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2

The following is not trying to give an answer how to overcome the numerical problem, but is thought to help the intuition. From the wikipedia-entry for tetration the remark about the Shell-Thron region implies, that the base $b=-2.5$ should give a divergent orbit for $z_{k+1}=b^{z_k}$, but unfortunately, the orbit beginning at $0,1,b,...$ runs early in ...

1

The inverse of $^ba = c$ $a = \sqrt [b]{c}_s$ ,for $b \in \Bbb N$ Where $\sqrt {(..)}_s$ is Super square root And $b=slog_a(c)$ Where $Slog(..)$ is Super logarithm Be careful that $^b(^{1/b}a) \not = a$

8

It is known as a tetration, and it is normally written as $^na$ where n is the height of the power tower. It is the forth hyperoperation. The zeroth hyperoperation is the successor function, and the first is the zeroth hyperoperation iterated, and so on A more general way to define the nth hyperoperation is, using the notation, $H_n(a,b)$ where n is the ...

10

A more general function that combines all those operators has been defined by Ackermann: $\varphi(m,n,p) = \begin{cases} \varphi(m, n, 0) = m + n \\ \varphi(m, 0, 1) = 0 \\ \varphi(m, 0, 2) = 1 \\ \varphi(m, 0, p) = m &\text{ for } p > 2 \\ \varphi(m, n, p) = \varphi(m, \varphi(m, n-1, p), p - 1) &\text{ for } n > 0 \text{ and } p > 0. \... 56 This operation${\rm Ops}(4)$is called tetration, from the greek root tetra meaning four; it's also sometimes called a "power tower". There are also many further generalizations of this type of sequence; Knuth's up-arrow notation gives$a^{a^{a^a}}=a\uparrow\uparrow4$, so that$a\uparrow\uparrow n$is the tetration operation. By adding more arrows you get ... -2 Solution: The method of this solution depends on "Completing the super square" of left side of the equation to make the base equal its power for getting super square root as following: By starting with.$x + a = b^x$Let$y = x + a$Substitute$y$in the equation$y = b^{y-a} \Rightarrow y = \frac {b^y}{b^a} \Rightarrow b^a y = b^y \Rightarrow \frac {...

1

Using the relation given in Wikipedia $$ssrt(a^ab^b)=\frac {\log(a^ab^b)}{W(\log(a^ab^b)}=\frac {a \log a + b \log b}{W(a \log a + b \log b)}$$ but as there is no nice formula for $W(x+y)$ this doesn't really get you where you want to go.

1

First, we have $$e^{\textstyle 10^{10^{2.8}}} = 10^{\textstyle \log_{10}(e) \cdot 10^{10^{2.8}}} = 10^{\textstyle 10^{\left( 10^{2.8} + \log_{10}\log_{10}(e) \right)}} = 10^{\textstyle 10^{10^{A}}}$$ where $$A = \log_{10}(10^{2.8} + \log_{10}\log_{10}(e)) \approx \log_{10}(630.96 - 0.36) \approx 2.7998$$ Now, setting $B=10^{10^A}$, do the same ...

1

In general, not necessarily. The left-hand side needs to be big enough. The smallest we can get is if we let $x \to -\infty$, which gives us $$10^{10^{10^{10^{x}}}} \to 10^{10^{10^{10^{-\infty}}}} \to 10^{10^{10^{0}}}=10^{10^{1}} = 10^{10}$$ so basically, we cannot reach down to $10^{10}$, but any number above that is large enough that we can find an $x$ ...

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