78*log(n,2)*sqrt(n) + 10000 = 2n
solve for n.
log(n,2) is equivalent to log(n) with base = 2.
+118608 Alan or Heureka may be able to help you. I do not think I can 
78*log(n,2)*sqrt(n) + 10000 = 2n
\(78*log_2n*\sqrt{n} + 10000 = 2n\\ 78*\sqrt{n}*log_2n+ 10000 = 2n\\\)
+118608 Thanks but the poster would probably like to see a more detailed reponse.
The Newton-Raphson method, or Newton Method, is a powerful technique for solving equations numerically. Like so much of the differential calculus, it is based on the simple idea of linear approximation. The Newton Method, properly used, usually homes in on a root with great efficiency.
n ≈ 565340.98322027286106109097.
The Newton-Raphson Method continues:
Example:solve -2 x + x^3 = 0 using Newton's method starting at x_0 = 2
Resut = x = 1.414213562373095...............
Symbolic form of Newton iteration:
x_(n + 1) = x_n - (x_n^3 - 2 x_n)/(3 x_n^2 - 2)
step | x | residual | derivative
0 | 2. | 4.00000 | 10.
1 | 1.6 | 0.896000 | 5.68
2 | 1.44225 | 0.115518 | 4.24029
3 | 1.41501 | 0.00319099 | 4.00677
4 | 1.41421 | 2.69193×10^-6 | 4.00001
5 | 1.41421 | 1.92151×10^-12 | 4.
6 | 1.41421 | 0 |
x_0 = 2.
x_1 = x_0 - (x_0^3 - 2 x_0)/(3 x_0^2 - 2)
x_1 = 2. - (0.4)
x_1 = 1.6
x_2 = x_1 - (x_1^3 - 2 x_1)/(3 x_1^2 - 2)
x_2 = 1.6 - (0.157746)
x_2 = 1.44225
x_3 = x_2 - (x_2^3 - 2 x_2)/(3 x_2^2 - 2)
x_3 = 1.44225 - (0.0272429)
x_3 = 1.41501
x_4 = x_3 - (x_3^3 - 2 x_3)/(3 x_3^2 - 2)
x_4 = 1.41501 - (0.000796401)
x_4 = 1.41421
x_5 = x_4 - (x_4^3 - 2 x_4)/(3 x_4^2 - 2)
x_5 = 1.41421 - (6.72981×10^-7)
x_5 = 1.41421
x_6 = x_5 - (x_5^3 - 2 x_5)/(3 x_5^2 - 2)
x_6 = 1.41421 - (4.80394×10^-13)
x_6 = 1.41421
precision | steps machine precision | 6 50 digits | 7 100 digits | 9
Exact Result: x = sqrt(2) ≈ 1.414213562373095048801688724
