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M32-14

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M32-14 [#permalink]

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New post 18 Jul 2017, 00:00
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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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New post 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, \(x=\sqrt{2}\)


Answer: C
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M32-14 [#permalink]

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New post 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?

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