document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=87050>Question 87050</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Osmium has a radioactive isotope that has a half life of 15.4 days.  If one gram of the isotope is present now, then amount A (in grams) present after t days is A=(1/2)^t/15.4\r<BR>\n" );
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document.write( "How many days will pass before there is 0.73 grams of the isotope present?  Round to the nearest day!  \r<BR>\n" );
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document.write( "*So I thought to put the 0.73 in for A, and help would be fabulous!  THANKS! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #63070 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=63070\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> you're off to a good start ... .73=(.5)^(t/15.4) ... taking log of both sides gives log(.73)=(t/15.4)log(.5)\r<BR>\n" );
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document.write( "log(.73)/log(.5)=t/15.4 ... 15.4(log(.73)/log(.5))=t ... t approximately equals 7 days </SPAN>\n" );
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