document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=201484>Question 201484</A>: <I><SPAN STYLE=\"word-break: break-all;\"> CAN SOMEONE HELP WITH THIS WORD PROBLEM?  Strontium-90 is a waste product from nuclear fission reactors. It's half-life is 28 years and its decay model is A=A(subscript 0)e^-0.024755t. Suppose a nuclear accident occurs and releases 50 grams of strontium-90 into the biosphere. How long will it take for strontium-90 to decay to a level of 20 grams? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #151695 by </B><A HREF=/tutors/aboutme.mpl?userid=scott8148><B>scott8148(6628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=scott8148><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=scott8148><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=151695\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 20 = 50 e^(-.024755 t)\r<BR>\n" );
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document.write( "dividing by 50 ___ .4 = e^(-.024755 t)\r<BR>\n" );
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document.write( "taking natural log ___ ln(.4) = -.024755 t\r<BR>\n" );
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document.write( "dividing by -.024755 ___ [ln(.4)] / -.024755 = t\r<BR>\n" );
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document.write( "you can do this on your calculator ___ it should have a natural log (ln) key </SPAN>\n" );
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