Infinite powers

Logic Level 1

If, x x x . . . = 2 x^{x^{x^{...}}}=2 , then what is x?(that is x^x^x and so on)

Answer up to two decimal places.


The answer is 1.41.

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1 solution

Marta Reece
Jun 21, 2017

x x x . . . = 2 x^{x^{x^{...}}}=2

ln x x x . . . = ln 2 \ln x^{x^{x^{...}}}=\ln 2

x x x . . . ln x = ln 2 x^{x^{x^{...}}}\ln x=\ln 2

2 ln x = ln 2 2\ln x=\ln 2

x = 1 2 ln 2 = ln 2 x=\frac12\ln 2=\ln\sqrt2

x = 2 x=\boxed{\sqrt2}

This is not complete. By your logic, x x x . . . = 4 x^{x^{x^{...}}}=4 has a solution of x = 2 x = \sqrt2 as well.

Pi Han Goh - 3 years, 11 months ago

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We define the recursive sequence

{ a 0 = 2 a n + 1 = 2 a n , n N \displaystyle \begin{cases} a_0=\sqrt{2} \\ a_{n+1}=\sqrt{2}^{\ a_n}, \ n \in \mathbb{N} \end{cases}

We can show inductively that a n {a_n} is increasing and bounded above by 2 2 , implying that a n {a_n} is indeed convergent. Let a n x a_n\to x . Clearly, 2 < x 2 \sqrt{2}<x\le 2 . But a n + 1 = 2 a n x a_{n+1}=\sqrt{2}^{a_n}\to x , as well.

Hence x = lim n a n = lim n a n + 1 = lim n 2 a n = 2 x \displaystyle x = \lim_{n\to\infty} a_n = \lim_{n\to\infty} a_{n+1} = \lim_{n\to\infty} \sqrt{2}^{\ a_n} = \sqrt{2}^{\ x} . Thus x x satisfies 2 x = x \sqrt{2}^{\ x}=x . Finally, we show that 2 x > x x ( 2 , 2 ) \sqrt{2}^x > x \ \forall \ x \in (\sqrt{2},2) . Thus, we have x = 2 x = 2 .

Zach Abueg - 3 years, 11 months ago

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Yup! Great! Kudos!

Pi Han Goh - 3 years, 11 months ago

Yeah, i think you first have to prove that the power tower converges.

maximos stratis - 3 years, 11 months ago

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