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Homework Help: Math Help Equations

  1. Aug 4, 2005 #1
    1) I need to express n in terms of X in this equation:

    [tex] X = \frac {10^n-1} {10^n} [/tex]

    I got to this so far but I don't know if I'm on the right path or not and I don't know how to continue:

    [tex] \log {X} + n = \log {(10^n-1)} [/tex]

    2) I have to prove this:

    [tex] \frac {1} {\sqrt{4-2\sqrt {3}}} = \frac {\sqrt{3}+1} {2} [/tex]

    Don't know how to continue from here:

    [tex] \frac {\sqrt{2} \sqrt {2+\sqrt {3}}} {2} [/tex]
     
  2. jcsd
  3. Aug 4, 2005 #2
    For number one:
    [tex] X = \frac {10^n-1} {10^n} [/tex]
    [tex] X = 1 - \frac {1} {10^n} [/tex]
    Which gives you:
    [tex] 1 - X = \frac {1} {10^n} [/tex]

    Can you go from here?
     
  4. Aug 4, 2005 #3

    saltydog

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    For (2), note that:

    [tex]1+\frac{\sqrt{3}}{2}=(a+b)^2[/tex]

    a and b has to be the expression you're trying to show right?
     
  5. Aug 4, 2005 #4

    EnumaElish

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    [tex] 2 = (\sqrt{3}+1) {\sqrt{4-2\sqrt {3}}} [/tex]
    [tex] 4 = (\sqrt{3}+1)^2 (4-2\sqrt {3}) [/tex]

    Now expand the square, then expand everything, then simplify to show left = right.
     
  6. Aug 4, 2005 #5
    This is the answer for the first one, isn't it?

    [tex] n = -\log {(1-X)} [/tex]

    It was easy, but I don't know why I couldn't do it. Thanks a lot.

    I'm having serious trouble with the second one

    I suppose you got those equalities by rearranging the terms in the first equation. The thing is that the problem just says simplify:

    [tex] \frac {1} {\sqrt{4-2\sqrt {3}}} [/tex]

    I'm sorry, it was my mistake. The second part is the answer given by the teacher.
     
  7. Aug 4, 2005 #6
    Also, do you mean I have to replace those equations in

    [tex] \sqrt{4+2\sqrt {3} [/tex]

    to get

    [tex] \sqrt {(\sqrt{3}+1)^2 (4-2\sqrt {3}) + [(\sqrt{3}+1) {\sqrt{4-2\sqrt {3}}}] \sqrt {3}} [/tex]
     
  8. Aug 4, 2005 #7

    VietDao29

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    Okay. Here is what you should do :
    You know that:
    [tex](\alpha + \beta) ^ 2 = \alpha ^ 2 + 2\alpha \beta + \beta ^ 2[/tex]
    You will try to arrange [tex]\aqrt{4 - 2 \sqrt{3}}[/tex] into something like: [tex](\alpha + \beta) ^ 2[/tex] Then you can easily take the square root of it.
    So you have [tex]\alpha ^ 2 + \beta ^ 2 = 4 \mbox{, and } \alpha \beta = -\sqrt{3}[/tex]
    Can you solve for [itex]\alpha[/itex], and [itex]\beta[/itex]?
    Then can you solve : [tex]\sqrt{(\alpha + \beta) ^ 2}[/tex]? Just remember that:
    [tex]\sqrt{A ^ 2} = |A|[/tex]
    Viet Dao,
     
    Last edited: Aug 4, 2005
  9. Aug 5, 2005 #8
    I finally did it. Thanks a lot Viet Dao. I would never have guessed I had to do that.
     
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