document.write( "Question 802244: Carbon-14 has a half-life of 5710 years. What is the age of a piece of wood if 32% of the carbon in the piece of wood is left? \n" ); document.write( "

Algebra.Com's Answer #483869 by ankor@dixie-net.com(22740) $\"\"$ $\"About$

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Carbon-14 has a half-life of 5710 years.
\n" ); document.write( "What is the age of a piece of wood if 32% of the carbon in the piece of wood is left?
\n" ); document.write( ":
\n" ); document.write( "Using the equation for radioactive decay:
\n" ); document.write( "A = Ao*2^(-t/h), where:
\n" ); document.write( "A = Amt after t time
\n" ); document.write( "Ao = Initial amt (t=0)
\n" ); document.write( "t = decay time
\n" ); document.write( "h = half-life of substance
\n" ); document.write( ":
\n" ); document.write( "Let initial amt = 1
\n" ); document.write( "1*2(-t/5710) = .32
\n" ); document.write( "use natural logs
\n" ); document.write( " $\"-t%2F5710\"$ ln(2) = ln(.32)
\n" ); document.write( " $\"-t%2F5710\"$ = $\"ln%28.32%29%2Fln%282%29\"$
\n" ); document.write( " $\"-t%2F5710\"$ = -1.644t = -1.644*-5710
\n" ); document.write( "t = 9,387 yrs age of the wood \n" ); document.write( "

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