The Tetra Operation 2

Algebra Level 2

Tetration is the operation notated as $n\uparrow\uparrow x$ which is equivalent to $\underbrace{\huge n^{n^{n^{.^{.^{.}}}}}}_{\normalsize x \space \text{number of} \space n\text{'s}}$

If $a \neq \infty$ , what is the smallest integer that is always larger than $n$ ?

$\large n \uparrow\uparrow \infty = a$

2 3 4 Varies, otherwise

a

if the variable is an integer 1

1 solution

Kaizen Cyrus
Feb 25, 2019

$\large \begin{aligned} n \uparrow\uparrow \infty &= a \\ n^{n^{n^{.^{.^{.}}}}} &= a \\ n^{\left( n^{n^{.^{.^{.}}}} \right)} &= a \\ n^{a} &= a \\ \sqrt[a]{n^{a}} &= \sqrt[a]{a} \\ n &= \sqrt[a]{a} \end{aligned}$

The highest value of $n$ , if it is equal to $\sqrt[a]{a}$ as pointed above, would be $\sqrt[e]{e}$ or $\approx 1.4447$ . Therefore, the smallest integer that is always larger than $n$ is $\boxed{2}$ .

The Tetra Operation 2

1 solution

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