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Thread: 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 $\displaystyle (x+9)^2$ must be greater than zero.

    $\displaystyle (x+9)^2 > 0$

    $\displaystyle x+9>0$

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

    Of course, your answer is correct if the problem means $\displaystyle \left[ {\log _8 \left( {x + 9} \right)} \right]^2.$
    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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