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.
Well, for starters to have valid argument for the logarithm we require
so we must ensure that this condition holds.
Now,
by the change of base formula. (I'm using base 10 here, but feel free to use the base of your choice. It doesn't affect the answer.)
Since log(x) is never negative:
Again, since log( ) is never negative, and the function is one to one:
Can you take this from here?
-Dan