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Math Help - Equations involving variables and Logarithms

  1. #1
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    Equations involving variables and Logarithms

    I have;

    log 2x = log 3+ log (x-1)

    and

    ln x + ln (x-2) - ln (x+4) = 0

    I can not figure out how to separate the variable from the constants.

    a\ log 2x = log 3+ log (x-1)
    b\ log 3+ log (x-1) - log 2x
    c\ (3x-3)/2x
    d\ log 3/2 (1-(1/x))
    c\ log 3/2 + log (1-(1/x))

    this approach doesn't seem to be going any place.
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    Re: Equations involving variables and Logarithms

    You can use the laws of logarithms to simplify your equations into one containing a single log

    \log_{b}(a) + \log_b(c) = \log_b(ac)

    \log_{b}(a) - \log_b(c) = \log_b \left(\dfrac{a}{c}\right)

    \log_b(a) = c \Leftrightarrow a = b^c


    If I take your first equation we can use the first law to say that \log(2x) = \log(3) + \log(x-1) \rightarrow \log \left(\dfrac{3(x-1)}{2x}\right) = 0.
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    Re: Equations involving variables and Logarithms

    Quote Originally Posted by e^(i*pi) View Post
    You can use the laws of logarithms to simplify your equations into one containing a single log

    \log_{b}(a) + \log_b(c) = \log_b(ac)

    \log_{b}(a) - \log_b(c) = \log_b \left(\dfrac{a}{c}\right)

    \log_b(a) = c \Leftrightarrow a = b^c


    If I take your first equation we can use the first law to say that \log(2x) = \log(3) + \log(x-1) \rightarrow \log \left(\dfrac{3(x-1)}{2x}\right) = 0.
    I do not see how this helps me at all. step c\ in my post is exactly what you conclude with. which is not the answer (x=3).
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    Re: Equations involving variables and Logarithms

    You have ended up with \displaystyle \log{\left[\frac{3(x - 1)}{2x}\right]} = 0. How do you undo a logarithm?
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    Re: Equations involving variables and Logarithms

    Quote Originally Posted by bkbowser View Post
    I do not see how this helps me at all. step c\ in my post is exactly what you conclude with. which is not the answer (x=3).
    What does \log(1)=~?
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    Re: Equations involving variables and Logarithms

    log (1) = 0
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    Re: Equations involving variables and Logarithms

    i think i'm trying to get a sort of zero (the input needs to be 1 not zero) of the rational function inside of the logarithm. clearly;

    (3x-3)/2x=1 when x = 3.

    but i can't come up with a generic way to evaluate for x with out just randomly inserting numbers or using a graphing program.

    i'm sorry if that wasn't as obvious as i thought it was in the first place or if i am incorrect and am doing the whole thing wrong.
    Last edited by bkbowser; July 5th 2011 at 12:46 PM. Reason: i phrased things inncorrectly.
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    Re: Equations involving variables and Logarithms

    In that case, simplify the RHS of your original equation to get \displaystyle \log{(2x)} = \log{[3(x - 1)]}.

    Can you go from here?
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    Re: Equations involving variables and Logarithms

    Quote Originally Posted by Prove It View Post
    In that case, simplify the RHS of your original equation to get \displaystyle \log{(2x)} = \log{[3(x - 1)]}.

    Can you go from here?
    no i can not.
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    Re: Equations involving variables and Logarithms

    Quote Originally Posted by Prove It View Post
    In that case, simplify the RHS of your original equation to get \displaystyle \log{(2x)} = \log{[3(x - 1)]}.
    Can you go from here?
    Quote Originally Posted by bkbowser View Post
    no i can not.
    Do you understand that if \log(X)=\log(Y)\text{ then }X=Y~?
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    Re: Equations involving variables and Logarithms

    Quote Originally Posted by Plato View Post
    Do you understand that if \log(X)=\log(Y)\text{ then }X=Y~?
    apparently not.

    3x-3 = 2x
    3x-2x = 3
    x = 3
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    Re: Equations involving variables and Logarithms

    Quote Originally Posted by bkbowser View Post
    apparently not.
    3x-3 = 2x
    3x-2x = 3
    x = 3
    Now you have done it correctly. Good!
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