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    Is x^x = 100 algebraically soluble?
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    (Original post by Bigcnee)
    Is x^x = 100 algebraically soluble?
    Nope not at ALEVEL knowledge atleast... :rolleyes:
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    (Original post by Bigcnee)
    Is x^x = 100 algebraically soluble?
    I think you would need to use the Newton-Raphson method, but that's an iterative method of solution rather than an analytical one, which I imagine is what you mean by "algebraically soluble" ??
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    (Original post by Fermat)
    I think you would need to use the Newton-Raphson method, but that's an iterative method of solution rather than an analytical one, which I imagine is what you mean by "algebraically soluble" ??
    Newton-Raphson method is not acceptable.
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    (Original post by Bigcnee)
    Is x^x = 100 algebraically soluble?
    The Gelfond-Schneider theorem states the following: if a is an algebraic number and a isn't 0 or 1, and b is an irrational algebraic number, then a^b is transcendental.

    So if x is the positive solution of x^x=100 then the GS theorem says that x can't be irrational algebraic because 100 isn't transcendental.

    So the only alternative is that x=p/q is rational. But if

    (p/q)^(p/q) = 100

    then

    (p/q)^p = 100^q.

    So p/q has to be an integer.

    Finally n^n = 100 hasn't a solution amongst the integers as 3^3 = 27 and 4^4 = 256 and so the solution x lies between 3 and 4.

    I'm not sure there's an easier solution to this problem without quoting the GS Theorem.
 
 
 
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