If loga 1 / 2 > 1 is known, the value range of a is obtained

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 ||