document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=762544>Question 762544</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Carbon dating. The amount of the carbon 14 present in animal bones after t year is given as P(t)=  Poe^-0.0012t. A bone has lost 30 % of its carbon 14. How old is the bone? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #464058 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=464058\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Carbon dating. The amount of the carbon 14 present in animal bones after t year is given as P(t)= Poe^-0.0012t. A bone has lost 30 % of its carbon 14. How old is the bone?<BR>\n" );
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document.write( "P(t)= Poe^-0.0012t<BR>\n" );
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document.write( "Solve: e^(-0.0012t) = 0.30<BR>\n" );
document.write( "Take the natural log:<BR>\n" );
document.write( "-0.0012t = ln(0.3)<BR>\n" );
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document.write( "-0.0012t = -1.2040<BR>\n" );
document.write( "t = 1003.31 years<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );