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

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If xy#0, is x^y > 0 ? [#permalink] New post 20 Jul 2014, 10:23
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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}

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[Reveal] Spoiler: OA

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Expert Post
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If xy#0, is x^y > 0 ? [#permalink] New post 20 Jul 2014, 10:23
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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.

Answer: B.

Try NEW Exponents and Roots PS question.
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DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: If xy#0, is x^y > 0 ? [#permalink] New post 20 Jul 2014, 11:12
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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}


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] New post 20 Jul 2014, 19:15
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] New post 21 Jul 2014, 02:57
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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] New post 21 Jul 2014, 04:12
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] New post 21 Jul 2014, 21:03
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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.

Answer is B
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Re: If xy#0, is x^y > 0 ? [#permalink] New post 23 Jul 2014, 09:48
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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.

Answer: B.

Kudos points given to correct solutions.

Try NEW Exponents and Roots PS question.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: If xy#0, is x^y > 0 ? [#permalink] New post 24 Jul 2014, 14:18
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
Re: If xy#0, is x^y > 0 ?   [#permalink] 24 Jul 2014, 14:18
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