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Math Help - More log stuff

  1. #16
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    Quote Originally Posted by SVXX View Post
    Ok, the last problem.

    |\log_{2}\left(\frac{4-x}{2x - 5}\right)| + \log_{2}x = 1

    ....
    Here is another attempt to solve the equation:

    1. On behalve of the summand \log_{2}x x must be greater than zero.

    2.
    \left| \log_{2}\left(\frac{4-x}{2x - 5}\right)\right| + \log_{2}x = 1

    \left|\log_{2}\left(\frac{4-x}{2x - 5}\right)\right| = \log_{2}\left(\frac2x\right)

    You have to consider 2 different cases:

    \log_{2}\left(\frac{4-x}{2x - 5}\right) > 0~\implies~ \frac{4-x}{2x - 5} > 1

    Then you must solve for x:

    \frac{4-x}{2x - 5} = \frac2x~\implies~ \boxed{x = \sqrt{10}}

    The negative solution isn't valid. See #1.

    Or

    \log_{2}\left(\frac{4-x}{2x - 5}\right) < 0~\implies~ - \log_{2}\left(\frac{4-x}{2x - 5}\right) > 0 ~\implies~ \frac{2x - 5}{4-x} > 1

    Then you have to solve for x:

    \frac{2x - 5}{4-x} = \frac2x ~\implies~ \boxed{x \approx 2.886}

    The negative solution isn't valid. See #1.

    Edit: To my great confusion both solutions don't satisfy the constraints to yield a fraction which is greater than 1.
    Last edited by earboth; May 12th 2008 at 07:03 AM.
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  2. #17
    Moo
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    Quote Originally Posted by SVXX View Post
    In that case, x = -1.38!
    I forgot something, I remembered that when taking my shower .. .This one doesn't work either because we have \log x, meaning that x has to be > 0...

    Sorry about that
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  3. #18
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    Quote Originally Posted by SVXX View Post
    Ok, the last problem.

    |\log_{2}(\frac{4-x}{2x - 5})| + \log_{2}x = 1
    ...
    To illustrate that your equation don't have real solutions I've sketched the graphs of

    f(x)=\left|\log_{2}\left(\frac{4-x}{2x - 5}\right)\right| + \log_{2}x

    and

    g(x)=1

    The graphs don't intersect. That means there doesn't exist an x so that the values of the functions are equal.
    Attached Thumbnails Attached Thumbnails More log stuff-svxx_betragfkt.gif  
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  4. #19
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    Quote Originally Posted by earboth View Post
    To illustrate that your equation don't have real solutions I've sketched the graphs of

    f(x)=\left|\log_{2}\left(\frac{4-x}{2x - 5}\right)\right| + \log_{2}x

    and

    g(x)=1

    The graphs don't intersect. That means there doesn't exist an x so that the values of the functions are equal.
    You're right, I went through the properties again..x cannot possibly be negative. That means these equations have no solution as per you peeps.
    That'll be interesting. Wonder what the teacher will say when I point that out .
    Anyways thanks for pointing out this huge blunder!
    I can write x doesn't belong to R then I guess.
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