document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=7136>Question 7136</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hi, I'm having a tad bit difficulty trying to work out this:  2n log n = 10^9 (log to the base 10).<BR>\n" );
document.write( "Now I know its probably easy to show but I can never get the right answer. Spent how long on it.  Heres wat i thought. n log n = (10^9/2). thats how far i got.  I tried using e, and ln and even did a square root.  And it was incorrect.  Please help it would be truely greatful. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #3948 by </B><A HREF=/tutors/aboutme.mpl?userid=khwang><B>khwang(438)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=khwang><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=3948\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  This question is above caclculus level and there is no exact value as an answer<BR>\n" );
document.write( " except some special forms such as n*log n = k*l0^k (k nonzero integers, e.g.<BR>\n" );
document.write( " n log n = 10, 200,3000 etc) <BR>\n" );
document.write( " <BR>\n" );
document.write( " The simplest way to solve it is to use Newton's method.\r<BR>\n" );
document.write( "\n" );
document.write( " Given 2 n log n = 10^9 ,(Here, don't use n as variable next time)<BR>\n" );
document.write( " To get rid of huge number, let x = log n, then we have<BR>\n" );
document.write( "  x 10^x = 10^9 or 2x 10^(x-9) = 1.<BR>\n" );
document.write( " Apply logon both sides, we have log 2x + x -9 = 0.\r<BR>\n" );
document.write( "\n" );
document.write( " Set f(x) = log  2x + x - 9		<BR>\n" );
document.write( " The derivative	f'(x) = 1/(2x*ln(10)) + 1		\r<BR>\n" );
document.write( "\n" );
document.write( " By observation, choose x = 8, use the recursive Newton's method:		<BR>\n" );
document.write( "  x_n+1 = x_n-f(x_n)/f'(x_n) [x_n means subscript]\r<BR>\n" );
document.write( "\n" );
document.write( " By the iterative table(I used Excel)<BR>\n" );
document.write( "			                                     Given function<BR>\n" );
document.write( " x_n	f(x_n)	           f'(x_n)	x_n-f(x_n)/f'(x_n)   2nlogn- 10^9<BR>\n" );
document.write( "  8	0.204119983	  1.027143405	7.801274115		600000000<BR>\n" );
document.write( "7.801274115 -0.005530352  1.027834843	7.806654699		-12653369.8<BR>\n" );
document.write( "7.806654699  0.000149664 1.027815658	7.806509084	       344674.5501<BR>\n" );
document.write( "7.806509084 -4.05043E-06 1.027816177	7.806513025     	-9326.411512<BR>\n" );
document.write( "7.806513025 1.09618E-07	1.027816163	7.806512919		252.4052805<BR>\n" );
document.write( "7.806512919 -2.96664E-09 1.027816164	7.806512922		-6.830934644<BR>\n" );
document.write( "7.806512922 8.02878E-11	1.027816164	7.806512921		0.184870243<BR>\n" );
document.write( "                                                  <BR>\n" );
document.write( " (the last column are error terms)<BR>\n" );
document.write( " From this table ,we see that the sequence x_n converges to 7.806512919<BR>\n" );
document.write( " and so n = 10^x = 10^(7.806512919) = 64049083.77\r<BR>\n" );
document.write( "\n" );
document.write( " Since the given number is very huge, small difference would cause large<BR>\n" );
document.write( " error. So, don't expect to get estimate values only up to hundredth or thousandth. \r<BR>\n" );
document.write( "\n" );
document.write( " If you have trouble understanding. Get a calculus book or search for<BR>\n" );
document.write( " Newton's method in the Web. Good luck!\r<BR>\n" );
document.write( "\n" );
document.write( " Kenny<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );