document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=250173>Question 250173</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In the formula A(t) = A0ekt, A is the amount of radioactive material<BR>\n" );
document.write( "remaining from an initial amount A0 at a given time t, and k is a negative<BR>\n" );
document.write( "constant determined by the nature of the material. A certain radioactive<BR>\n" );
document.write( "isotope decays at a rate of 0.3% annually. Determine the half-life of this<BR>\n" );
document.write( "isotope, to the nearest year. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #182255 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=182255\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> .997 = e^k ___ ln(.997) = k\r<BR>\n" );
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document.write( ".5 = e^[t ln(.997)]\r<BR>\n" );
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document.write( "ln(.5) = t ln(.997)\r<BR>\n" );
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document.write( "[ln(.5)] / [ln(.997)] = t\r<BR>\n" );
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document.write( "231 = t (approx) </SPAN>\n" );
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