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Math Help - Equation problem

  1. #1
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    Equation problem

    Hello!

    Solve this equation for x: \sqrt3^x+1=2^x

    Obviously, the solution is x=2. This can be proven by graphing \sqrt3^x+1 and 2^x.

    But I can't find any proper ALGEBRAIC way to solve this equation. Any idea?

    Thank you.
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  2. #2
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    Anyone?
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Try to explore the function f(x)=(sqrt3)^x+1-2^x
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  4. #4
    A Plied Mathematician
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    See here for a related thread.
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  5. #5
    MHF Contributor Unknown008's Avatar
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    Maybe iteration?

    2^x = \sqrt{3} ^x + 1

    x_{n+1} = \dfrac{ln(\sqrt{3}^{x_n} + 1)}{ln2}

    Take a positive value as a start. Then, take the result you get in the same formula again and again until you get consistent results.

    For example, take the first as x = 1.5

    x_{1} = \dfrac{ln(\sqrt{3}^{1.5} + 1)}{ln2} = 1.713478979...

    Use this again,

    x_{2} = \dfrac{ln(\sqrt{3}^{1.713478979...} + 1)}{ln2} = 1.833140309...

    Again and again, until you get 1.999 which is enough to say that x = 2

    EDIT: There was a LaTeX error which took me some time to spot... and before I knew it, Ackbeet posted the link...
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  6. #6
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    Thank you very much!
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