document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1075479>Question 1075479</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A fossilized leaf contains 15% 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. Solve the problem.<BR>\n" );
document.write( "  A.  35,828 <BR>\n" );
document.write( "  B.  15,299 <BR>\n" );
document.write( "  C.  1311 <BR>\n" );
document.write( "  D.  21,839  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #690147 by </B><A HREF=/tutors/aboutme.mpl?userid=jorel1380><B>jorel1380(3719)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jorel1380><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=jorel1380><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=690147\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The half-life of a specimen is defined as the length of time it takes to lose half of its original mass. In this case, there is only 15% of the normal carbon-14 left. So:<BR>\n" );
document.write( ".15=(.5)^t/5600 where t is age of the fossil, in years.\r<BR>\n" );
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document.write( "First, calculate for the single value:<BR>\n" );
document.write( ".15=.5^t<BR>\n" );
document.write( "log 0.15=log 0.5^t<BR>\n" );
document.write( "log 0.15=t log 0.5<BR>\n" );
document.write( "t=log 0.15/log 0.5=2.7369655941662061664165804855416<BR>\n" );
document.write( "Then multiply by 5600:<BR>\n" );
document.write( "5600 x 2.7369655941662061664165804855416=15,327 years as the approximate age of the fossil. &#9786;&#9786;&#9786;&#9786; </SPAN>\n" );
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