If loga 1 / 2 > 1 is known, the value range of a is obtained
1=loga(a)
Then log a (1 / 2) > log a (a)
If 0
If loga (a ^ 2 + 1) < loga2a < 0, calculate the value range of A
loga(a²+1)
If loga (a ^ 2 +)
(a-1)²=a²-2a+1>=0
a²+1>=2a
And loga (A & # 178; + 1)
If loga (a + 2)
If 00 a + 2 > 2A > 1
1 / 20 2A > 0 a + 2 is obtained
If loga (1 / 2) 0, and a ≠ 1), find the value range of A
0 < a < 1/2
(2 / 3)
loga(2/3)
When 0 < x <, 4 ^ ×< loga x, then the value range of A
4^x<4^1/2=2
loga x>2=loga a^2
0
loga(x) 0
R
When 0 < x ≤ 1 / 2, 4 ^ x < loga ^ x, then the value range of A
0
f(x)=loga | loga x|(0
(1) First, x > 0, then | log a x | 0, then x is not equal to 1
(2) Log a | log a x | > 1, launch | log a x ||