數學知識交流---解方程

(1)亦有ALL solutions！

2011-07-27 07:26:17 補充：
(1) Solve 2^(5x) - 10 = 0.

2^(5x) - 10 = 0
(e^(ln 2))^(5x) = 10
e^(5x ln 2) = 10
x ln 32 = ln 10 + 2nπi
x = (ln 10 + 2nπi)/(ln 32)


(2) Solve x^8 - 1 = 0.

x^8 - 1 = 0
x^8 = 1
x^8 = cos 2nπ + i sin 2nπ
x = (cos 2nπ + i sin 2nπ)^(1/8)
x = cos nπ/4 + i sin nπ/4
x = cos 0 + i sin 0 or cos π/4 + i sin π/4 or cos π/2 + i sin π/2 or cos 3π/4 + i sin 3π/4 or cos π + i sin π or cos 5π/4 + i sin 5π/4 or cos 3π/2 + i sin 3π/2 or cos 7π/4 + i sin 7π/4
x = 1 or (√2)/2 + i(√2)/2 or i or - (√2)/2 + i(√2)/2 or - 1 or - (√2)/2 - i(√2)/2 or - i or (√2)/2 - i(√2)/2

1. 2^(5x) - 10 = 0
2^(5x) = 10
5x log 2 = log 10 = 1
x = 1/(5 log 2).
2.
(x^4)^2 - 1^2 = 0
(x^4 + 1)(x^4 - 1) = 0
(x^4 + 1)(x^2 + 1)(x^2 -1) = 0
(x^4 + 1)(x^2 + 1)(x + 1)(x - 1) = 0
so the real roots are 1 and - 1.


2011-07-21 18:08:25 補充：
For x^2 + 1 = 0, x = +/- sqrt (-1), that is +/- i.
For x^4 + 1 = 0, x = +/- i^(1/4), +/- i^(3/4).

2011-07-21 18:15:48 補充：
For x^4 + 1 = 0, x in complex form : cos (pi/8) +/- i sin(pi/8) and sin (pi/8) +/- cos (pi/8). Total 4 complex roots.