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 :

If log_(0.3)(x-1)ltlog_(0.09)(x-1) , then x lies in the interval

If log_(0.3)(x-1)ltlog_(0.09)(x-1) ,then x lies in the interval

If log_(0.3)(x-1)

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"_(0.3)(x-1)lt"log"_(0.09)(x-1) , then find the interval in which x will lie.

If log_(0.3) (x-1) lt log _(0.09) (x-1) , " then " x ne 1 lies in : a)(1,2) b)(0,1) c) (1,infty) d) (2,infty)

(log)_(0.3)(x-1)<(log)_(0.09)(x-1)thenx in the lies in the interval- a.(2,oo) b.(1,2) c.(-2,-1) d.(-oo,2)