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Tower of exponents solution approach unique to exponents?

  1. Feb 4, 2016 #1
    I saw a YouTube video presenting what seemed like a clever solution to ##x^{x^{x^{.^{.}}}} = 2## (which is to say: an infinite tower of exponents of x = 2). He said to consider just the exponents and ignore the base and realize that those exponents themselves become a restatement of the whole left side of the equation. Equate them to 2 and you get ##x^2=2## and thus ##x= \sqrt{2}##.

    I thought that was clever and tried to see if this could work for other operations like multiplication (##x\cdot x\cdot x \cdot\ ...=2##), division (##\frac{x}{\frac{x}{\frac{x}{...}}}=2##), addition (##x+x+x+...=2##) and subtraction(##x-x-x-...=2##), but realized it didn't. Any idea what's special about the exponent operation that allows for this clever "substitution of infinite terms" type of solution?
     
  2. jcsd
  3. Feb 4, 2016 #2

    mfb

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    2016 Award

    Staff: Mentor

    The exponent tower can have a limit, the other processes do not (apart from boring special cases).

    That should work as well:
    $$x\sqrt{x\sqrt{x\sqrt \dots}}$$
     
  4. Feb 5, 2016 #3

    Samy_A

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    Even for the infinite tower of exponents the trick will only work if ##e^{-e}<x<e^{1/e}##. As ##e^{1/e}=1.444...## and ##\sqrt 2=1.414...##, it works for ##\sqrt 2##.
    https://thatsmaths.files.wordpress.com/2013/01/powertowerlambert.pdf
     
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