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# If xy#0, is x^y > 0 ?

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Math Expert
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If xy#0, is x^y > 0 ? [#permalink]

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20 Jul 2014, 11:23
Expert's post
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Difficulty:

65% (hard)

Question Stats:

47% (02:29) correct 53% (01:08) wrong based on 187 sessions

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If $$xy\neq{0}$$, is $$x^y>0$$?

(1) $$\sqrt[x]{32}=\frac{1}{2}$$

(2) $$\sqrt[-y]{16}=\frac{1}{2}$$

Kudos for a correct solution.

[Reveal] Spoiler: OA

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Math Expert
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Kudos [?]: 82624 [1] , given: 10108

If xy#0, is x^y > 0 ? [#permalink]

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20 Jul 2014, 11:23
1
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Expert's post
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SOLUTION

If $$xy\neq{0}$$, is $$x^y>0$$?

(1) $$\sqrt[x]{32}=\frac{1}{2}$$.

Rewrite as $$32^{\frac{1}{x}}=2^{-1}$$;
Since $$32=2^5$$, then we'd have that $$2^{\frac{5}{x}}=2^{-1}$$;
Equate the powers of 2 on both sides: $$\frac{5}{x}=-1$$.
Solving gives $$x=-5$$.

Now, if $$y$$ is even, then $$x^y=(-5)^{even}>0$$ but if $$y$$ is not even, say if it's 1, then $$x^y=(-5)^1<0$$. Not sufficient.

(2) $$\sqrt[-y]{16}=\frac{1}{2}$$

Rewrite as $$16^{\frac{1}{-y}}=2^{-1}$$;
Since $$16=2^4$$, then we'd have that $$2^{\frac{4}{-y}}=2^{-1}$$;
Equate the powers of 2 on both sides: $$\frac{4}{-y}=-1$$.
Solving gives $$y=4$$.

Since given that $$x$$ is not zero (from $$xy\neq{0}$$), then $$x^y=x^{4}>0$$. Sufficient.

Try NEW Exponents and Roots PS question.
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Re: If xy#0, is x^y > 0 ? [#permalink]

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20 Jul 2014, 12:12
1
KUDOS
If $$xy\neq{0}$$, is $$x^y>0$$?

(1) $$\sqrt[x]{32}=\frac{1}{2}$$

(2) $$\sqrt[-y]{16}=\frac{1}{2}$$

Sol: Given $$xy\neq{0}$$

St 1 $$\sqrt[x]{32}=\frac{1}{2}$$ can be written as (2)^5/x= 2^-1
or x=-5

Now if y=2 then x^y>0 but if y=1 then x^y =-5<0. Since 2 answers are possible so St 1 is not sufficient

St2 can written as (2)^-4/y=2^-1

or y=4.

Now we know $$x\neq{0}$$ and y=4 so for any value of x, x^y >0

Ans is B
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Re: If xy#0, is x^y > 0 ? [#permalink]

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20 Jul 2014, 20:15
WoundedTiger wrote:
If $$xy\neq{0}$$, is $$x^y>0$$?

(1) $$\sqrt[x]{32}=\frac{1}{2}$$

(2) $$\sqrt[-y]{16}=\frac{1}{2}$$

Sol: Given $$xy\neq{0}$$

St 1 $$\sqrt[x]{32}=\frac{1}{2}$$ can be written as (2)^5/x= 2^-1
or x=-5

Now if y=2 then x^y>0 but if y=1 then x^y =-5<0. Since 2 answers are possible so St 1 is not sufficient

St2 can written as (2)^-4/y=2^-1

or y=4.

Now we know $$x\neq{0}$$ and y=4 so for any value of x, x^y >0

Ans is B

Right.

