Correct Answer - B
We have, `x^("log"_(5)x) gt 5`
Clearly, it is meaningful for` x gt 0`.
Now, two case arise:
CASE I When `x gt 1`
In this case, we have
`x^("log"_(5)x) gt 5`
`rArr "log"_(5)x gt "log"_(x)5 " "[because a^(x) gt N rArr x gt "log"_(a) N, "if" a gt 1]`
`rArr "log"_(5)x gt (1)/("log"_(5)x)`
`rArr ("log"_(5)x -1)("log"_(5)x + 1)gt 0`
`rArr "log"_(5) x-1 gt 0 rArr x gt 5 rArr x in (5, oo)`
CASE II When `0 lt x lt 1`
In this case, we have
`x^("log"_(5)x) gt 5`
`rArr "log"_(5)x lt "log"_(x) 5 " " [because a^(x) gt N rArr x lt "log"_(a) N, "if" 0 lt a lt 1]`
`rArr "log"_(5)x gt (1)/("log"_(5)x)`
`rArr ("log"_(5)x)^(2)-1 gt 0 " " [because "log"_(5)x lt 0]`
`rArr ("log"_(5)x-1)("log"_(5)x+1) gt 0`
`rArr ("log"_(5)x+1)rArr x lt 5^(-1) rArr x in (0, 1//5)`
Hence, ` x in (0, 1//5) cup (5, oo)`
