document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=707616>Question 707616</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the half-life of nitrogen-16 is 7 seconds. how long does it take for 100mg of nitrogen-16 be reduced to 6.25mg? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #435776 by </B><A HREF=/tutors/aboutme.mpl?userid=jsmallt9><B>jsmallt9(3758)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jsmallt9><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=435776\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This is an exponential growth/decay problem. A general equation for these is:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+A%5B0%5D%2Ar%5Et ALIGN=MIDDLE  ALT=\"A+=+A%5B0%5D%2Ar%5Et\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "where<BR>\n" );
document.write( "t = number of units of time<BR>\n" );
document.write( "A = the amount after t units of time<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A%5B0%5D ALIGN=MIDDLE  ALT=\"A%5B0%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" > = the initial amount (IOW: The amount at t = 0)<BR>\n" );
document.write( "r = the factor of change for 1 unit of time. If r > 1 then the equation is for growth and if 0 < r < 1 then the equation is for decay.<br><BR>\n" );
document.write( "In your problem:<BR>\n" );
document.write( "t = the number of units of 7 seconds that have passed.<BR>\n" );
document.write( "A = 6.25<BR>\n" );
document.write( "r = 1/2 (since the amount decreases by 1/2 every 7 seconds)<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A%5B0%5D ALIGN=MIDDLE  ALT=\"A%5B0%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" > = 100<BR>\n" );
document.write( "So your equation is:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=6.25+=+100%2A%281%2F2%29%5Et ALIGN=MIDDLE  ALT=\"6.25+=+100%2A%281%2F2%29%5Et\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "Now we solve for t. We start by isolating the base, 1/2, and its exponent by dividing each side by 100:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=6.25%2F100+=+%281%2F2%29%5Et ALIGN=MIDDLE  ALT=\"6.25%2F100+=+%281%2F2%29%5Et\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "which simplifies to:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0.0625+=+%281%2F2%29%5Et ALIGN=MIDDLE  ALT=\"0.0625+=+%281%2F2%29%5Et\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "If the left side is a power of 1/2 then we can solve this by hand. If not, then we will need to use exponents. It will easier to see if the left side is a power of 1/2 if we rewrite it as a fraction:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=625%2F10000+=+%281%2F2%29%5Et ALIGN=MIDDLE  ALT=\"625%2F10000+=+%281%2F2%29%5Et\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Clearly 25 will go into both 625 and 10000:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=25%2F400+=+%281%2F2%29%5Et ALIGN=MIDDLE  ALT=\"25%2F400+=+%281%2F2%29%5Et\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Clearly 25 will go again into both 25 and 400:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F16+=+%281%2F2%29%5Et ALIGN=MIDDLE  ALT=\"1%2F16+=+%281%2F2%29%5Et\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "If we know (or try out) some powers of 1/2 we should quickly find that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F16+=+%281%2F2%29%5E4 ALIGN=MIDDLE  ALT=\"1%2F16+=+%281%2F2%29%5E4\"   DISABLED_style= \"max-width:90%; height:auto;\" >. So t = 4. Now remember that t is how many sets of 7 seconds that have passed. So the amount reduces from 100 mg to 6.25 mg not in 4 seconds but in 4*7 or 28 seconds.\r<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );