方程Log2为底(X平方-4)>2+log2为底(x+2)的解是多少

Question

如题所述

Answer 1

log(2)[x²－4]>2＋log(2)[x＋2]
log(2)[x²－4]>log(2)[4(x＋2)]
则：
x²－4>4(x＋2)>0
1、x²－4>4x＋8
x²－4x－12>0
(x－6)(x＋2)>0
得：x<－2或x>6
2、4(x＋2)>0
x>－2
则原不等式的解集是：{x|x>6}

Answer 2

log(2)(x^2-4)> 2+ log(2)(x+2)
= log(2) (4(x+2) )
case 1:
x^2-4 > 4(x+2)
x^2-4x-12 >0
(x-6)(x+2)>0
x>6 or x<-2
case 2:
log(2)(x^2-4) is defined for
x^2-4 >0
x>2 or x<-2
case 3:
log(2)(x+2) is defined for
x+2>0
x>-2
case 1 and case 2 and case 3
"x>6 or x<-2" and "x>2 or x<-2" and "x>-2"
ie x>6

Answer 3

首先x^2-4>0 x+2>0
解得x>2
log2 (x^2-4)>2+log2 (x+2)
即log2 (x^2-4)>log2 4+log2 (x+2)
log2 (x^2-4)>log2 (4x+8)
∴x^2-4>4x+8
x^2-4x-12>0
解得x>6或x<-2
与x>2取交集得，x>6