78*log(n,2)*sqrt(n) + 10000 = 2n

solve for n.

log(n,2) is equivalent to log(n) with base = 2.

Guest Jun 2, 2017

#1**+1 **

Alan or Heureka may be able to help you. I do not think I can

78*log(n,2)*sqrt(n) + 10000 = 2n

\(78*log_2n*\sqrt{n} + 10000 = 2n\\ 78*\sqrt{n}*log_2n+ 10000 = 2n\\\)

Melody
Jun 2, 2017

#2

#5**0 **

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

Guest Jun 2, 2017

edited by
Guest
Jun 2, 2017