document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1015192>Question 1015192</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please help me answer this word problem: \r<BR>\n" );
document.write( "\n" );
document.write( "The rate at which a radioactive isotope disintegrates is proportional to the amount present. If a 30g sample will contain only 20g after ten minutes, what is the half-life of this isotope, correct to the nearest tenth of a minute? Answer this question by first finding the amount y of this isotope as a function of time t(in minutes) in the form y=y0e^kt. \r<BR>\n" );
document.write( "\n" );
document.write( "Thank-you, your help is always appreciated! <BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #631576 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=631576\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The rate at which a radioactive isotope disintegrates is proportional to the amount present. If a 30g sample will contain only 20g after ten minutes, what is the half-life of this isotope, correct to the nearest tenth of a minute? Answer this question by first finding the amount y of this isotope as a function of time t(in minutes) in the form <BR>\n" );
document.write( "y = yo*e^(kt)<BR>\n" );
document.write( "------<BR>\n" );
document.write( "20 = 30*e^(k*(1/6))<BR>\n" );
document.write( "----<BR>\n" );
document.write( "e^(k/6) = 2/3<BR>\n" );
document.write( "-----<BR>\n" );
document.write( "k/6 = ln(2/3) = -0.4055<BR>\n" );
document.write( "-----<BR>\n" );
document.write( "k = -2.4328<BR>\n" );
document.write( "------<BR>\n" );
document.write( "Substitute for \"k\" in the formula:<BR>\n" );
document.write( "y = yo*e^(-2.4328t)<BR>\n" );
document.write( "---<BR>\n" );
document.write( "Note: Half-life means y = yo/2<BR>\n" );
document.write( "-----<BR>\n" );
document.write( "Then e^(-2.4328t) = 1/2<BR>\n" );
document.write( "----<BR>\n" );
document.write( "-2.4328t = ln(1/2) = -0.6932<BR>\n" );
document.write( "----<BR>\n" );
document.write( "time = -0.6932/-2.4328 = 0.2849 hrs  = 17min 5.7sec<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" );