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Math Help - I can't find what I'm doing wrong in this inequality

  1. #1
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    Angry I can't find what I'm doing wrong in this inequality

    I am trying to solve
    log_{0.5}\frac{x+3}{x-1} \geq 0.

    \frac{log\frac{x+3}{x-1}}{log_{0.5}} \geq 0

    Because log 0.5 is always negative, the entire numerator must be negative as well for the entire expression to be positive. Therefore,

     0 < \frac{x+3}{x-1} \leq 1.

    For  0 < \frac{x+3}{x-1} I reached the conclusion that  x < -3 or  x > 1

    But I'm stuck at \frac{x+3}{x-1} \leq 1
    if I multiply the denominator over it will be x + 3 < x -1, which makes no sense to me
    Please, if someone could point out where I made my error, I've spent a lot of time on this question and the due date is approaching fast. Thank you!
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  2. #2
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    \log_{.5}{x} is only negative for x>1, it is positive for 0<x<1
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  3. #3
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    Yes that's what I already have. If log_{0.5}x is positive on 0 < x < 1 then log_{0.5}\frac{x+3}{x-1} is positive for 0 < (x+3)/(x-1) < 1 but that is the inequality that I'm stuck on.
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  4. #4
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    sorry I made an error in latex, it's fixed now
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  5. #5
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    Quote Originally Posted by xxlvh View Post

    But I'm stuck at \frac{x+3}{x-1} \leq 1
    if I multiply the denominator over it will be x + 3 < x -1, which makes no sense to me
    it makes no sense, why did you multiply then? you can't, because don't know that the sign of x-1 is, so you need to proceed from a different way.

    start by leaving that inequality with a LHS in function of x and with a RHS in zero, what i meant is you to leave it as f(x)\le0, and then you'll be able to solve it as you did it before.
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  6. #6
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    So: \frac{x+3}{x-1}-1 \leq 0
    Would the critical values that I need to inspect be -3 and 1?
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  7. #7
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    no, clear the clutter first: \frac4{x-1}\le0 and now we require that x-1<0.
    Last edited by Krizalid; September 19th 2009 at 11:53 AM.
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  8. #8
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    Ah! I see, that brings it to x - 1 < 0 is the answer then x < -1?
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  9. #9
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    yes, but just for that inequality, not for the whole one you wrote before.

    you split two inequalities, we have then two solution sets, and those solution sets must intersect so that you can get the full interval where the logarithm is negative.
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  10. #10
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    Correction I meant to type x < 1 on that last post.
    Okay then back to  0 < \frac{x+3}{x-1} \leq 1
    the lhs inequality resulted with x < -3 and x > 1 and the rhs inequality resulted in x < 1... does that just turn it to x < -3?
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