document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=81826>Question 81826</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If I had the equation, 60.6=58e^-k(1), how would I solve it using Newton's Law of Cooling? I am clueless when it comes to Newton's Law of Cooling, so could you please explain that to me as well. Thank you in advance. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #58623 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=58623\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 60.6=58e^-k(1), how would I solve it using Newton's Law of Cooling?<BR>\n" );
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document.write( "There is only one variable, k, so that is what you are solving for.<BR>\n" );
document.write( "1.044828 = e^(-k)<BR>\n" );
document.write( "Take the natural log of both sides to get;<BR>\n" );
document.write( "-k = 0.043852<BR>\n" );
document.write( "k= 0.043852<BR>\n" );
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document.write( "As for Newton's Law of Cooling, I suggest you use Google <BR>\n" );
document.write( "to find a readable explanation.<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
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