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

Expert's post

1

This post was BOOKMARKED

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. _________________

Re: If xy#0, is x^y > 0 ? [#permalink]
21 Jul 2014, 02:57

1

This post received 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.

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.

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

Expert's post

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. _________________

Re: If xy#0, is x^y > 0 ? [#permalink]
24 Oct 2015, 10:52

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