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# M32-14

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Math Expert
Joined: 02 Sep 2009
Posts: 50009

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18 Jul 2017, 00:00
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Difficulty:

45% (medium)

Question Stats:

58% (00:52) correct 42% (01:08) wrong based on 33 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. $$2$$

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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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GMAT 1: 690 Q49 V35
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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?
Manager
Joined: 12 Jun 2016
Posts: 217
Location: India
WE: Sales (Telecommunications)

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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?

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Joined: 02 Sep 2009
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26 Sep 2017, 05:43
1
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?

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https://gmatclub.com/forum/new-tough-an ... l#p1029228
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26 Sep 2017, 06:25
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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GMAT 1: 710 Q47 V41
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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
Re: M32-14 &nbs [#permalink] 10 Mar 2018, 14:30
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# M32-14

Moderators: chetan2u, Bunuel

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