document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=47894>Question 47894</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The amount of a radioactive tracer remaining after t days is given A=Alower 0e^-0.058t, where A0 is the starting amount at the beginning of the time period. How many days will it take for one-half of the orgininal amount to decay? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #31690 by </B><A HREF=/tutors/aboutme.mpl?userid=Nate><B>Nate(3500)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Nate><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=31690\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A0/2 = A0*e^(-0.058t)<BR>\n" );
document.write( "1/2 = e^(-0.058t)<BR>\n" );
document.write( "ln(1/2) = -0.058t<BR>\n" );
document.write( "-ln(1/2)/0.058 = t<BR>\n" );
document.write( "About 12 days </SPAN>\n" );
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