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# M32-14

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Math Expert
Joined: 02 Sep 2009
Posts: 43348

Kudos [?]: 139651 [0], given: 12794

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17 Jul 2017, 23:00
Expert's post
3
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BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

59% (00:42) correct 41% (01:34) wrong based on 22 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 [?]: 139651 [0], given: 12794

Math Expert
Joined: 02 Sep 2009
Posts: 43348

Kudos [?]: 139651 [0], given: 12794

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17 Jul 2017, 23:00
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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Joined: 02 Dec 2015
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08 Aug 2017, 10:11
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

Manager
Joined: 12 Jun 2016
Posts: 224

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Location: India
WE: Sales (Telecommunications)

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26 Sep 2017, 04:35
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?

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Kudos [?]: 49 [0], given: 152

Math Expert
Joined: 02 Sep 2009
Posts: 43348

Kudos [?]: 139651 [1], given: 12794

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26 Sep 2017, 04:43
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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?

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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Manager
Joined: 12 Jun 2016
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26 Sep 2017, 05:25
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?

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] 26 Sep 2017, 05:25
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# M32-14

Moderators: chetan2u, Bunuel

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