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Thread: logs question

  1. #1
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    logs question

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

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

    $\displaystyle log_2{512} = log_2{x^2} $

    $\displaystyle 512 = x^2 $

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

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

    $\displaystyle log_2{512} = log_2{x^2} $

    $\displaystyle 512 = x^2 $

    $\displaystyle \sqrt{512} = x $

    so x = 22.6 , is this correct?

    appreciate any input.

    thanks
    $\displaystyle 16 = 2^4$
    $\displaystyle 32 = 2^5$

    $\displaystyle 9 = (log_2{x})^2$

    $\displaystyle log_2{x} = \pm 3$

    $\displaystyle x = 2^{\pm 3}$

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

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

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

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

    You want

    $\displaystyle log_2{512} = (log_2{x})^2 $ ,

    which is where the

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