document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1059769>Question 1059769</A>: <I><SPAN STYLE=\"word-break: break-all;\"> To the nearest year find the half-life of plutonium -239 given that it decays according to the formula A= A0^e-0.0000284t  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #674841 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=674841\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A= A0^e-0.0000284t <BR>\n" );
document.write( "A=(1/2)*Ao<BR>\n" );
document.write( "Therefore 1/2Ao=Ao^e(-0.0000284t)<BR>\n" );
document.write( "A0 cancel<BR>\n" );
document.write( "take logs of both sides<BR>\n" );
document.write( "log (1/2)=-0.0000284t, because ln of e^(t)=t.  Ln and e annihilate each other.<BR>\n" );
document.write( "-0.693/-0.0000284=24,401 years=t<BR>\n" );
document.write( "That compares favorably with quoted 25,000 years.<BR>\n" );
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