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Thread: solve inequality

  1. #1
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    solve inequality

    $\displaystyle \log_{x}(\frac{5}{2}-\frac{1}{x})>1$
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  2. #2
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    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.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by perash View Post
    $\displaystyle \log_{x}(\frac{5}{2}-\frac{1}{x})>1$
    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
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  4. #4
    MHF Contributor red_dog's Avatar
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    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 $\displaystyle x\in\left(\frac{2}{5},\frac{1}{2}\right)\cup(1,2)$
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by red_dog View Post
    First of all we have to put the conditions for the existance of logarithm and for the base:
    $\displaystyle x\neq 1$
    Ooh! I didn't think to look at that!

    (The story of my day, basically. )

    -Dan
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