You can
put this solution on YOUR website! 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
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