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Thread: Solve for x

  1. #1
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    Solve for x

    $\displaystyle
    \frac{1}{2^x}-\frac{1}{x^2}=0$

    $\displaystyle \frac{x^2-2^x}{2^x(x^2)}=0
    $

    I dont know what to do from here.
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  2. #2
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    Quote Originally Posted by JimmyRP View Post
    $\displaystyle
    \frac{1}{2^x}-\frac{1}{x^2}=0$

    $\displaystyle \frac{x^2-2^x}{2^x(x^2)}=0
    $

    I dont know what to do from here.
    two "obvious" solutions are x = 2 and x = 4 ... there is one more that is negative, but you'll have to use a calculator to find it.
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  3. #3
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    Quote Originally Posted by skeeter View Post
    two "obvious" solutions are x = 2 and x = 4 ... there is one more that is negative, but you'll have to use a calculator to find it.
    could you expand on how you would get the negative. I want to learn I don't want just the answers :P
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  4. #4
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    Quote Originally Posted by JimmyRP View Post
    could you expand on how you would get the negative. I want to learn I don't want just the answers :P
    you would have to graph the function $\displaystyle x^2 - 2^x$ with a calculator and find the zeros.

    note that the negative value cannot been found using elementary algebraic techniques.
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  5. #5
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    could you point me to the forum where i could learn these techniques?
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