Infinite Tower

Calculus Level 1

$\large { x }^{ { x }^{ { x }^{ { x }^{ \ldots } } } }=2\\ x=?$

Note: x can't equal a negative root.

87

\sqrt { 2 }

4

{ x }^{ x }

{ 2 }^{ x }

2

{ x }^{ 2 }

5

4 solutions

Viki Zeta
Jul 22, 2016

$x^{x^{x^{x^{x^{\ldots}}}}} = 2 \\ \implies x^{(x^{x^{x^{x^{\ldots}}}})} = 2 - \fbox{1}\\ \text{Let 'y' represent the base of power 'x'.} \\ \implies y^{(x^{x^{x^{x^{x^{\ldots}}}}})} = 2 \\ \implies y^2 = 2 \text{ (From } \fbox{1} \text{)}\\ \implies y = \sqrt[]{2} \\ \implies x = \sqrt[]{2}$

This is the correct method though you're missing the most vital part. The convergence is the important of all in this sums. See Here .

Aditya Narayan Sharma - 4 years, 10 months ago

Alex Wang
Jul 21, 2016

X^x^x^x^x....=2

x^x^x^x^x........=2

X^2=2

X= $\sqrt{2}$

Why u didn't considered the negative root?

will jain - 4 years, 10 months ago

What if the question was $\large { x }^{ { x }^{ { x }^{ { x }^{ \ldots } } } }=4$ .

Then, $x^4=4$ which would mean that $x=\sqrt{2}$ .

Now if $x=\sqrt{2}$ , what is $\large { x }^{ { x }^{ { x }^{ { x }^{ \ldots } } } }$ . Is it $2$ or $4$ ?

Janardhanan Sivaramakrishnan - 4 years, 10 months ago

It doesn't work for the number 4. It is like saying x+2=x+4. It only works between $e^{-e}$ and $e^{1/e}$

Lok Lok Li - 2 years, 7 months ago

Lok Lok Li
Nov 9, 2018

${^{\infty}x}=2$

$x^{{^{\infty}x}}=2$

$x^2=2$

$x=\sqrt{2}$

Power tower notation I see?

alex wang - 2 years, 6 months ago

Gregor Sauron
Jul 22, 2016

x^(x^(x^(x^(.........))) = a Then because there are infinetly many x-es we can write x^a = a Hence the solution is: a√(a) In this case √2

Infinite Tower

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