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Math Help - Logarithmic Equation is undefined (help).

  1. #1
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    Logarithmic Equation is undefined (help).

    Solving a Logarithmic Equation

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


    the answers for this is x=-4 , x=3

    x=-4 is undefined , however when we combine it
    it becomes

    log((x+2)(X-1))=1 and substitute x=-4 its defined .

    Sense log(x+2)+log(x-1)=log((x+2)(X-1)) why there is a difference??
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  2. #2
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    e^(i*pi)'s Avatar
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    Re: Logarithmic Equation is undefined (help).

    If y = \log[f(x)] then f(x) > 0 in order for a real solution to exist. In other words you discard -4 as a solution because it's not in the domain of the original equation

    It works in \log[(x+2)(x-1)] because both x+2 and x-1 are negative when x=-4 then their product is positive.
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    Re: Logarithmic Equation is undefined (help).

    Quote Originally Posted by e^(i*pi) View Post
    If y = \log[f(x)] then f(x) > 0 in order for a real solution to exist. In other words you discard -4 as a solution because it's not in the domain of the original equation

    It works in \log[(x+2)(x-1)] because both x+2 and x-1 are negative when x=-4 then their product is positive.
    But why do both give different answers while they are they same??

    according to log(A.B)=log(A)+log(B)

    log(x+2)+log(x-1)=log((x+2)(X-1))
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  4. #4
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    Re: Logarithmic Equation is undefined (help).

    Quote Originally Posted by omar1 View Post
    But why do both give different answers while they are they same??

    according to log(A.B)=log(A)+log(B)

    log(x+2)+log(x-1)=log((x+2)(X-1))
    In going from log(x+2)+log(x-1)=1 to log[(x+2)(x-1)]=1 you have changed the original equation and picked up an extraneous solution. A 'solution' found in this way must always be substituted into the original equation, checked and rejected (with reason given) if it does not satisfy the original equation.
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    Re: Logarithmic Equation is undefined (help).

    Thank you very much for helping me!
    Mr. fantastic and e^(i*pi) .
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