Can anyone make an algebraic formula out of the boxed expression? i’m trying to make a formula that gives you the correct answer if you input ‘a’ into it. i’ve used tetration, super square roots, and super logarithms but cant find a solution.

i am giving out 100 points and the brainliest to the person that can solve this.

answer:

step-by-step explanation:

{\displaystyle {}^{n}x}{}^{n}x, for n = 2, 3, 4, …, showing convergence to the infinitely iterated exponential between the two dots

in mathematics, tetration (or hyper-4) is an operation based on iterated, or repeated, exponentiation. it is the next hyperoperation after exponentiation, but before pentation. the word was coined by reuben louis goodstein from tetra- (four) and iteration.

under the definition as repeated exponentiation, the notation {\displaystyle {^{n}a}}{\displaystyle {^{n}a}} means {\displaystyle {a^{a^{\cdot ^{\cdot ^{a{\displaystyle {a^{a^{\cdot ^{\cdot ^{a, where n copies of a are iterated via exponentiation, right-to-left, i.e. the application of exponentiation {\displaystyle n-1}n-1 times. n is called the "height" of the function, while a is called the "base," analogous to exponentiation. it would be read as "the nth tetration of a".

tetration is also defined recursively as

{\displaystyle {^{n}a}: ={\begin{cases}1& {\text{if }}n=0\\a^{\left(^{(n-1)}a\right)}& {\text{if }}n> 0\end{cases}}}{\displaystyle {^{n}a}: ={\begin{cases}1& {\text{if }}n=0\\a^{\left(^{(n-1)}a\right)}& {\text{if }}n> 0\end{cases}}},

allowing for attempts to extend tetration to non-natural numbers suc