document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=693912>Question 693912</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How long would it take 100,000 grams of radioactive iodine, which has a half-life of 60 days, to decay to 25,000 grams?  Use the formaula N=N (1/2)^t, where N is the final amount of the substance, N is the initial amount, and t represents the number of half-lives. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #428023 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=428023\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> How long would it take 100,000 grams of radioactive iodine, which has a half-life of 60 days, to decay to 25,000 grams? Use the formaula N(t)= N (1/2)^t, where N is the final amount of the substance, N is the initial amount, and t represents the number of half-lives.<BR>\n" );
document.write( "25000 = 100000*(1/2)^t<BR>\n" );
document.write( "----<BR>\n" );
document.write( "(1/4) = (1/2)^t<BR>\n" );
document.write( "---<BR>\n" );
document.write( "t = 2 half-lives = 2*60 days = 120 days<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" );