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