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18 Jul 2017, 00:00
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C
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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 2, then what is the value of \(x\)?
A. \(\frac{1}{2}\)
B. \(\sqrt[4]{2}\)
C. \(\sqrt{2}\)
D. \(\sqrt{3}\)
E. \(2\)
A. \(\frac{1}{2}\)
B. \(\sqrt[4]{2}\)
C. \(\sqrt{2}\)
D. \(\sqrt{3}\)
E. \(2\)
18 Jul 2017, 00:00
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Official Solution:
If \(x>0\) and the expression \(x^{x^{x^{...}}}\), where the given expression extends to an infinite number of exponents, equals 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: \(x^{(x^{x^{...}})}\). Given that the expression extends to an infinite number of exponents, the expression inside the parentheses would also equal 2. This allows us to substitute the value inside the parentheses with 2 and rewrite the given expression as \(x^2=2\). Hence, \(x = \sqrt{2}\).
Answer: C
If \(x>0\) and the expression \(x^{x^{x^{...}}}\), where the given expression extends to an infinite number of exponents, equals 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: \(x^{(x^{x^{...}})}\). Given that the expression extends to an infinite number of exponents, the expression inside the parentheses would also equal 2. This allows us to substitute the value inside the parentheses with 2 and rewrite the given expression as \(x^2=2\). Hence, \(x = \sqrt{2}\).
Answer: C
Xavipersonal
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Joined: 02 Dec 2015
Last visit: 02 May 2018
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Location: Denmark
Schools: IE Sept18 Intake (A)
GMAT 1: 690 Q49 V35
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08 Aug 2017, 11:11
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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?
How can the product of infinite x be equal to 2?
