Question 7136: Hi, I'm having a tad bit difficulty trying to work out this: 2n log n = 10^9 (log to the base 10).
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.
Answer by khwang(438)   (Show Source):
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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