document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=146096>Question 146096</A>: <I><SPAN STYLE=\"word-break: break-all;\"> This is a problem for exponential decay. I am just not understanding it.\r<BR>\n" );
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document.write( "A radioactive substance has a half-life of 2 years. There is 36 g to begin with. Write and equation for y=mass left after x=half-years.\r<BR>\n" );
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document.write( "Then solve for 100 years. (how much of the 36 g will be left?) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #106646 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=106646\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A radioactive substance has a half-life of 2 years. <BR>\n" );
document.write( "There is 36 g to begin with. <BR>\n" );
document.write( "Write and equation for y=mass left after x=half-years. <BR>\n" );
document.write( "Then solve for 100 years. (how much of the 36 g will be left?)<BR>\n" );
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document.write( "A(t) = Ao(1/2)^(t/2)<BR>\n" );
document.write( "Comment: That t/2 exponent means you should multiply by 1/2 every 2-years.<BR>\n" );
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document.write( "A(100) = 36(1/2)^(100/2)<BR>\n" );
document.write( "A(100) = 36(1/2)^50<BR>\n" );
document.write( "A(100) = 36*0.00000000000000088817842<BR>\n" );
document.write( "A(100) = 0.0000000000000319744...grams<BR>\n" );
document.write( "Or<BR>\n" );
document.write( "A(100) = 3.197 x 10^-14 grams<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H </SPAN>\n" );
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