1.

`(log)_(0. 3)(x-1)

Answer» Correct Answer - A
We have,
`"log"_(0.3) (x-1) lt "log"_(0.09) (x-1)`
Clearly, it is defined for ` x gt 1`
Now,
`"log"_(0.3)(x-1) lt "log"_(0.09) (x-1)`
`rArr "log"_(0.3)(x-1) lt "log"_((0.3))^(2) (x-1)`
`rArr "log"_(0.3)(x-1) lt (1)/(2) "log"_(0.3) (x-1)`
`rArr "log"_(0.3)(x-1) lt 0 rArr x -1 gt (0.3)^(0) rArr x gt 2 rArr x in (2, oo)`


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