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Math Help - Problem involving logs

  1. #1
    Ant
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    Problem involving logs

    2\log_{3} y = c-\log_{3} x

    where y = 2 and x = 4

    show that:

    y=4x^{-1/2}

    I've been able to find a value for C when y=2 and x=4 but this doesn't seem to help me at all. any help is really appreciated!
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by Ant View Post
    2\log_{3} y = c-\log_{3} x

    where y = 2 and x = 4

    show that:

    y=4x^{-1/2}

    I've been able to find a value for C when y=2 and x=4 but this doesn't seem to help me at all. any help is really appreciated!
    a \log b=\log b^a

    \log a+\log b=\log ab

    --> 2\log_{3} y=\log_3 y^2

    ----> \log_3 y^2+\log_3 x=c

    ------> \log_3 xy^2=c

    --------> xy^2=3^c

    etc
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  3. #3
    Ant
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    hi, thanks a lot for responding so fast!

    \log_3{xy}^2 = C
    <br />
\log_{3}xy^2 = log_3{3^c}

    at this point am I correct in saying that the log to base 3's cancel, leaving xy^2 = 3^c?
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  4. #4
    Moo
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    Quote Originally Posted by Ant View Post
    hi, thanks a lot for responding so fast!

    \log_3{xy}^2 = C
    <br />
\log_{3}xy^2 = log_3{3^c}

    at this point am I correct in saying that the log to base 3's cancel, leaving xy^2 = 3^c?
    Yes !
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  5. #5
    Ant
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    okay thanks, so the rest of the problem will follow like this:

    xy^2 = 3^c

    y = \frac{\sqrt{3^c}}{\sqrt{x}}

    y = \sqrt{3^c} * x^{-1/2}

    Where C can be found:

    xy^2=3^c

    16=3^c

    lg{16}=C * lg{3}

    \frac{lg{16}}{lg{3}}=c
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