document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1108262>Question 1108262</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A certain radioactive isotope has leaked into a small stream. Three hundred days after the&#8203; leak, 13&#8203;% of the original amount of the substance remained. Determine the&#8203; half-life of this radioactive isotope. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #723272 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=723272\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A=Aoe^(-kt)<BR>\n" );
document.write( "A/Ao=0.13<BR>\n" );
document.write( "0.13=e^(-k*300)<BR>\n" );
document.write( "ln both sides<BR>\n" );
document.write( "-2.040=-300k<BR>\n" );
document.write( "k=2.040/300=0.0068<BR>\n" );
document.write( "==================<BR>\n" );
document.write( "Now do it when A/Ao=1/2 and solve for t<BR>\n" );
document.write( "ln (1/2)=e^-(.0068*t)<BR>\n" );
document.write( "-0.693=-0.0068t<BR>\n" );
document.write( "t=101.91 days<BR>\n" );
document.write( "This makes sense, because 3 half lives is 300 days<BR>\n" );
document.write( "Three half lives leaves about 1/8 left, and 13% is about 1/8. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );