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Thread: 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:

    $\displaystyle \log{(2x-2)}-\log{(x^2-1)}
    =\log{2(x+1)}, x>1
    $

    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 $\displaystyle 0<x<1$ then the $\displaystyle log(x^2-1)$ 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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    $\displaystyle \displaystyle \log{(2x-2)} - \log{(x^2-1)} = \log{[2(x-1)]} - \log{[(x-1)(x + 1)]}$

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

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

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

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

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

    Be careful . . .

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

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


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


    And all three inequalities are satisfied when $\displaystyle x > 1.$

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