document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=684727>Question 684727</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the number of grams of a certain radioactive substance present after t hours is given by the equation Q=Q(0)e^-0.45t, where Q(0) represents the initial number of grams. how long will it take 2500 grams to be REDUCED to 1250 grams? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #424186 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=424186\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the number of grams of a certain radioactive substance present after t hours is given by the equation Q=Q(0)e^-0.45t, where Q(0) represents the initial number of grams. how long will it take 2500 grams to be REDUCED to 1250 grams?<BR>\n" );
document.write( "--------<BR>\n" );
document.write( "Solve for \"t\":<BR>\n" );
document.write( "1250 = 2500*e^(-0.45t)<BR>\n" );
document.write( "----<BR>\n" );
document.write( "e^(-0.45t) = 0.5<BR>\n" );
document.write( "---<BR>\n" );
document.write( "Take the log of both sides:<BR>\n" );
document.write( "-0.45t = ln(0.5) = -0.6931<BR>\n" );
document.write( "t = 1.54 hours<BR>\n" );
document.write( "=================<BR>\n" );
document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );