document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1122713>Question 1122713</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The half life of a certain radioactive materials is 3 hours. How many kilograms of this material will be left after 48 hours if there are initially 10 kilograms? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #738859 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=738859\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> P=Po*e^(-kt); P/Po=1/2<BR>\n" );
document.write( "1/2=e^(-3k)<BR>\n" );
document.write( "ln both sides<BR>\n" );
document.write( "-0.693=-3k<BR>\n" );
document.write( "k=0.231\r<BR>\n" );
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document.write( "Let t=48<BR>\n" );
document.write( "P=10*e^(-0231*48); that will be e^(-11.088)<BR>\n" );
document.write( "P=1.529 x 10^(-4) kg \r<BR>\n" );
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document.write( "Check<BR>\n" );
document.write( "48 hours is 16 half lives <BR>\n" );
document.write( "2^(-16)=1.5 x 10^(-5) same as above fraction of the original amount.<BR>\n" );
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