document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=423151>Question 423151</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The decay rate of a certain chemical is 9.5% per year. What is it's half life? Use the exponential decay model P(t)=P/subscript/0e^-kt where k is the decay rate and P/subscript/0 is the original amount of chemical.. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #295225 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=295225\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  The decay rate of a certain chemical is 9.5% per year. What is it's half life? Use the exponential decay model P(t)=P/subscript/0e^-kt where k is the decay rate and P/subscript/0 is the original amount of chemical..<BR>\n" );
document.write( "-------------------------<BR>\n" );
document.write( "P(t) = Po*e^(-kt)<BR>\n" );
document.write( "----<BR>\n" );
document.write( "(0.5)Po = Po*e^(-0.095t)<BR>\n" );
document.write( "----<BR>\n" );
document.write( "e^(-0.095t)= 0.5<BR>\n" );
document.write( "----<BR>\n" );
document.write( "Take the natural log of both sides.<BR>\n" );
document.write( "-0.095t = ln(0.5)<BR>\n" );
document.write( "t = 7.296 years (half-life of the chemical)<BR>\n" );
document.write( "=========================<BR>\n" );
document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );