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Math Help - Restrictions for logarithms

  1. #1
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    Restrictions for logarithms

    I missed a lesson on the product and quotient laws for logarithms and am having a little trouble figuring out the restrictions.

    For example:

    Log (x^2+7x+12) - Log (x^2-9)

    = Log (x+3)(x+4) - Log (x-3)(x+3)

    = Log [(x+3)(x+4)]/(x-3)(x+3)]

    = Log (x+4)/(x-3)

    How would I go about finding the restrictions on this logarithm?

    I know in total, there are three possibilities: x>-3, x>-4, x>3

    And in the back of the book, it shows that x>3 is the only restriction.

    What I tried was to plug in each restriction into each separate log to see if it worked, but when I tried it that way I found that both x>3 and x>-4 worked, and that conflicts with the answer in the back of the book.

    All help greatly appreciated!
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  2. #2
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    The domain of a log is (0,\infty).
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  3. #3
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    Quote Originally Posted by youngb11 View Post
    = Log (x+4)/(x-3)
    Firstly x\neq 3, this should be clear.

    Taking x<3 will give you a negative result inside the logarithm, refer to post #2 for the reason this can't be.
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  4. #4
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    Quote Originally Posted by pickslides View Post
    Firstly x\neq 3, this should be clear.
    That's what I was thinking, but the back of the book doesn't include that as a restriction.

    Quote Originally Posted by pickslides View Post
    Taking x<3 will give you a negative result inside the logarithm, refer to post #2 for the reason this can't be.
    So to find the restrictions, I have to take all the possible restrictions and the final answer will include the restrictions which will not result in a negative inside the logarithm?
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  5. #5
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    Whatever x you plug in, you MUST have a value of f(x)>0 where f(x) is the expression inside the log.
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  6. #6
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    \displaystyle\log{\left(\frac{x^2+7x+12}{(x^2-9)}\right)}

    \displaystyle\log{\left(\frac{(x+3)(x+4)}{(x-3)(x+3)}\right)}\Rightarrow\log{\left(\frac{x+4}{x-3}\right)}

    x\neq \pm 3, -4

    \displaystyle \frac{x+4}{x-3}>0
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  7. #7
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    Quote Originally Posted by youngb11 View Post
    That's what I was thinking, but the back of the book doesn't include that as a restriction.
    If x>3 is the solution

    x= 3 \subset x\leq 3
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  8. #8
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    Once you check the intervals, you will have (,)\cup (,)
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