Answer :

case I

When a > 1


let x1,x2 (0,)


We have, x1<x2


loge x1 < loge x2


f(x1) < f(x2)


So, f(x) is increasing in (0,)


case II


When 0 < a < 1


f(x) = loga x


when a<1 log a < 0


let x1<x2


log x1 < log x2


[]


f(x1) > f(x2)


So, f(x) is decreasing in (0,)


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