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505-555 (Easy)|   Algebra|   Exponents|      
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Bunuel
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Is x = 1? (1) x = x^3^0 (2) x^2 - 2x + 1 = 0 06 Feb 2015, 07:21
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70% (00:55) correct 30% (01:02) wrong based on 434 sessions

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Is x = 1?

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

(2) x^2 - 2x + 1 = 0
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B
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Is x = 1? (1) x = x^3^0 (2) x^2 - 2x + 1 = 0 06 Feb 2015, 07:29
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(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.

The answer is B.
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Is x = 1? (1) x = x^3^0 (2) x^2 - 2x + 1 = 0 06 Feb 2015, 08:08
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Bunuel
Is x = 1?

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

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

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

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

Answer B.
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Is x = 1? (1) x = x^3^0 (2) x^2 - 2x + 1 = 0 09 Feb 2015, 02:18
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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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Is x = 1? (1) x = x^3^0 (2) x^2 - 2x + 1 = 0 09 Feb 2015, 05:55
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Bunuel
Is x = 1?

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

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

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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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Is x = 1? (1) x = x^3^0 (2) x^2 - 2x + 1 = 0 20 Dec 2016, 07:10
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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 ?

What if we had :

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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Is x = 1? (1) x = x^3^0 (2) x^2 - 2x + 1 = 0 21 Apr 2023, 05:16
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