Infinite power towers will never fail!

Algebra Level 4

Solve for $x$ : $\large \underbrace{x^{x^{x^{x^{x^{x^{\dots}}}}}}}_{\text{infinite }x \text{'s}} = 6$

\sqrt[6]{6}

and

-\sqrt[6]{6}

-\sqrt[6]{6}

There are no solutions

\sqrt[6]{\dfrac{1}{6}}

\sqrt[6]{6}

\dfrac{1}{6}

1 solution

James Watson
Sep 1, 2020

If we look at the equation $y=x^{x^{x^{x^{x^{\dots}}}}}$ , we see that this there is a solution for $x$ if $e^{-1} \leq y \leq e$ . Since $6$ does not fit this inequality, we can conclude that there are $\green{\boxed{\text{no values of } x \text{ that satisfy the equation}}}$

Here is a video on this if you are confused :)

Addendum: You may have solved this equation and found that $\sqrt[6]{6}$ is a solution. However, this is wrong because if you plug this in for $x$ it returns a number that is not $6$ (the number being $1.62424384 \dots$ ). Since this number is not $6$ , the solution $\sqrt[6]{6}$ is not a valid solution.

I solved it the way you would solve continued fractions (and got the wrong answer):

$6=x^{x^{x^{x^{x^\ldots}}}}=x^6\Rightarrow x=\boxed{\sqrt[6]{6}}$ $\Large\color{#D61F06}\textbf{××××××××××}$

Lâm Lê - 9 months, 1 week ago

yes but does that number when substituted for $x$ return $6$ ?

James Watson - 9 months, 1 week ago

Wait, why wouldn't that work?

alex wang - 9 months, 1 week ago

This wouldn't work because $6$ is not in the range of the function $x^{x^{x^{x^{\dots}}}}$ . Because of this, there is no number that can be plugged in and return $6$ , therefore there are no solutions.

James Watson - 9 months, 1 week ago

☺ Since I watched this video before, I was able to answer the question correctly. ☺

Ömer Ertürk - 5 months, 1 week ago

Infinite power towers will never fail!

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