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Math Help - Find The Domain Of The Logarithmic Function

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    Find The Domain Of The Logarithmic Function

    f(x)= log8 (x + 9)^2
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    Quote Originally Posted by J_Pettway View Post
    f(x)= log8 (x + 9)^2
    You can only take the log of positive numbers, so (x+9)^2 must be greater than zero.

    (x+9)^2 > 0

    x+9>0

    x>-9
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    Quote Originally Posted by colby2152 View Post
    You can only take the log of positive numbers, so (x+9)^2 must be greater than zero.
    (x+9)^2 > 0, x+9>0, x>-9
    Because \log _8 \left( {x + 9} \right)^2  = 2\log _8 \left| {x + 9} \right|
    The domain is \left\{ {x \in \mathbb{R}:x \ne  - 9} \right\}.
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    Quote Originally Posted by Plato View Post
    Because \log _8 \left( {x + 9} \right)^2  = 2\log _8 \left| {x + 9} \right|
    The domain is \left\{ {x \in \mathbb{R}:x \ne  - 9} \right\}.
    Wouldn't log properties say \log _8 \left( {x + 9} \right)^2  = 2\log _8  (x + 9) ???
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    Quote Originally Posted by colby2152 View Post
    Wouldn't log properties say \log _8 \left( {x + 9} \right)^2  = 2\log _8  (x + 9) ???
    No.
    Realize that \left( {x + 9} \right)^2  = \left| {x + 9} \right|^2 .

    Of course, your answer is correct if the problem means \left[ {\log _8 \left( {x + 9} \right)} \right]^2.
    That is the way most computer algebra systems would read it.

    But I don't think that is what was meant.
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