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Math Help - Logarithms

  1. #1
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    Logarithms

    Solve for y in terms of x
    6log_2 x + log_2 y=3\Longrightarrow log_2 x^6 + log_2 y=3
    Then I took the anti log of both sides.
    Which gave me x^6 + y=2^3\Longrightarrow y=\frac{8}{x^6}
    Is that the right answer?
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  2. #2
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    You need to think again.
    \log _2 \left( {x^6 } \right) + \log _2 \left( y \right) = \log _2 \left( {x^6 y} \right)<br />
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  3. #3
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    Hello, OzzMan!

    Solve for y in terms of x.

    . . 6\log_2 x + \log_2 y\:=\:3\quad \Rightarrow\quad \log_2(x^6) + \log_2 y\:=\:3

    Then I took the anti log of both sides,
    . . which gave me x^6 + y\:=\:2^3 . . . . no


    but you got the right answer . . . How did you do that?

    We have: . \log_2(x^6) + \log_2(y) \;=\;3\quad\Rightarrow\quad \log_2(x^6y) \;=\;3

    Take antilogs: . x^6y \;=\;2^3\quad\Rightarrow\quad x^6y \;=\;8\quad\Rightarrow\quad y \;=\;\frac{8}{x^6}

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  4. #4
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    log_2(x^6y)=3
    \Rightarrow x^6y=2^3
    \Rightarrow y=\frac{8}{x^6}
    Last edited by OzzMan; February 8th 2008 at 03:36 PM.
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  5. #5
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    Is it possible to do this another way and still solve for y in terms of x? This is probably the only other way possible (unless I'm wrong):

    6Log_2\;x + Log_2\;y=3
    \Rightarrow Log_2(xy)=\frac{1}{2}
    \Rightarrow xy=2^{\frac{1}{2}}
    \Rightarrow y=\frac{2^\frac{1}{2}}{x}

    Different answer though. You think both are correct?
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  6. #6
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    Quote Originally Posted by OzzMan View Post
    Is it possible to do this another way and still solve for y in terms of x? This is probably the only other way possible (unless I'm wrong):

    6Log_2\;x + Log_2\;y=3
    \Rightarrow Log_2(xy)=\frac{1}{2} You didn't divide the \log_2(y) by 6
    \Rightarrow xy=2^{\frac{1}{2}}
    \Rightarrow y=\frac{2^\frac{1}{2}}{x}

    Different answer though. You think both are correct?
    I've marked the line where you made a mistake.
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