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If the expression x^x^x^(...), where the given expression expression

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Bunuel
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If the expression x^x^x^(...), where the given expression expression [#permalink] New post   20 Jan 2017, 12:51
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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. 1/2

B. \(\sqrt[4]{2}\)

C. \(\sqrt{2}\)

D. \(\sqrt{3}\)

E. 2


Official Solution is HERE.

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C
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Bunuel
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Re: If the expression x^x^x^(...), where the given expression expression [#permalink] New post   18 Jul 2017, 00:02
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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 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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vitaliyGMAT
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Re: If the expression x^x^x^(...), where the given expression expression [#permalink] New post   20 Jan 2017, 13:21
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Bunuel wrote:
If 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. 1/2

B. \(\sqrt[4]{2}\)

C. \(\sqrt{2}\)

D. \(\sqrt{3}\)

E. 2


Hi, nice question.

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

\(lnx^{x^{x}^{...}} = ln2\)

\(x^{x^{x}^{...}}*lnx = ln2\)

\(2*lnx = ln2\)

\(lnx = \frac{1}{2} * ln2\)

\(lnx = ln \sqrt{2}\)

\(x = \sqrt{2}\)

Answer C
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GMATAcademy
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Re: If the expression x^x^x^(...), where the given expression expression [#permalink] New post   20 Jan 2017, 13:35
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Because the pattern continues infinitely, the exponent to which the first x is raised is identical to the overall expression.

We can use this information to write an equation:

(overall expression) = x^(overall expression) = 2

or just:

x^(overall expression) = 2

Since we know the overall expression equals 2, we can plug in:

x^(2) = 2

C is the only answer that satisfies the equation: when we square root 2, we get 2.
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Re: If the expression x^x^x^(...), where the given expression expression [#permalink] New post   20 Jan 2017, 23:28
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Bunuel wrote:
If 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. 1/2

B. \(\sqrt[4]{2}\)

C. \(\sqrt{2}\)

D. \(\sqrt{3}\)

E. 2



Soln:

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

We are given that X raised to \(x^{x^{x}^{...}}\) = 2;

Since the powers of X extends to infinity, we can retain the base 'X' and replace the whole power by 2. This reduces the equation to a simple expression of:

\(x^{2}\) = 2


Hence X =\(\sqrt{2}\)
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Re: If the expression x^x^x^(...), where the given expression expression [#permalink] New post   31 Jan 2017, 05:46
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Can somebody please explain why the answer is E and not C?
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Re: If the expression x^x^x^(...), where the given expression expression [#permalink] New post   31 Jan 2017, 05:57
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Bunuel wrote:
If 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. 1/2

B. \(\sqrt[4]{2}\)

C. \(\sqrt{2}\)

D. \(\sqrt{3}\)

E. 2



Hi..

\(x^{x^{x}^{...}}=2\)....
Since it is to infinite series, \(x^{x^{x}^{...}}=x^2=2\),
Or \(x=\sqrt{2}\)
C..

Bunuel, pl check the OA, must be a typo error
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Bunuel
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Re: If the expression x^x^x^(...), where the given expression expression [#permalink] New post   31 Jan 2017, 06:11
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chetan2u wrote:
Bunuel wrote:
If 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. 1/2

B. \(\sqrt[4]{2}\)

C. \(\sqrt{2}\)

D. \(\sqrt{3}\)

E. 2



Hi..

\(x^{x^{x}^{...}}=2\)....
Since it is to infinite series, \(x^{x^{x}^{...}}=x^2=2\),
Or \(x=\sqrt{2}\)
C..

Bunuel, pl check the OA, must be a typo error


Yes, it's C. Edited the OA.
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Bunuel
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Re: If the expression x^x^x^(...), where the given expression expression [#permalink] New post   18 Jul 2017, 00:09
Expert Reply
Bunuel wrote:
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. 1/2

B. \(\sqrt[4]{2}\)

C. \(\sqrt{2}\)

D. \(\sqrt{3}\)

E. 2


Official Solution is HERE.


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Re: If the expression x^x^x^(...), where the given expression expression [#permalink] New post   18 Jul 2017, 00:55
Bunuel wrote:
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. 1/2

B. \(\sqrt[4]{2}\)

C. \(\sqrt{2}\)

D. \(\sqrt{3}\)

E. 2


Official Solution is HERE.



Let y = \(x^{x^{x}^{...}}\) = 2
So, y = x^y = 2
Taking log on both sides
\(log y = y log x\)
\(log 2 = 2 log x\)
\(log x = 1/2 log 2 = log (\sqrt{2})\)
\(x = \sqrt{2}\)
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sasidharrs
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Re: If the expression x^x^x^(...), where the given expression expression [#permalink] New post   19 Jul 2017, 20:36
does GMAT tests natural logorithms?
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Bunuel
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Re: If the expression x^x^x^(...), where the given expression expression [#permalink] New post   19 Jul 2017, 21:48
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sasidharrs wrote:
does GMAT tests natural logorithms?


You don't need this staff at all. If you check official solution above you'll see that this question can be solved without logarithms.
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Re: If the expression x^x^x^(...), where the given expression expression [#permalink] New post   19 Jul 2017, 23:27
sasidharrs wrote:
does GMAT tests natural logorithms?


Bunuel gave very good solution and now I also feel that it can be solved without using natural log.
I don't know why I used log , probably because we always use log in such types of problem but it can be solved easily by using the method explained by bunuel.

It goes like this :
\(y = x^x^x^x^{....} = 2\)
\(So, x^2 = 2\)
\(x = \sqrt{2}\)
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Re: If the expression x^x^x^(...), where the given expression expression [#permalink] New post   26 Jul 2017, 03:52
Interesting one - plays on wording and what we make of it.
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?

Basically the power of x ^ x^x...so the exponent = 2. It also says x =2. So by plugging base and exponent we have x^2 = 2 (cause x =2; stated earlier in question). Which is root 2.

A. 1/2

B. 2√424

C. 2√2

D. 3√3

E.
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Re: If the expression x^x^x^(...), where the given expression expression [#permalink] New post   18 Jun 2018, 17:10
Bunuel wrote:
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



Hi, can you explain why:

Since the expression extends to an infinite number of exponents, then the expression in brackets would also equal to 2.

How did you come up with this? Why can't it equal say 3 or 0 or 4923.3 or any other number.
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Re: If the expression x^x^x^(...), where the given expression expression [#permalink] New post   18 Jun 2018, 20:12
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jasonfodor wrote:
Bunuel wrote:
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



Hi, can you explain why:

Since the expression extends to an infinite number of exponents, then the expression in brackets would also equal to 2.

How did you come up with this? Why can't it equal say 3 or 0 or 4923.3 or any other number.


We know that \(x^{x^{x^{...}}}=2\), where x's are infinite. Infinite - 1 is also infinite, so \(x^2=2\).
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Re: If the expression x^x^x^(...), where the given expression expression [#permalink] New post   22 Jun 2018, 14:27
the powers of X extends to infinity, we can retain the base 'X' and replace the whole power by 2.
I am sorry how did you come up with that rule , I read all the comments and all of you say the same thing, but i am not able to get it yet.
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Re: If the expression x^x^x^(...), where the given expression expression [#permalink] New post   13 Oct 2019, 06:36
Could anyone explain the logic of why one need to consider x^2 ?
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Re: If the expression x^x^x^(...), where the given expression expression [#permalink] New post   28 Jan 2023, 03:33
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Re: If the expression x^x^x^(...), where the given expression expression [#permalink]
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