a2logaaax
xx
aa
x2
x1
2
∵a
x1
ax2∴a
x1
x1
a∴aaaa
x2
x1
aax11即x2aa
∴logaaaxloga10
aa
2
∴fx1fx2
故fx为减函数2解：1当a1时，令uax∈ayu22u1u122y在u∈a上是增函数，∴当ua即axax1时ymi
14∴a22a114解得a3或a52当0a1时，类似可求得a∴a3或a
13131a1a
fr