If x and y are positive integers, what is the value of x^y?

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If x and y are positive integers, what is the value of  x^y ?

(1)  y^x= 9

(2)  x^{(2y)}= 64

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yashikaaggarwal · Answer

1) Y^x = 9
Case 1: 9^1 = 9
Case 2: 3^2 = 9 
(Insufficient)

2) X^2y = 64
Prime factors of 64 = 2^6
X^2y = 2^6
On comparing both sides. 
X = 2
Y = 3
Therefore x^y = 8
(Sufficient)

IMO B

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Merc · Answer

Premise - X,Y = POSITIVE INTEGERS 
X^Y=? 
 
STAT1 - Y^X =9 
CASES - 
- X=2, Y=3 OR X=1,Y=9 
Both give different values for X^Y , 
hence not sufficient

STAT 2 
X^2y=2^6
now possible cases with both x and y integers are
x=2 Y= 3 : and X=8 , Y =1 
Both give same value for X^Y=8
hence sufficient 
ANSWER B

THANKYOU
MERC

Soumya200011 · Answer

Given x and y are positive integers.
stmt 1: y^x=9
possibilities-
y=3, x=2
y=9, x=1
stmt 1 - insufficient because we do not have a definite answer.
stmt 2: x^(2y)=64
x^(2y)=64 --> x^y=8 (because x and y are +ve)
possibilities-
x=2, y=3
x=8, y=1
stmt 2- insufficient because we do not have a definite answer.

Combining both statements,
we see an overlap of values,
x=2, y=3.
Therefore answer, C

RaghavKhanna · Answer

Given:   x and y are positive integers => x,y ->   Statement 1:  
y^x = 9
(i) 9^1 = 9 
=> y = 9, x = 1 
=> x^y = 1^9 = 1

(ii) 3^2 = 9 
=> y = 3, x = 2
=> x^y = 2^3 = 8

=>  Statement 1 is insufficient

Statement 2:  
x^2y = 64
(i) x^2y = 8^2
=> x = 8, y = 1
=> x^y = 8

(ii) x^2y = 2^6
=> x = 2, y = 3
=> x^y = 8

No other possibility is there.

=>  Statement 2 is sufficient

Answer: B

QuantMadeEasy · Answer

Step 1: Understanding statement 1 alone 
 y^x =9 =  3^2 
Hence, possible values are
When  y^x =9; x=1, y = 8 
When  y^x= 3^2 ; x = 2, y = 3
Insufficient

Step 2: Understanding statement 2 alone  
 x^{2y} =64 =  2^6 = 8^2 
When  x^{2y} =  2^6 ,=; x= 2, y = 3,  x^y  = 8
When  x^{2y} = 8^2 ; x = 8, y = 1,  x^y  = 8
Sufficient

IMO B

GMATWhizTeam · Answer

Solution 
 Step 1: Analyse Question Stem 
 • x and y are positive integers.
• We need to find the value of  x^y

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE 
 Statement 1:   y^x = 9 
 • There can be following cases:
 o Case 1:  y^x = 9^1 
  In this case, x = 1 and y = 9
 • Therefore,  x^y =1^9 = 1    
 o Case 2:  y^x = 3^2 
  In this case, x = 2 and y = 3
 • Therefore,  x^y =2^3 = 8    
• We are getting two different results. 
Hence, statement 1 is NOT sufficient and we can eliminate answer Options A and D.

Statement 2:   x^{(2y)} =64 
• We have,  x^{(2y)} =64⟹ (x^y)^2 = 64 ⟹ x^y = 8   x^y  cannot be negative or -8]
Hence statement 2 is sufficient.
Thus, the correct answer is  Option B.

minustark · Answer

1) When y = 9 and x = 1, x^y = 1. Again when y = 3, x = 2, x^y = 2^3 = 8. Two different answers. Noot sufficient.

2) x^2y = 64. 64 can be expressed as 64^1 (which is not possible here as y cannot be 1/2). 8^2 (possible when x = 8, y =1, x^y = 8), or 4^3 (not possible as y cannot be 3/2). Can ignore (-8)^2 as x and y have to be positive integers. So one unique value found. Sufficient.

B is the answer.

exc4libur · Answer

(1) insufic
y^x=9: {3^2,9^1}
x^y: {2^3=8}{1^9=1}

(2) sufic
x^2y=64={64^1,8^2,4^3,2^6}
2y={1,2,3,6}
y={1/2,1,3/2,3}={1,3}
x={8,2}
x^y: {8^1=8}{2^3=8}

ans (B)

NeoNguyen1989 · Answer

C is the answer 
(1) not sufficient. There are two scenarios; x=1 and y=9 or x=2 and y=3
(2) not sufficient. There are two scenarios; x = 2 and y =3 or x=8 and y=1

combine (1) and (2), we have x=2 and y=3

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If x and y are positive integers, what is the value of x^y?

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