x^((log)5x)>5 implies x in (0,oo) (b) [...

`x^((log)_5x)>5` implies `x in (0,oo)` (b) [2,2.5] (c) (2,2.5) (d) (0,2.5)

` x in (0, infty)`

` x in (0, 1//5) cup (5, infty)`

` x in (1, infty)`

` x in (1, 2)`

Text Solution

Verified by Experts

The correct Answer is:

`x^(log_(5) x) gt 5`
Taking logarithm with base 5, we have
`(log_(5)x) (log_(5)x) gt 1`
` or (log_(5)x-1) (log_(5)x+1) gt 0`
` or log_(5) x gt 1 or log_(5) x lt - 1`
`rArr x gt 5 or x lt 1//5`
Also we must have ` x gt 0` . Thus,
` x in (0, 1//5) cup (5, infty)`.

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