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Math Help - Logarithm Inequality. Help!

  1. #1
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    Logarithm Inequality. Help!

    Log x ((x+3)/(x-1)) > Log x x ??

    how to find x that satisfy this inequality?
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    Re: Logarithm Inequality. Help!

    Quote Originally Posted by psw1412 View Post
    Log x ((x+3)/(x-1)) > Log x x ??

    how to find x that satisfy this inequality?
    Logarithm Inequality. Help!-untitled2.gif

    Solve the inequality and choose the interval in which x > 0
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    Re: Logarithm Inequality. Help!

    Quote Originally Posted by votan View Post
    Click image for larger version. 

Name:	untitled2.gif 
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    Solve the inequality and choose the interval in which x > 0

    is it correct??
    Attached Thumbnails Attached Thumbnails Logarithm Inequality. Help!-20130922_210751_320x240.jpg  
    Last edited by psw1412; September 22nd 2013 at 11:12 AM.
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    Re: Logarithm Inequality. Help!

    Quote Originally Posted by psw1412 View Post
    is it correct??
    NO! It should be:
    \\\frac{3-x}{x-1}>x\\\frac{3-x}{x-1}-x>0\\\frac{3+2x-x^2}{x-1}>0\\\frac{(3-x)(1+x)}{x-1}>0.

    That means x<-1\text{ or }1<x<3
    Thanks from psw1412
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    Re: Logarithm Inequality. Help!

    thank you.
    one more question. How could you write that inequality from codecogs.com in a few rows? Every time I press "enter" it continues in the same row. Like this: Logarithm Inequality. Help!-codecogseqn.png

    2x=4
    x=2
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    Re: Logarithm Inequality. Help!

    Quote Originally Posted by Plato View Post
    NO! It should be:
    \\\frac{3-x}{x-1}>x\\\frac{3-x}{x-1}-x>0\\\frac{3+2x-x^2}{x-1}>0\\\frac{(3-x)(1+x)}{x-1}>0.

    That means x<-1\text{ or }1<x<3
    The argument of log cannot be negative. Also, the numerator is x + 3, not 3 - x
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    Re: Logarithm Inequality. Help!

    Quote Originally Posted by votan View Post
    The argument of log cannot be negative. Also, the numerator is x + 3, not 3 - x
    I was not addressing the OP.
    If I were then I would have quoted it.
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    Re: Logarithm Inequality. Help!

    Quote Originally Posted by psw1412 View Post
    is it correct??
    Do not multiply both sides by the denominator. It is OK in equality provided the value of x does not make the denominator 0.
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