document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1048706>Question 1048706</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How much time is needed for a sample of Pd-100 to lose 93.75% of its original amount? Pd-100 has a half-life of 3.634 days.Give an exponential model  as a function t. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #688490 by </B><A HREF=/tutors/aboutme.mpl?userid=jorel1380><B>jorel1380(3719)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jorel1380><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=jorel1380><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=688490\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let n be the original amount of pd-100. If it loses 93.75% of its' original amount, then the amount left equals 100-93.75=6.25%. So:<BR>\n" );
document.write( ".0625=(.5)^t/3.634<BR>\n" );
document.write( ".5^4=.5^t/3.634<BR>\n" );
document.write( "t/3.634=4<BR>\n" );
document.write( "t=14.536 or<BR>\n" );
document.write( "log .0625=log .5^t/3.634<BR>\n" );
document.write( "log .0625=t/3.634*log .5<BR>\n" );
document.write( "(log .0625/log .5)*3.634=t<BR>\n" );
document.write( "t=14.536 days. &#9786;&#9786;&#9786;&#9786; </SPAN>\n" );
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