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

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If xy#0, is x^y > 0 ? [#permalink]

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New post 20 Jul 2014, 11:23
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If xy#0, is x^y > 0 ? [#permalink]

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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.

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

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New post 20 Jul 2014, 12:12
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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]

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New post 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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New post 21 Jul 2014, 03:57
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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]

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New post 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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New post 21 Jul 2014, 22:03
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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]

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New post 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.

Answer: B.

Kudos points given to correct solutions.

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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]

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