If `log_(0.3)(x-1) lt log_(0.09)(x-1)`, then x lies in the interval :

If (log)_(0. 3)(x-1)<(log)_(0. 09)(x-1), then x lies in the interval (2,oo) (b) (1,2) (-2,-1) (d) None of these

If log_(0.5)(x-2)ltlog_(0.25)(x-2) then x lies is the interval (A) (-3,-2) (B) (2,3) (C) (3,oo) (D) none of these

If log_(0.04) (x-1)>=log_(0.2) (x-1) then x belongs to the interval

If "log"_(4)(3x^(2) +11x) gt 1 , then x lies in the interval

If log_(3)x-log_(x)27 lt 2 , then x belongs to the interval

If log_(2)(x-2)ltlog_(4)(x-2) ,, find the interval in which x lies.

|log_(3) x| lt 2

If a!=0 and log_x (a^2+1)lt0 then x lies in the interval (A) (0,oo) (B) (0,1) (C) (0,a) (D) none of these

If log_(cosx) sinx>=2 and x in [0,3pi] then sinx lies in the interval

If log_(1/3)((3x-1)/(x+2)) is less than unity, then 'x' must lie in the interval :