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# M32-14

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Math Expert
Joined: 02 Sep 2009
Posts: 41698

Kudos [?]: 124659 [0], given: 12079

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18 Jul 2017, 00:00
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Difficulty:

45% (medium)

Question Stats:

50% (00:28) correct 50% (01:07) wrong based on 10 sessions

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If $$x>0$$ and the expression $$x^{x^{x^{...}}}$$, where the given expression extends to an infinite number of exponents, equals to 2, then what is the value of $$x$$?

A. $$\frac{1}{2}$$
B. $$\sqrt[4]{2}$$
C. $$\sqrt{2}$$
D. $$\sqrt{3}$$
E. $$2$$
[Reveal] Spoiler: OA

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Kudos [?]: 124659 [0], given: 12079

Math Expert
Joined: 02 Sep 2009
Posts: 41698

Kudos [?]: 124659 [0], given: 12079

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18 Jul 2017, 00:00
Official Solution:

If $$x>0$$ and the expression $$x^{x^{x^{...}}}$$, where the given expression extends to an infinite number of exponents, equals to 2, then what is the value of $$x$$?

A. $$\frac{1}{2}$$
B. $$\sqrt[4]{2}$$
C. $$\sqrt{2}$$
D. $$\sqrt{3}$$
E. $$2$$

Re-write as: $$x^{(x^{x^{...}})}$$. Since the expression extends to an infinite number of exponents, then the expression in brackets would also equal to 2. Thus we can replace the expression in brackets with 2 and rewrite the given expression as $$x^2=2$$. So, $$x=\sqrt{2}$$

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Kudos [?]: 124659 [0], given: 12079

Intern
Joined: 02 Dec 2015
Posts: 8

Kudos [?]: 1 [0], given: 31

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08 Aug 2017, 11:11
I don't really understand the following Since the expression extends to an infinite number of exponents, then the expression in brackets would also equal to 2.

How can the product of infinite x be equal to 2?

Kudos [?]: 1 [0], given: 31

Re: M32-14   [#permalink] 08 Aug 2017, 11:11
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# M32-14

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