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Math Help - Logarithmic equation help

  1. #1
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    Logarithmic equation help

    log(x-1) - log(x+6) = log(x-2) - log(x-3)

    Is this right?

    (x-1)/(x+6) = (x-2)/(x-3)

    (x-3)(x-1)/(x+3)(x+6) = (x-2)(x+6)/(x+3)(x+6)

    (x-1)/(x-6) = (x-2)/(x-3)

    x^2 + 4x - 12 = x^2 - 4x - 3

    x = 9/8
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  2. #2
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    Looks right. BUT
    log(x-2)=log(9/8-2)=log(-7/8) which is not defined.

    So, something's wrong !

    -O
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by NeedHelp18 View Post
    log(x-1) - log(x+6) = log(x-2) - log(x-3)

    Is this right?

    (x-1)/(x+6) = (x-2)/(x-3)

    (x-3)(x-1)/(x+3)(x+6) = (x-2)(x+6)/(x+3)(x+6)

    (x-1)/(x-6) = (x-2)/(x-3)
    huh?

    shouldn't that line be (x - 1)(x - 3) = (x - 2)(x + 6)? ...(which would yield (x - 1)/(x + 6) = (x - 2)/(x - 3) ...but that's not what we would do)

    expand both sides:

    x^2 - 4x ~{\color{red}+}~ 3 = x^2 + 4x - 12

    now solve
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  4. #4
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    Hello, NeedHelp18!


    \log(x-1) - \log(x+6) \:=\: \log(x-2) - \log(x-3)

    Is this right?

    \log\left[\frac{x-1}{x+6}\right] \:=\:\log\left[\frac{x-2}{x-3}\right]

    \frac{x-1}{x+6} \:=\:\frac{x-2}{x-3}


    \frac{{\color{red}\rlap{////////////}}(x-3)(x-1)}{{\color{red}\rlap{////////////}}(x+3)(x+6)} \:{\color{red}\rlap{///}}=\: \frac{{\color{red}\rlap{////////////}}(x-2)(x+6)}{{\color{red}\rlap{////////////}}(x+3)(x+6)} .
    . . . What was all that about?

    \frac{x-1}{x+6} \:=\:\frac{x-2}{x-3}

    x^2 + 4x - 12 \:= \:x^2 - 4x {\color{red}\rlap{\;///}}- 3 . . . . no

    x \:= \:\frac{9}{8}

    We have: . \log(x-1) - \log(x+6) \:=\:\log(x-2)-\log(x-3)

    Rearrange terms: . \log(x-1) + \log(x-3) \:=\:\log(x-2) + \log(x+6)

    And we have: . \log\bigg[(x-1)(x-3)\bigg] \:=\:\log\bigg[(x-2)(x+6)\bigg]

    Exponentiate: . (x-1)(x-3) \:=\:(x-2)(x+6) \quad\Rightarrow\quad x^2-4x+3 \:=\:x^2+4x-12

    . . which simplifies to: . 8x \:=\:15 \quad\Rightarrow\quad x \:=\:\frac{15}{8}


    But this answer turns out to be extraneous.

    The equation has no solution.

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