document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=323985>Question 323985</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose that you have 12 grams of a substance, which decays exponentially and has a half-life of 23 years.  How long will it be before there is only 1.5 grams left? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #231868 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=231868\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Suppose that you have 12 grams of a substance, which decays exponentially and has a half-life of 23 years. How long will it be before there is only 1.5 grams left?<BR>\n" );
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document.write( "A(t) = Ao*(1/2)^(t/23)<BR>\n" );
document.write( "1.5 = 12*(1/2)^(t/23)<BR>\n" );
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document.write( "(1/2)^(t/23) = 0.125<BR>\n" );
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document.write( "Take the log of both sides:<BR>\n" );
document.write( "(t/23)*log(1/2) = log(0.125)<BR>\n" );
document.write( "t/23 = 3<BR>\n" );
document.write( "---<BR>\n" );
document.write( "t = 69 years<BR>\n" );
document.write( "===================<BR>\n" );
document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );