Question 7136
 This question is above caclculus level and there is no exact value as an answer
 except some special forms such as n*log n = k*l0^k (k nonzero integers, e.g.
 n log n = 10, 200,3000 etc) 
 
 The simplest way to solve it is to use Newton's method.

 Given 2 n log n = 10^9 ,(Here, don't use n as variable next time)
 To get rid of huge number, let x = log n, then we have
  x 10^x = 10^9 or 2x 10^(x-9) = 1.
 Apply logon both sides, we have log 2x + x -9 = 0.

 Set f(x) = log  2x + x - 9		
 The derivative	f'(x) = 1/(2x*ln(10)) + 1		

 By observation, choose x = 8, use the recursive Newton's method:		
  x_n+1 = x_n-f(x_n)/f'(x_n) [x_n means subscript]

 By the iterative table(I used Excel)
			                                     Given function
 x_n	f(x_n)	           f'(x_n)	x_n-f(x_n)/f'(x_n)   2nlogn- 10^9
  8	0.204119983	  1.027143405	7.801274115		600000000
7.801274115 -0.005530352  1.027834843	7.806654699		-12653369.8
7.806654699  0.000149664 1.027815658	7.806509084	       344674.5501
7.806509084 -4.05043E-06 1.027816177	7.806513025     	-9326.411512
7.806513025 1.09618E-07	1.027816163	7.806512919		252.4052805
7.806512919 -2.96664E-09 1.027816164	7.806512922		-6.830934644
7.806512922 8.02878E-11	1.027816164	7.806512921		0.184870243
                                                  
 (the last column are error terms)
 From this table ,we see that the sequence x_n converges to 7.806512919
 and so n = 10^x = 10^(7.806512919) = 64049083.77

 Since the given number is very huge, small difference would cause large
 error. So, don't expect to get estimate values only up to hundredth or thousandth. 

 If you have trouble understanding. Get a calculus book or search for
 Newton's method in the Web. Good luck!

 Kenny