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Math Help - Quick log restriction clarification

  1. #1
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    Quick log restriction clarification

    Hey, I have an exam tomorrow afternoon and I have a quick question I'd like to clear up. In the following example:

    \log{(2x-2)}-\log{(x^2-1)}<br />
=\log{2(x+1)}, x>1<br />

    Why is the restriction x>1? In all the previous examples in the book, they show x>0 as you cannot log negatives. But in this example, they show x>1. I figure x cannot equal 1, but why cannot it be 0<x<1?
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  2. #2
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    If 0<x<1 then the log(x^2-1) bit will be undefined, since the value within the log function will be less than 0.
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  3. #3
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    \displaystyle \log{(2x-2)} - \log{(x^2-1)} = \log{[2(x-1)]} - \log{[(x-1)(x + 1)]}

    \displaystyle = \log{\left[\frac{2(x-1)}{(x-1)(x + 1)}\right]}

    \displaystyle = \log{\left(\frac{2}{x+1}\right)}

    \displaystyle \neq \log{[2(x+1)]}...
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  4. #4
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    Hello, youngb11!

    \log(2x-2)-\log(x^2-1) \:=\:\log[2(x+1)],\;\; x>1

    \text{Why is the restriction }x>1\,?
    \text{In all the previous examples in the book, they show }x>0
    \text{as you cannot log negatives.}

    Be careful . . .

    It is not \,x that must be positive,
    . . It is the log argument that must be positive.

    The restrictions are: . \begin{Bmatrix} 2x - 2 &>& 0 & [1] \\ x^2-1 &>& 0 & [2] \\ 2(x+1) &>& 0 & [3] \end{Bmatrix}


    \begin{array}{ccccccccc}\text{From [1]:} & 2x - 2 \:>\:0 &\Rightarrow&  2x \:> \:2 &\Rightarrow& x \:> \:1  \\ \\[-3mm]<br />
\text{From [2]:} & x^2 - 1 \:>\:0 &\Rightarrow& x^2 \:>\:1 &\Rightarrow& |x| \:>\:1 \\ \\[-3mm]<br />
\text{From [3]:} & 2(x+1) \:>\:0 &\Rightarrow& x+1 \:>\:0 &\Rightarrow& x \:>\: \text{-}1 \end{array}


    And all three inequalities are satisfied when x > 1.

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