+0

# Math equation

0
414
6

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
+94105
+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
+1

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

n ≈ 565,341. using Newton-Raphson method.

Guest Jun 2, 2017
#3
+94105
+1

Thanks but the poster would probably like to see a more detailed reponse.

Melody  Jun 2, 2017
#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
#4
+92648
+1

I'm afraid I'll have to join Melody, here.....I don't know how to solve this "by hand"

For what it's worth......here's the graphical solution :

https://www.desmos.com/calculator/9w8cvwjb3y

CPhill  Jun 2, 2017
#6
0

CPhill and the guest who solved the question. Thank you so much.

I didn't know anything about Newton-Raphson method at all. It was a good learning experience.

Guest Jun 2, 2017