
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\n" ); document.write( "Let Ao = 1, then remaining amt (A) after t yrs: 1 -.12 = .88 (lost 12%)
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\n" ); document.write( "1*2^(-t/5730) = .88
\n" ); document.write( "using nat logs
\n" ); document.write( "ln[2^(-t/5730)] = ln(.88)
\n" ); document.write( "log equiv of exponents
\n" ); document.write( " $\"-t%2F5730\"$ *ln(2) = ln(.88)
\n" ); document.write( " $\"-t%2F5730\"$ *.693 = -.128
\n" ); document.write( " $\"-t%2F5730\"$ = $\"%28-.128%29%2F.693\"$
\n" ); document.write( " $\"-t%2F5730\"$ = -.1844
\n" ); document.write( "t = -5730 * -.1844
\n" ); document.write( "t = 1057 yrs old
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\n" ); document.write( "Check solution on a calc: enter 2^(-1057/5730) = .8799 ~.88 or 88% remaining \n" ); document.write( "

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