1. solve inequality

$\log_{x}(\frac{5}{2}-\frac{1}{x})>1$

2. 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.

3. Originally Posted by perash
$\log_{x}(\frac{5}{2}-\frac{1}{x})>1$
Well, for starters to have valid argument for the logarithm we require
$\frac{5}{2} - \frac{1}{x} > 0$

$\frac{1}{x} < \frac{5}{2}$

$x > \frac{2}{5}$
so we must ensure that this condition holds.

Now,
$log_{x}(\frac{5}{2}-\frac{1}{x})>1$

$\frac{log(\frac{5}{2}-\frac{1}{x})}{log(x)} > 1$
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:
$log(\frac{5}{2}-\frac{1}{x}) > log(x)$

Again, since log( ) is never negative, and the function $10^x$ is one to one:
$\frac{5}{2} - \frac{1}{x} > x$

Can you take this from here?

-Dan

4. First of all we have to put the conditions for the existance of logarithm and for the base:
$\displaystyle\frac{5}{2}-\frac{1}{x}>0$ and $x>0, \ x\neq 1\Rightarrow x\in\left(\frac{2}{5},\infty\right)-\{1\}$
If $\displaystyle x\in\left(\frac{2}{5},1\right)$ then
$\displaystyle\frac{5}{2}-\frac{1}{x}.
If $x\in(1,\infty)\Rightarrow\frac{5}{2}-\frac{1}{x}>x\Rightarrow x\in(1,2)$.
Then $x\in\left(\frac{2}{5},\frac{1}{2}\right)\cup(1,2)$

5. Originally Posted by red_dog
First of all we have to put the conditions for the existance of logarithm and for the base:
$x\neq 1$
Ooh! I didn't think to look at that!

(The story of my day, basically. )

-Dan