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Math Help - Exponential and Logarithmic Equations

  1. #1
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    Talking Exponential and Logarithmic Equations

    Solve the logarithmic equation algebraically and approximate the result

    Ln x + Ln(x - 2) = 1

    I solved it to approximately 2.928

    I want to ask why the negative number is extraneous because I'm having a difficult time grasping what is going on with the 2 variables being found on the graph, and graphing it wise. Seriously...why are negative answers extraneous???

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by ugkwan View Post
    Solve the logarithmic equation algebraically and approximate the result

    Ln x + Ln(x - 2) = 1

    I solved it to approximately 2.928

    I want to ask why the negative number is extraneous because I'm having a difficult time grasping what is going on with the 2 variables being found on the graph, and graphing it wise. Seriously...why are negative answers extraneous???

    Thanks in advance!
    The left hand side can be written as Ln(x(x - 2)) using the usual log rule. Therefore you have the quadratic equation x^2 - 2x - e = 0. The gaps are left for you to fill, as is the exact solution. Note that only values of x that satisfy x > 2 will stisfy the original equation because the log of a negative number is not real.
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