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Bunuel
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Solution



Given
    • x^y=256

To find
    • Number of the different values of y^x

Approach and Working out

    • We know that 2^8 = 256
      o Hence, x= 2 and y =8
         y ^ x = 8^2 = 64

    • We know that -2^8 = 256
      o Hence, x= -2 and y =8
         y ^ x = 8^-2 = 1/64

    • 4^4 = 256
      o Hence, x= 4 and y =4
         y ^ x = 4^4 = 256

    • -4^4 = 256
      o Hence, x= -4 and y =4
         y ^ x = 4^-4 = 1/256

    • 16^2 = 256
      o Hence, x= 16 and y =2
         y ^ x = 2^16


    • -16^2 = 256
      o Hence, x= -16 and y =2
         y ^ x = 1/2^16

    • 256^1 = 256
      o Hence, x= 256 and y =1
         y ^ x = 1^256 = 1

So, total 7 values are possible.

Thus, option D is the correct answer.
Correct Answer: Option D
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I doubt that explanation is correct though your answer choice is fortunately correct.

For example.
2 is one of the factor of 64, but 2 is not the value of y^x in any case.

Archit3110
Bunuel
If x and y are integers and \(x^y = 256\), what is the number of the different values of \(y^x\)?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

Are You Up For the Challenge: 700 Level Questions
\(x^y = 256\)
x=2 and y = 8
so \(y^x\) = 8^2 ; 64
64 = 2^6 ; total factors = 7
IMO D
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Archit3110
Bunuel
If x and y are integers and \(x^y = 256\), what is the number of the different values of \(y^x\)?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

Are You Up For the Challenge: 700 Level Questions
\(x^y = 256\)
x=2 and y = 8
so \(y^x\) = 8^2 ; 64
64 = 2^6 ; total factors = 7
IMO D

Archit3110

How are factors and values of \(y^x\) same ?
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