If xy 0, is x^y 0 ?

Question

If  xy
eq{0} , is  x^y>0 ?

(1)  \sqrt {32}=\frac{1}{2}

(2)  \sqrt {16}=\frac{1}{2}

Bunuel · Accepted Answer

SOLUTION

If  xy
eq{0} , is  x^y>0 ?

(1)  \sqrt {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 {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
eq{0} ), then  x^y=x^{4}>0 . Sufficient.

Answer: B.

Try NEW Exponents and Roots  PS question .

WoundedTiger · Answer

If  xy
eq{0} , is  x^y>0 ?

(1)  \sqrt {32}=\frac{1}{2}

(2)  \sqrt {16}=\frac{1}{2}

Sol: Given  xy
eq{0}

St 1  \sqrt {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
eq{0}  and y=4 so for any value of x, x^y >0

Ans is B

hamzakb · Answer

Right.

I was initially mistaking y for -4 But given -y has been stated in the question, y=4 and hence result always positive

DesiGmat · Answer

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.

rajkamalr · Answer

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

justbequiet · Answer

(1)  \sqrt {32}=\frac{1}{2}

\sqrt {2^5}=\frac{1}{2}

x=-5

-5^y>0

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

(2)  \sqrt {16}=\frac{1}{2}

\sqrt {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
eq{0} . This eliminates the possibility that  x=0

Sufficient.

Answer is B

cs2209 · Answer

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

dineshril · Answer

Given xy eq{0} i.e x eq{0} & y eq{0} To Find is x^y>0 => x^y>0 possible if => Case 1 x > 0 and y > 0 or y < 0 => Case 2 x < 0 , y= MUST be EVEN (+ve or -ve) => Thus if 'y' is EVEN 'x' can be >0 or <0 then x^y>0 always true So the Ques can be Re-fraced as is 'y' EVEN Stat 1 \sqrt {32}=\frac{1}{2} => 32=(\frac{1}{2})^x => 32=2^{-x} => 2^5=2^{-x} => Equating powers of 2 we have -x=5 => or x=-5 => Thus 'x' is <0 but no information of 'y'. Therefore NOT SUFFICIENT Stat 2 \sqrt {16}=\frac{1}{2} => 16^{\frac{1}{-y}}=2^{-1} ; => Since 16=2^4 , Thereforet 2^{\frac{4}{-y}}=2^{-1} ; => Equating powers of 2 we have \frac{4}{-y}=-1 . => Thus y=4 i.e y=even => Since x eq{0} and y=even . Therefore SUFFICIENT Therefore "B" Regards Dinesh

PoonamS · Answer

For x^y > 0, we need
either
 x > 0
or
 y is even.

S1: x is negative. So, x^y can be positive or negative depending on the y being even or odd.
not sufficient.

S2: y is even.
So, no matter what the value of x is, x^y is >0.
sufficient.

Ans B

GmatPoint · Answer

﻿Given that xy  
e\ 0 
Using statement 1:

﻿\sqrt {32}\ =\ \left(\frac{1}{2}ight)\ ,\ \left(32ight)^{\frac{1}{x}}=\ \frac{1}{2} 
﻿ 2^{\frac{5}{x}}=\ \frac{1}{2\ }=\ 2^{-1} 
﻿x = -5.
﻿Using statement 2:
 ﻿\sqrt {16}=\ \frac{1}{2} 
 ﻿\left(16ight)^{-\frac{1}{y}}=\ \left(2ight)^{-4} 
﻿y = 4.
﻿Hence substituting the values of x, y in  x^y 
﻿We get this value to be equal to 625.
﻿Hence this is greater than zero
﻿

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If xy 0, is x^y > 0 ?

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