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Math Help - find x

  1. #1
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    find x

    $\log_[x]{81} - \log_[2x]{36}=2$
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  2. #2
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    how i can use latex
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  3. #3
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    Quote Originally Posted by perash View Post
    how i can use latex

    Hello,

    have a look here: http://www.mathhelpforum.com/math-he...-tutorial.html

    EB
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  4. #4
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    Quote Originally Posted by perash View Post
    $\log_[x]{81} - \log_[2x]{36}=2$
    Hello,

    I assume that you know x=e^{\ln(x)}
    and
    \log_{b}{(x)}=\frac{\ln(x)}{\ln(b)}

    \log_{x}{(81)}-\log_{2x}{(36)}=2 \Longleftrightarrow \frac{\ln(81)}{\ln(x)}-\frac{\ln(36)}{\ln(2x)}=2

    With \ln(2x)=\ln(x)+\ln(2) the LCD is: \ln(x) \cdot (\ln(x)+\ln(2))

    \frac{4\ln(3) \cdot (\ln(x)+\ln(2))}{\ln(x) \cdot (\ln(x)+\ln(2))}-\frac{(2\ln(2)+2\ln(3)) \cdot \ln(x)}{\ln(x) \cdot (\ln(x)+\ln(2))}=2

    After a few simple transformations you should get:

    (From this point the Latex doesn't work properly. I'll post the rest as plain text)

    (\ln(x))^2+\ln(4/3)*\ln(x) - 2*\ln(3)*\ln(2)=0
    I'm using now the logarithms in decimal form. Substitute: y=ln(x) and you'll get:

    y^2+0.287682y-5.5079002=0 \Longleftrightarrow  y=1.0986123 \vee y=-1.3862943

    Now re-substitute using x=e^y and you'll get: x = 3 \ \vee \  x = \frac{1}{4}.

    Probably there is a more elegant and easier way to solve this equation - but I couldn't find it.

    EB

    PS.: Obviously the Latex is back to duty!
    Last edited by earboth; February 8th 2007 at 04:52 AM.
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  5. #5
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    find x,y if

    [tex]\sqrt{x}+y=3[\math]
    [\sqrt{y}+x=5]
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  6. #6
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    Quote Originally Posted by perash View Post
    [tex]\sqrt{x}+y=3[\math]
    [\sqrt{y}+x=5]
    If you have a new question, please post it in a new thread.

    \sqrt{x} + y = 3 (Your "\" on the last LaTeX bracket is backward.)
    \sqrt{y} + x = 5

    Take the first equation:
    y = 3 - \sqrt{x}

    Then insert this into the second equation:
    (*) \sqrt{3 - \sqrt{x}} + x = 5

    \sqrt{3 - \sqrt{x}} = 5 - x <-- Square both sides.

    3 - \sqrt{x} = 25 - 10x + x^2

    \sqrt{x} = -x^2 + 10x - 22 <-- Squaring again.

    x = x^4 - 20x^3 + 144x - 440x + 484

    x^4 - 20x^3 + 144x^2 - 441x + 484 = 0

    I can't find a way to factor this thing, so I'm going to suggest a numerical solution. I get that x = 4 and x = 5.8 or so. (The other two solutions don't work.) Thus y = 1 and y = 0.591681084 or so, respectively.

    -Dan
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  7. #7
    ocean
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    A=1/2bh solve for b

    Quote Originally Posted by perash View Post
    $\log_[x]{81} - \log_[2x]{36}=2$
    A=1/2bh,solve for b
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