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Math Help - logs question

  1. #1
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    logs question

    \frac{ log_2{16} + log_2{32} }{log_2{x}} = log_2{x}

     \frac{log_2{512}}{log_2{x}} = log_2{x}

     log_2{512} = log_2{x^2}

     512 = x^2

     \sqrt{512} = x

    so x = 22.6 , is this correct?

    appreciate any input.

    thanks
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Tweety View Post
    \frac{ log_2{16} + log_2{32} }{log_2{x}} = log_2{x}

     \frac{log_2{512}}{log_2{x}} = log_2{x}

     log_2{512} = log_2{x^2}

     512 = x^2

     \sqrt{512} = x

    so x = 22.6 , is this correct?

    appreciate any input.

    thanks
    16 = 2^4
    32 = 2^5

    9 = (log_2{x})^2

    log_2{x} = \pm 3

    x = 2^{\pm 3}

    Therefore x = 8 \: , \: x = \frac{1}{8}
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  3. #3
    Junior Member woof's Avatar
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    Quote Originally Posted by Tweety View Post
    \frac{ log_2{16} + log_2{32} }{log_2{x}} = log_2{x}

     \frac{log_2{512}}{log_2{x}} = log_2{x}

     log_2{512} = log_2{x^2}
    PROBLEM is here, as

    (log_2{x})*(log_2{x})\not=log_2{x^2}

    You want

     log_2{512} = (log_2{x})^2 ,

    which is where the

     9 = (log_2{x})^2 written in the above post, comes from.
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