數學.....指數解方程 F.4

1. 2[1/(3x-1)]^(-3) =54
[1/(3x-1)]^(-3) = 27
(3x – 1)^3 = 3^3
3x – 1 = 3
3x = 4
x = 4/3
2. [2/(5x+3)]^(-2) =25
[(5x + 3)/2]^2 = 25
(5x + 3)^2 = 100
(5x + 3)^2 – 10^2 = 0
(5x + 3 + 10)(5x + 3 – 10) = 0
(5x + 13)(5x – 7) = 0
x = -2.6 or x = 1.4

2010-01-23 12:16:22 補充：
(1) 2[(1/3)x –1]^(-3) = 54
[(1/3)x –1]^(-3) = 27 = 3^3
(1/3)x –1 = 1/3
(1/3)x = 4/3
x = 4
(2) [(2/5)x + 3]^(-2) = 25 = 5^2
[(2/5)x + 3]^2 = 1/5^2
(2/5)x + 3 = +/- 1/5
(2/5)x = -16/5 or -14/5
x = - 7 or -8 (not only -7)