document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1014988>Question 1014988</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please help me answer this word problem: \r<BR>\n" );
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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" );
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document.write( "Thank-you, your help is always appreciated! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #631326 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=631326\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> y=yoe^kt<BR>\n" );
document.write( "20=30e^10k<BR>\n" );
document.write( "(2/3)=e^10k<BR>\n" );
document.write( "ln of both sides, then divide by 10 without rounding.<BR>\n" );
document.write( "ln(2/3)=10k<BR>\n" );
document.write( "k=-0.04055<BR>\n" );
document.write( "===================<BR>\n" );
document.write( "Now for half life<BR>\n" );
document.write( "y=(1/2)yo, so<BR>\n" );
document.write( "(1/2)=e^(-0.04055)t<BR>\n" );
document.write( "ln of both sides, then divide by coefficient of t<BR>\n" );
document.write( "-0.693/-0.04055<BR>\n" );
document.write( "t=17.09 minutes.  This makes intuitive sense, since 10 minutes is not the half life but is getting close to it.<BR>\n" );
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