Log(base x)[5/2 -1/x] > 1

[5/2 -1/x] > x^1

2.5 -1/x > x

Multiply both sides by x,

2.5x -1 > x^2 --------no change in sense of inequality since x is positive

x^2 -2.5x +1 < 0

2x^2 -5x +2 < 0

(2x -1)(x -2) < 0

x < 1/2 -----------(1)

or x < 2 -----------(2)

And, since there are no real logs of negative numbers, then,

5/2 -1/x > 0

5x -2 > 0

x > 2/5

x > 0.4 ------------(3)

Investigate those three critical points.

i) In the interval (0.4, 1/2)

say x = 0.45,

Log(base 0.45)[5/2 -1/(0.45)] >? 1

0.2778 >? (0.45)^1

No, so this interval is not a solution.

ii) In the interval (1/2, 2)

say x = 1.5,

Log(base 1.5)[5/2 -1/(1.5)] >? 1

1.8333 >? (1.5)^1

Yes, so this interval is a solution.

iii) In the interval (2, infinity)

say x = 5,

Log(base 5)[5/2 -1/5)] >? 1

2.3 >? (5)^1

No, so this interval is not a solution.

Therefore, x = (1/2, 2), or 1/2 < x < 2 ------------answer.