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M32-14

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M32-14  [#permalink]

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New post 18 Jul 2017, 00:00
1
1
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

59% (01:12) correct 41% (01:35) wrong based on 22 sessions

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Re M32-14  [#permalink]

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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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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]

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New post 26 Sep 2017, 05:35
Hello Bunuel,

I too have the same question as Xavipersonal. How does the expression in bracker of \(x^{(x^{x^{...}})}\) end up as 2? Is this some kind of a rule?

Thanks in advance for your answer!
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Re: M32-14  [#permalink]

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New post 26 Sep 2017, 05:43
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susheelh wrote:
Hello Bunuel,

I too have the same question as Xavipersonal. How does the expression in bracker of \(x^{(x^{x^{...}})}\) end up as 2? Is this some kind of a rule?

Thanks in advance for your answer!


Let me ask you: does the expression in brackets differ from the expression which equals to 2 in any way?

Similar questions:
https://gmatclub.com/forum/if-the-expre ... 98647.html
https://gmatclub.com/forum/new-tough-an ... l#p1029228
https://gmatclub.com/forum/find-the-val ... 38049.html
https://gmatclub.com/forum/find-the-val ... 75403.html
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M32-14  [#permalink]

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New post 26 Sep 2017, 06:25
1
Got it Bunuel! Thank you so much. I absolutely need to read the question very closely. I seem to be missing a lot when reading/solving 600-700 level questions :(

I now understand that this is how it pans out

Given = \(x^{x^{x^{...}}}\) = 2 \(-----> (1).\)

= \(x^{(x^{x^{...}})}\) = 2

Since the terms in the bracket goes on for infinity, it is actually the same as (1). Substituting

= \(x^2\) = 2

The question says \(X > 0\)

So, finally we get x = \(\sqrt{2}\)

Thanks also for the similar questions for practice!


Bunuel wrote:
susheelh wrote:
Hello Bunuel,

I too have the same question as Xavipersonal. How does the expression in bracker of \(x^{(x^{x^{...}})}\) end up as 2? Is this some kind of a rule?

Thanks in advance for your answer!


Let me ask you: does the expression in brackets differ from the expression which equals to 2 in any way?

Similar questions:
https://gmatclub.com/forum/if-the-expre ... 98647.html
https://gmatclub.com/forum/new-tough-an ... l#p1029228
https://gmatclub.com/forum/find-the-val ... 38049.html
https://gmatclub.com/forum/find-the-val ... 75403.html

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Re: M32-14  [#permalink]

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New post 10 Mar 2018, 14:30
options A and E can be immediately eliminated as we know A will quickly tend to much less than 2 and E to much more. I got right of root 3 as well on sight - it just looks unlikely

If you express root 2 as 2^1/2, the rooted exponents quickly tend towards 1. 2^1 = 2
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Re: M32-14  [#permalink]

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New post 24 Mar 2019, 04:29
Bunuel,

Is there a significance of the info that x>0 in this question?

Thanks.
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Re: M32-14   [#permalink] 24 Mar 2019, 04:29
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