I was initially mistaking y for -4 But given -y has been stated in the question, y=4 and hence result always positive
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Re: If xy#0, is x^y > 0 ? [#permalink]

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21 Jul 2014, 03:57
1
KUDOS
From the statement given in the argument.

xy != 0
Hence we can conclude neither X nor Y = 0

Now from statement 2,
we can find out the value of Y i.e. 4
No matter X is positive or negative, the value of x^y would always be greater than 0.
Hence Statement B is sufficient to answer the question.
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Re: If xy#0, is x^y > 0 ? [#permalink]

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21 Jul 2014, 05:12
Bunuel wrote:

If $$xy\neq{0}$$, is $$x^y>0$$?

(1) $$\sqrt[x]{32}=\frac{1}{2}$$

(2) $$\sqrt[-y]{16}=\frac{1}{2}$$

Kudos for a correct solution.

Statement 1 says that x is a positive number and x cannot be negative. Hence a positive number raised to any will always be positive. Hence sufficient.

Statement 2 says that y is negative. x^y sign still depends on sign of x or on whether y is even or not. Hence insufficient.

According to me... ANS : A
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Re: If xy#0, is x^y > 0 ? [#permalink]

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21 Jul 2014, 22:03
1
KUDOS
Bunuel wrote:

If $$xy\neq{0}$$, is $$x^y>0$$?

(1) $$\sqrt[x]{32}=\frac{1}{2}$$

(2) $$\sqrt[-y]{16}=\frac{1}{2}$$

Kudos for a correct solution.

(1) $$\sqrt[x]{32}=\frac{1}{2}$$

$$\sqrt[x]{2^5}=\frac{1}{2}$$

$$x=-5$$

$$-5^y>0$$

We do not know whether the exponent will be + or - therefore insufficient.

(2) $$\sqrt[-y]{16}=\frac{1}{2}$$

$$\sqrt[-y]{2^4}=\frac{1}{2}$$

$$y=4$$

$$x^4>0$$

Any number increased to an even power is always positive.
We are also told that $$xy\neq{0}$$. This eliminates the possibility that $$x=0$$

Sufficient.

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Re: If xy#0, is x^y > 0 ? [#permalink]

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23 Jul 2014, 10:48
SOLUTION

If $$xy\neq{0}$$, is $$x^y>0$$?

(1) $$\sqrt[x]{32}=\frac{1}{2}$$.

Rewrite as $$32^{\frac{1}{x}}=2^{-1}$$;
Since $$32=2^5$$, then we'd have that $$2^{\frac{5}{x}}=2^{-1}$$;
Equate the powers of 2 on both sides: $$\frac{5}{x}=-1$$.
Solving gives $$x=-5$$.

Now, if $$y$$ is even, then $$x^y=(-5)^{even}>0$$ but if $$y$$ is not even, say if it's 1, then $$x^y=(-5)^1<0$$. Not sufficient.

(2) $$\sqrt[-y]{16}=\frac{1}{2}$$

Rewrite as $$16^{\frac{1}{-y}}=2^{-1}$$;
Since $$16=2^4$$, then we'd have that $$2^{\frac{4}{-y}}=2^{-1}$$;
Equate the powers of 2 on both sides: $$\frac{4}{-y}=-1$$.
Solving gives $$y=4$$.

Since given that $$x$$ is not zero (from $$xy\neq{0}$$), then $$x^y=x^{4}>0$$. Sufficient.

Kudos points given to correct solutions.

Try NEW Exponents and Roots PS question.
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Re: If xy#0, is x^y > 0 ? [#permalink]

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24 Jul 2014, 15:18
Since x and y is not 0, we need to know if y is even.
Statement 1 --> 2^{\frac{5}{x}}=2^{-1} ---> x=-5
But we do not know if y is even or odd

Statement 2 --> 2^{\frac{-4}{x}}=2^{-1} ---> y= 4
Since y is even, x^{y} is even as long as x is not 0 -> sufficient
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Re: If xy#0, is x^y > 0 ? [#permalink]

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24 Oct 2015, 11:52
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Re: If xy#0, is x^y > 0 ?   [#permalink] 24 Oct 2015, 11:52
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