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Inequalities and Logs

  1. Sep 26, 2006 #1
    Check this out:

    [tex]1 < 2 [/tex]

    [tex]\Rightarrow \frac{1}{4} < \frac{1}{2}[/tex]

    [tex]\Rightarrow (\frac{1}{2})^2 < \frac{1}{2}[/tex]

    [tex]\Rightarrow \log(\frac{1}{2})^2 < \log(\frac{1}{2}) [/tex]

    [tex]\Rightarrow 2\cdot\log(\frac{1}{2}) < \log(\frac{1}{2})[/tex]

    [tex]\Rightarrow 2 < 1[/tex]

    What happened? What did I do wrong?
  2. jcsd
  3. Sep 26, 2006 #2


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    what is log(1/2)?
  4. Sep 26, 2006 #3
    I meant log base 10 of 1/2. I don't think it matters though...
  5. Sep 26, 2006 #4
    it's still the same problem. As shmoe suggested, try calculating log(1/2) :tongue2:
  6. Sep 27, 2006 #5


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    You divided both sides of the equation by log(1/2). What happens to an inequality when you multiply or divide by sides by a ____________? Once again, what is log(1/2)?
  7. Sep 27, 2006 #6

    matt grime

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    That wasn't what you were supposed to think. I hope the other questions have cleared up what was intended. This problem is usually the first 'trick' they play on you with logs, and the first one they explain.
  8. Sep 27, 2006 #7
    I got it! I am so smart.:biggrin:
  9. Sep 27, 2006 #8


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    But you still haven't told us what log(1/2) is. And why that is important.
  10. Sep 27, 2006 #9
    Its a negative number and in the last step I divide both sides of log(1/2) which changes the sense of the inequality.
  11. Jun 3, 2009 #10
    Thanks for this little thread. some of us are a bit slow on the uptake....
    I was working on a different problem and this helped jog me into seeing that multiplication/division by a negative number constitutes changing the sign on both sides of the equation.

    By the way, it does matter slightly what base is used. If the base is < 1, then log(x), where x < 1, will actually be positive. Then when you divide out the logs you won't change the inequality.
    However, the results come out the same because you have to change the inequality at an earlier stage, namely when you first take logs. Taking logs with a base < 1 of both sides reverses the inequality.

    0.25 < 0.5
    but for 0.1^x = 0.25, and 0.1^y = 0.5; x > y (when raising a number < 1 to a power, the larger the power the smaller the result)
    Last edited: Jun 3, 2009
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