document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=566011>Question 566011</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Potassium 42 has a decay rate of approximately 5.5% per hour, assuming an exponential decay model, fine the number of hours it will take for the original quantity of potassium 42 to b halved? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #366057 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=366057\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Potassium 42 has a decay rate of approximately 5.5% per hour, assuming an exponential decay model, find the number of hours it will take for the original quantity of potassium 42 to b halved?<BR>\n" );
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document.write( "y = ab^x<BR>\n" );
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document.write( "(1/2)a = a*b^x<BR>\n" );
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document.write( "1/2 = b^x<BR>\n" );
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document.write( "1/2 = 0.945^x<BR>\n" );
document.write( "x = log(1/2)/log(0.945)<BR>\n" );
document.write( "--<BR>\n" );
document.write( "x = 12.25 years<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
document.write( "================ </SPAN>\n" );
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