document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1077705>Question 1077705</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Substances X and Y decompose at a rate proportional to the amount present.<BR>\n" );
document.write( "Tests shows that substance X loses one half of its mass every 16 hours and<BR>\n" );
document.write( "substance Y loses one half of its mass every 21 hours. At this moment there are<BR>\n" );
document.write( "3.0 kg of X and 3.0 kg of Y. When will there be three times as much of Y remaining<BR>\n" );
document.write( "as there is of X? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #692176 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=692176\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> half the mass lost in 16 hours is (1/2)=e^(-16k)<BR>\n" );
document.write( "-ln2=-16k<BR>\n" );
document.write( "k=ln2/16=0.04332 for X<BR>\n" );
document.write( "k=ln2/21=0.03301 for Y<BR>\n" );
document.write( "We want e^(-0.04332t)/e^(-0.03301t)=1/3<BR>\n" );
document.write( "3e^(-0.04332t)=e^(-0.03301t)<BR>\n" );
document.write( "do ln both sides<BR>\n" );
document.write( "ln3-0.04332t=-0.03301t<BR>\n" );
document.write( "ln3=0.01031t<BR>\n" );
document.write( "t=106.56 hours<BR>\n" );
document.write( "In that time, X will have 3e^(-4.616)=0.02967 kg<BR>\n" );
document.write( "Y will have 3e^(-3.517)=0.0890 kg<BR>\n" );
document.write( "That ratio is 3.00 to 1. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );