document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1004004>Question 1004004</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A fossilized leaf contains 12% 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 #620593 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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/cgi-bin/show-face.mpl?user=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=620593\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> We use the equation<BR>\n" );
document.write( "A(t) = Ao*e^(-t/5600)<BR>\n" );
document.write( "Here, A(t) = .12Ao so<BR>\n" );
document.write( ".12Ao = Ao*e^(-t/5600)<BR>\n" );
document.write( ".12 = e^(-t/5600)<BR>\n" );
document.write( "Take the ln of both sides and get<BR>\n" );
document.write( "-2.12 = -t/5600<BR>\n" );
document.write( "t = 11873 years </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );