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Thread: Logharitmic inequality

  1. #1
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    Logharitmic inequality

    $\displaystyle log_x|y-2|>log_{|y-2|} x$, Can somebody help me?
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    Re: Logharitmic inequality

    Quote Originally Posted by greg1987 View Post
    $\displaystyle log_x|y-2|>log_{|y-2|} x$, Can somebody help me?
    It looks like you have dumped us into the middle of a problem.
    What is the exact statement of the entire problem?
    There must be more to the question.
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  3. #3
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    Re: Logharitmic inequality

    hi

    try using: $\displaystyle \log_a b=\frac{1}{\log_b a}$

    then use substitution.

    hope this helps.
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    Re: Logharitmic inequality

    We've got to find a set of points (x,y) (and mark it on a plane) which satisfies this inequality.That's all.
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    Re: Logharitmic inequality

    Quote Originally Posted by anonimnystefy View Post
    hi

    try using: $\displaystyle \log_a b=\frac{1}{\log_b a}$

    then use substitution.

    hope this helps.
    ok so i've got $\displaystyle \log_x |y-2|>\frac{1}{\log_x |y-2|}$
    Now I need to multiply the denominator?
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  6. #6
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    Re: Logharitmic inequality

    you cannot multiply because you don't know if the sign of the denominator is + or -.

    did you try using substitution?it will make it a little easier to see what you can do.
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  7. #7
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    Re: Logharitmic inequality

    I don't know what should I substitute, some clue?
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  8. #8
    Member anonimnystefy's Avatar
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    Re: Logharitmic inequality

    try $\displaystyle t=\log_x |y-2|$
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    Re: Logharitmic inequality

    so -1<t<0 or t>1?
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  10. #10
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    Re: Logharitmic inequality

    that's pretty much right.

    what do you get when you sub?
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    Re: Logharitmic inequality

    rewrite by meaning:

    x^a = (y-2); (y -2)^(1/a) = x

    is same as:

    a > 1/a solve for a
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  12. #12
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    Re: Logharitmic inequality

    $\displaystyle t=\log_x |y-2|>1$ or $\displaystyle -1<t=\log_x |y-2|<0$ and how can I determine x and y?
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    Re: Logharitmic inequality

    Quote Originally Posted by greg1987 View Post
    We've got to find a set of points (x,y) (and mark it on a plane) which satisfies this inequality.That's all.
    Well then you have included that.
    You know that $\displaystyle x>0~\&~x\ne 1$, WHY?
    For the same reason $\displaystyle y\ne 2$.
    If $\displaystyle y>e+2$ then $\displaystyle \log_x|y-2|>0$ WHY?
    What about $\displaystyle y<2-e~?$

    There is a start.
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  14. #14
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    Re: Logharitmic inequality

    Here is a plot (starting from 100,000 random points), both axes are [-5,5].
    Attached Thumbnails Attached Thumbnails Logharitmic inequality-view.jpg  
    Last edited by ZeroDivisor; Oct 8th 2011 at 08:45 AM.
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