document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1022772>Question 1022772</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If something decays according to the equation y=ae^-0.0974t, where t- is in days.  After 15 days, 20 grams of the compound remain. What was the original amount of the compound?  Please show and explain the steps.  Thanks  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #638375 by </B><A HREF=/tutors/aboutme.mpl?userid=Alan3354><B>Alan3354(69443)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Alan3354><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=Alan3354><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=638375\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If something decays according to the equation y=ae^-0.0974t, where t- is in days.  After 15 days, 20 grams of the compound remain. What was the original amount of the compound?<BR>\n" );
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document.write( "a = amount 15 days earlier.<BR>\n" );
document.write( "y = amount now<BR>\n" );
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document.write( "y = a*e^(-0.0974*15) = 20<BR>\n" );
document.write( "a = 20/e^(-0.0974*15)<BR>\n" );
document.write( "a =~ 86.205 grams </SPAN>\n" );
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