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Math Help - Solve for x.

  1. #1
    Affinity
    Guest

    Solve for x.

    ln(x^2 - 5)=0

    I did it this way.. e^(ln[x^2 - 5])=e^0

    x^2 - 5=1

    x^2 = 6

    which x = the square root of 6.

    Would anyone kindly check this to see if I did it correctly? Thank you!
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  2. #2
    Member
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    you work is correct.
    but u missed a solution.
    not that if x^2=6 then we have x=\pm\sqrt{6}.
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  3. #3
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    Quote Originally Posted by Affinity View Post
    ln(x^2 - 5)=0

    I did it this way.. e^(ln[x^2 - 5])=e^0

    x^2 - 5=1

    x^2 = 6

    which x = the square root of 6.

    Would anyone kindly check this to see if I did it correctly? Thank you!
    If \ln (x^2-5)=0
    Thus,
     x^2-5=1
    Thus,
    x^2=6
    Thus,
    x=\pm \sqrt{6}
    Both of these are in domain of function
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