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If x^y = 1 then what is the value of x?

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If x^y = 1 then what is the value of x?  [#permalink]

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New post 16 Sep 2018, 08:07
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If \(x^y = 1\), then what is the value of x?

(1) x < 0
(2) y is even integer

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If x^y = 1 then what is the value of x?  [#permalink]

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New post 16 Sep 2018, 08:33
OA: E

Given: \(x^y = 1\)

Statement 1: \(x < 0\)

taking \(x = -1\), and \(y = 2\) , It satisfies the \(x^y = 1\) as \((-1)^2\) is equal to \(1\).

taking \(x = -2\), and \(y = 0\) , It satisfies the \(x^y = 1\) as \((-2)^0\) is equal to \(1\).

Statement \(1\) alone is insufficient as we are not getting a unique value of \(x\).

Statement 2: \(y\) is even integer

taking \(x = 1\), and \(y = 2\) , It satisfies the \(x^y = 1\) as \((1)^2\) is equal to \(1\).

taking \(x = 100\), and \(y = 0\) , It satisfies the \(x^y = 1\) as \((100)^0\) is equal to \(1\).

Statement \(2\) alone is insufficient as we are not getting a unique value of \(x\).

Combining \((1)\) and \((2)\) will also be insufficient.
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If x^y = 1 then what is the value of x?  [#permalink]

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New post 17 Sep 2018, 02:59
Most important element to answer this question is to know that any number raise to "0" is 1. (e.g. 1^0 =1 or -1^0=1)

Now let's work this out.

x^y=1, x=?

St 1: x<0 (x is -ve)

Case 1: -1^2 = 1
Case 2: -2^0 = 1

Therefore X could be -1,-2,-3 or any -ve number if the power is 0. Insufficient AD/BCE

St 2: Y is even

(Note: 0 is an even number)

Case 1: x=1, y=2
1^2=1
Case 2: x=2, y=0
2^0=1

Hence B is insufficient. AD/BCE

St1 + St2

x^y=1 (x<0, y is even)

Case 1: x=-1 and y=2
=> -1^2=1

Case 2: x=-2 and y=0
=> -2^0=1

From statement 1 and 2, we have proved that x can have multiple values. AD/BCE

Therefore the answer is E

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Re: If x^y = 1 then what is the value of x?  [#permalink]

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New post 17 Sep 2018, 13:51
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GMATPrepNow wrote:
If \(x^y = 1\), then what is the value of x?

(1) x < 0
(2) y is even integer

\({x^y} = 1\)

\(? = x\)

Let´s ALGEBRAICALLY BIFURCATE statements (1) and (2) together, to be "shielded" that the right answer is (E) immediately!

\(\left( {1 + 2} \right)\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {x,y} \right) = \left( { - 1,0} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = - 1 \hfill \\
\,{\text{Take}}\,\,\left( {x,y} \right) = \left( { - 2,0} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = - 2 \hfill \\
\end{gathered} \right.\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: If x^y = 1 then what is the value of x?  [#permalink]

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New post 17 Sep 2018, 13:59
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ArjunJag1328 wrote:
Most important element to answer this question is to know that any number raise to "0" is 1. (e.g. 1^0 =1 or -1^0=1)

Hi ArjunJag1328!

You are correct for any real number... except zero.

More explicitly:

\({0^0}\,\,{\text{is}}\,\,{\text{not}}\,\,{\text{defined}}\)

Curiosity: there are some cases in Mathematics that it is CONVENIENT to admit this expression is also equal to 1 (when you are dealing with multi-indices, for example).

But this is just a "local convention" to help you present, in that context, formulas in a much more pleasant way...

I hope you find this interesting!

Regards,
Fabio.

P.S.: in the GMAT you will NOT be asked to evaluate 0^0 , of course.
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Re: If x^y = 1 then what is the value of x?  [#permalink]

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New post 23 Sep 2018, 14:08
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GMATPrepNow wrote:
If \(x^y = 1\), then what is the value of x?

(1) x < 0
(2) y is even integer


Target question: What is the value of x?

Given: x^y = 1
If x^y = 1, then there are 3 possible cases:
case i: x = 1, and y = any value (e.g., 1^9 = 1)
case ii: x = -1, and y = an even integer (e.g., (-1)^4 = 1)
case iii: x = any non-zero value, and y = 0 (e.g., 7^0 = 1)


Statement 1: x < 0
Let's TEST some values.
There are several values of x and y that satisfy statement 1. Here are two:
Case a: x = -1, and y = 2. Notice that x^y = (-1)^2 = 1. In this case, the answer to the target question is x = -1
Case b: x = -3, and y = 0. Notice that x^y = (-3)^0 = 1. In this case, the answer to the target question is x = -3
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: y is even integer
Let's TEST some values.

PRO-TIP: When testing values, always check to see if you can reuse previous values.
In this case, we can reuse BOTH cases:
Case a: x = -1, and y = 2. Notice that x^y = (-1)^2 = 1. In this case, the answer to the target question is x = -1
Case b: x = -3, and y = 0. Notice that x^y = (-3)^0 = 1. In this case, the answer to the target question is x = -3
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
IMPORTANT: Notice that I was able to use the same counter-examples to show that each statement ALONE is not sufficient. So, the same counter-examples will satisfy the two statements COMBINED.
In other words,
Case a: x = -1, and y = 2. Notice that x^y = (-1)^2 = 1. In this case, the answer to the target question is x = -1
Case b: x = -3, and y = 0. Notice that x^y = (-3)^0 = 1. In this case, the answer to the target question is x = -3
Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer: E

Cheers,
Brent
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Re: If x^y = 1 then what is the value of x? &nbs [#permalink] 23 Sep 2018, 14:08
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