document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1139086>Question 1139086</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A certain radioactive isotope has leaked into a small stream. One hundred days after the​ leak, 14% of the original amount of the substance remained. Determine the​ 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 #756865 by </B><A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><B>ankor@dixie-net.com(22740)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><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=ankor@dixie-net.com><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=756865\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A certain radioactive isotope has leaked into a small stream.<BR>\n" );
document.write( " One hundred days after the​ leak, 14% of the original amount of the substance remained.<BR>\n" );
document.write( " Determine the​ half-life of this radioactive isotope. <BR>\n" );
document.write( ":<BR>\n" );
document.write( "The radio-active decay formula: A = Ao*2^(-t/h), where:<BR>\n" );
document.write( "A = amt of substance remains after t time<BR>\n" );
document.write( "Ao = initial amt<BR>\n" );
document.write( "t = time of decay<BR>\n" );
document.write( "h = half-life of substance<BR>\n" );
document.write( ":<BR>\n" );
document.write( " Ao = 1,<BR>\n" );
document.write( " A = .14<BR>\n" );
document.write( " t = 100 days<BR>\n" );
document.write( "1*2^(-100/h) = .14<BR>\n" );
document.write( "using natural logs<BR>\n" );
document.write( "ln(2^(-100/h) = ln(.14)<BR>\n" );
document.write( "log equiv of exponents<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-100%2Fh ALIGN=MIDDLE  ALT=\"-100%2Fh\"   DISABLED_style= \"max-width:90%; height:auto;\" >*ln(2) = ln(.14)<BR>\n" );
document.write( ":<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-100%2Fh ALIGN=MIDDLE  ALT=\"-100%2Fh\"   DISABLED_style= \"max-width:90%; height:auto;\" >= <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ln%28.14%29%2Fln%282%29 ALIGN=MIDDLE  ALT=\"ln%28.14%29%2Fln%282%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "find the antilog<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-100%2Fh ALIGN=MIDDLE  ALT=\"-100%2Fh\"   DISABLED_style= \"max-width:90%; height:auto;\" > = -2.8365<BR>\n" );
document.write( "h = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-100%29%2F%28-2.8365%29 ALIGN=MIDDLE  ALT=\"%28-100%29%2F%28-2.8365%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "h = 35.25 days is the half life of the substance\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );