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# Is x = 1?

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Math Expert
Joined: 02 Sep 2009
Posts: 58453

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06 Feb 2015, 07:21
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Difficulty:

15% (low)

Question Stats:

74% (00:54) correct 26% (01:04) wrong based on 219 sessions

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Is x = 1?

(1) $$x = x^{3^0}$$

(2) x^2 - 2x + 1 = 0

Kudos for a correct solution.

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Re: Is x = 1?  [#permalink]

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06 Feb 2015, 07:29
1
(1) $$x = x^{3^0}$$ Insufficient as $$3^0 = 1$$ so $$x = x^1$$and x can be any value

(2) x^2 - 2x + 1 = 0 Sufficient as this factors into $$(x-1)(x-1)=0$$ so x must equal 1.

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Re: Is x = 1?  [#permalink]

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06 Feb 2015, 08:08
Bunuel wrote:
Is x = 1?

(1) $$x = x^{3^0}$$

(2) x^2 - 2x + 1 = 0

Kudos for a correct solution.

statement 1 yields x=x which is an identity. The equation holds for any value.

statement 2: $$(x-1)^2$$=0 and x=1

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Re: Is x = 1?  [#permalink]

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09 Feb 2015, 02:18
1
Yep I also agree.

[1] gives x=x, which can be anything. NS
[2] gives (x-1)(x-1)=0 --> x=1 or x=1. S
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Re: Is x = 1?  [#permalink]

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09 Feb 2015, 05:55
Bunuel wrote:
Is x = 1?

(1) $$x = x^{3^0}$$

(2) x^2 - 2x + 1 = 0

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

Be careful with statement (1): when the parentheses are around the first exponent, you must treat the first exponent as a base and raise it to the power of the second exponent. Thus x=x^3^0 can be simplified as x^1, so statement (1) says x = x; NOT SUFFICIENT. Statement (2) gives a quadratic in which b^2=4ac, so there is only one solution; SUFFICIENT, as whatever solution we get will definitively answer the question. (In case you’re curious, the quadratic solved does say that x = 1). (B)
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20 Dec 2016, 07:10
The way the question is right now, so no parenthesis, what operation do we do first? x^3^0 PEMDAS do we start with 3^0 ?

1) (x^3)^0

would we get X^(3*0) so X^0=1 ?

The official solution talks about parenthesis, but there aren't any in the statements. What if there are no parenthesis? What is the order?
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Re: Is x = 1?  [#permalink]

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21 Oct 2018, 02:37
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Re: Is x = 1?   [#permalink] 21 Oct 2018, 02:37
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