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Math Help - Exponents as logs?

  1. #1
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    Exponents as logs?

    x^log x = 100x

    How do I solve this? My book says the answers are 100 and .1 but how do I prove this. There is NOTHING in the book that answers this. Thanks
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  2. #2
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    Quote Originally Posted by satx View Post
    x^log x = 100x

    How do I solve this? My book says the answers are 100 and .1 but how do I prove this. There is NOTHING in the book that answers this. Thanks
    1. x = 10^{\log(x)}

    2. Your equation becomes:

    x^{\log( x)} = 100 \cdot x

    \left(10^{\log(x)}\right)^{\log(x)} = 10^2 \cdot 10^{\log(x)}

    \left(\log(x)\right)^2-\log(x)-2=0

    3. This is a quadratic equation in \log(x) . Solve for \log(x). You should come out with \log(x) = 2 \vee \log(x)=-1.

    4. Calculate x.
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  3. #3
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    Hello, satx!

    Solve for x\!:\;\;x^{\log x} \:=\: 100x
    Note that x \neq 0.

    Divide by x\!:\;\;x^{\log x - 1} \:=\:100

    \text{Take logs, base }x\!:\;\;\log_x\left(x^{\log x - 1}\right) \:=\:\log_x(100) \quad\Rightarrow\quad (\log x - 1)\cdot\underbrace{\log_x(x)}_{\text{This is 1}} \:=\:\log_x(100)

    We have: . \log x - 1 \:=\:\log_x100

    Change-of-base Formula: . \log x - 1 \:=\:\frac{\log 100}{\log x} \quad\Rightarrow\quad \log x - 1\:=\:\frac{2}{\log x}

    Multiply by \log x\!:\;\;(\log x)^2 - \log x \:=\:2 \quad\Rightarrow\quad (\log x)^2 - \log x - 2 \:=\:0


    Factor: . (\log x + 1)(\log x - 2) \:=\:0

    And we have: . \begin{array}{cccccccccccc}<br />
\log x + 1 \;=\:0 & \Rightarrow & \log x \:=\:\text{-}1 & \Rightarrow & x \:=\:10^{-1} \:=\:\boxed{\frac{1}{10}}\\ \\[-3mm]<br />
\log x - 2 \:=\:0 & \Rightarrow & \log x \:=\:2 & \Rightarrow & x \:=\:10^2 \:=\:\boxed{100}\end{array}

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