Logarithm is meaningful, the base numbers are a>0 and a≠ 1, and the true number >; 0,x & gt0
x^[loga(x)]<; Answer? x?
Classification discussion:
( 1)
a & gt 1,
loga[x^(loga(x))]<; loga(a? x? )
【loga(x)】? -2 loga(x)-3 & lt; 0
[loga(x)+ 1][loga(x)-3]& lt; 0
- 1 & lt; loga(x)& lt; three
1/a & lt; X< answer? The solution set of inequality is (1/a, a? )
(2)
0 & lta & lt 1,
loga[x^(loga(x))]>; loga(a? x? )
【loga(x)】? -2 loga(x)-3 & gt; 0
[loga(x)+ 1][loga(x)-3]& gt; 0
loga(x)& lt; -1 or loga(x) > 3
X> 1/a or 0
The solution set of inequality is (0, a? )U( 1/a,+∞)