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Re: x^y<0 [#permalink]
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Given that \(𝑥^𝑦 < 0\), this implies that

QA Though there are certain restrictions on the possible values of 𝑦 , for the expression \(𝑥^𝑦\) to be negative, base 𝑥 should be negative. Hence, Quantity A is negative.

QB We can observe that for \(𝑥^𝑦\) to be negative 𝑦 cannot be even. However, it can be odd or in certain cases a fraction, (as even power always gives nonnegative answer), but nothing about the sign of 𝑦 can be concluded. Hence, Quantity B can be positive or negative.

D is the answer. Very tricky question to solve rather conceptually

I suggest you to follow the link at the bottom of the question above >>> Inequalities theory
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x^y<0 [#permalink]
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We know that \(x^y\) is negative.

But this cannot be accomplished if \(y\) is negative or positive, as negative \(y\) only gives us a fractional answer.

\(x\) has to be negative.

But if \(y\) odd, \(x^y\) is negative, irrespective of whether \(y\) is negative or positive. If \(y\) is even, \(x^y\) is positive.

Since \(x\) has to be negative and \(y\) has to be a positive or negative odd number, there is no clear discernible relationship between the

Therefore, the answer is D.
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Re: x^y<0 [#permalink]
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x^y<0

From the above we can deduce:

That x must be some negative number (can be fraction, integer etc)

and that y can be either positive or negative, but y may not be even

Since we have no other way of narrowing down the values of x and y, there is no possible way to determine their relationship.
You could have x=-100, y=1 and x=-100 y=-101
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Re: x^y<0 [#permalink]
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