document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=194375>Question 194375</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A fossilized leaf contains 27% of its normal amount of carbon 14. How old is the fossil (to the nearest year)? Use 5600 years as the half-life of carbon 14. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #145860 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=145860\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A fossilized leaf contains 27% of its normal amount of carbon 14. How old is the fossil (to the nearest year)? Use 5600 years as the half-life of carbon 14.<BR>\n" );
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document.write( "A(t) = A(o)(1/2)^(t/5600<BR>\n" );
document.write( "0.27Ao = Ao*(1/2)^(t/5600<BR>\n" );
document.write( "(1/2)^(t/5600) = 0.27<BR>\n" );
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document.write( "Take the log to get:<BR>\n" );
document.write( "(t/5600)log(0.5) = log(0.27)<BR>\n" );
document.write( "t/5600 = 1.8889<BR>\n" );
document.write( "t = 10578.22 years<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );