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Math Help - A variable raised to a power that contains that variable, all equal to one?

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    A variable raised to a power that contains that variable, all equal to one?

    I put the following problem into Wolfram Alpha and found that the solutions are the square root of e and 1. Can somebody explain how you'd solve it to get both of those answers? I only got the square root of e. Here's the problem:

    x^(ln(x) - 0.5) = 1

    It makes sense to me that 1 works as well, but am I missing some kind of rule that would have made me get 1 when I solved it the first time around?
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    Re: A variable raised to a power that contains that variable, all equal to one?

    Quote Originally Posted by MRich520 View Post
    x^(ln(x) - 0.5) = 1
    I am glad that x=1 works for you.

    Now do you agree that (\sqrt{e})^0=1~?
    If you do then what does \ln(\sqrt{e})=~?
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    Re: A variable raised to a power that contains that variable, all equal to one?

    Quote Originally Posted by Plato View Post
    I am glad that x=1 works for you.

    Now do you agree that (\sqrt{e})^0=1~?
    If you do then what does \ln(\sqrt{e})=~?

    \ln(\sqrt{e})=1/2

    Eh...I'm not sure if that was the answer you were asking for. Mind clarifying?
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    Re: A variable raised to a power that contains that variable, all equal to one?

    Quote Originally Posted by MRich520 View Post
    \ln(\sqrt{e})=1/2
    Eh...I'm not sure if that was the answer you were asking for. Mind clarifying?
    Well what is \ln(\sqrt{e})-0.5~?
    And what is \sqrt{e} to that power?
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    Re: A variable raised to a power that contains that variable, all equal to one?

    Hello, MRich520!

    \text{Solve for }x\!:\;\;x^{(\ln x - \frac{1}{2})} \:=\: 1

    Take logs: . \ln\left(x^{(\ln x - \frac{1}{2})}\right) \:=\:\ln 1

    We have: . \left(\ln x - \tfrac{1}{2}\right)\cdot \ln x \:=\:0


    And we have two equations to solve:

    . . \begin{array}{ccccccc}\ln x - \tfrac{1}{2} \:=\:0 & \Rightarrow & \ln x \:=\:\tfrac{1}{2} & \Rightarrow & x \:=\:e^{\frac{1}{2}} \\ \\ \ln x \:=\:0 & \Rightarrow & x \:=\:e^0 & \Rightarrow & x \:=\:1 \end{array}

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    Re: A variable raised to a power that contains that variable, all equal to one?

    Quote Originally Posted by Soroban View Post
    Hello, MRich520!


    Take logs: . \ln\left(x^{(\ln x - \frac{1}{2})}\right) \:=\:\ln 1

    We have: . \left(\ln x - \tfrac{1}{2}\right)\cdot \ln x \:=\:0


    And we have two equations to solve:

    . . \begin{array}{ccccccc}\ln x - \tfrac{1}{2} \:=\:0 & \Rightarrow & \ln x \:=\:\tfrac{1}{2} & \Rightarrow & x \:=\:e^{\frac{1}{2}} \\ \\ \ln x \:=\:0 & \Rightarrow & x \:=\:e^0 & \Rightarrow & x \:=\:1 \end{array}

    Thanks!

    I should know that from my classes .

    And sorry Plato I forgot to respond
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