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Thread: a log question

  1. #1
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    a log question

    How do I answer this log equation?

    $\displaystyle log_{2} x + log_{x} 2= 2
    $
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  2. #2
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    Quote Originally Posted by shawli View Post
    How do I answer this log equation?

    $\displaystyle log_{2} x + log_{x} 2= 2
    $
    Use the change of base rule on both parts

    $\displaystyle \log_b(a) = \frac{\log_c(a)}{\log_c(b)}$

    So that $\displaystyle \log_2(x) = \frac{\ln (x)}{\ln(2)}$
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  3. #3
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    Hello, shawli!

    Another approach . . .

    Thereom: .$\displaystyle \log_ba \:=\:\frac{1}{\log_ab}$



    $\displaystyle \log_2 x + \log_x 2\:=\: 2$

    We are given: .$\displaystyle \log_2x + \log_x2 \:=\:2$


    Use theorem: .$\displaystyle \log_2x + \frac{1}{\log_2x} \:=\:2$


    Multiply by $\displaystyle \log_2x\!:\;\;(\log_2x)^2 + 1 \:=\:2\log_2x \quad\Rightarrow\quad (\log_2x)^2 - 2\log_2x + 1 \:=\:0$


    Factor: .$\displaystyle \left(\log_2x - 1\right)^2 \:=\:0\quad\Rightarrow\quad \log_2x -1 \:=\:0 $


    Therefore: . $\displaystyle \log_2x \:=\:1 \quad\Rightarrow\quad x \:=\:2$

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  4. #4
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    Thanks!
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