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

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- Dec 22nd 2007, 09:39 AMperashsolve inequality
$\displaystyle \log_{x}(\frac{5}{2}-\frac{1}{x})>1$

- Dec 22nd 2007, 11:00 AMticbol
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. - Dec 22nd 2007, 11:01 AMtopsquark
Well, for starters to have valid argument for the logarithm we require

$\displaystyle \frac{5}{2} - \frac{1}{x} > 0$

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

$\displaystyle x > \frac{2}{5}$

so we must ensure that this condition holds.

Now,

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

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

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:

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

Again, since log( ) is never negative, and the function $\displaystyle 10^x$ is one to one:

$\displaystyle \frac{5}{2} - \frac{1}{x} > x$

Can you take this from here?

-Dan - Dec 22nd 2007, 11:12 AMred_dog
First of all we have to put the conditions for the existance of logarithm and for the base:

$\displaystyle \displaystyle\frac{5}{2}-\frac{1}{x}>0$ and $\displaystyle x>0, \ x\neq 1\Rightarrow x\in\left(\frac{2}{5},\infty\right)-\{1\}$

If $\displaystyle \displaystyle x\in\left(\frac{2}{5},1\right)$ then

$\displaystyle \displaystyle\frac{5}{2}-\frac{1}{x}<x\Rightarrow x\in\left(\frac{2}{5},\frac{1}{2}\right)$.

If $\displaystyle x\in(1,\infty)\Rightarrow\frac{5}{2}-\frac{1}{x}>x\Rightarrow x\in(1,2)$.

Then $\displaystyle x\in\left(\frac{2}{5},\frac{1}{2}\right)\cup(1,2)$ - Dec 22nd 2007, 11:35 AMtopsquark