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If x and y are integers and x^y=256, what is the number of the differe
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Updated on: 25 Jun 2017, 22:16
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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
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Originally posted by aazt on 25 Jun 2017, 09:01.
Last edited by Bunuel on 25 Jun 2017, 22:16, edited 1 time in total.
Renamed the topic and edited the question.




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Re: If x and y are integers and x^y=256, what is the number of the differe
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25 Jun 2017, 20:19
aazt wrote: If x and y are integers and \(x^y\)=256, what is the number of the different values of \(y^x\)?
Options: (A) 4 (B) 5 (C) 6 (D) 7 (E) 8 Hi, 256 has only 2 as a factor. 256 =\(2^8=4^4=16^2=256^1\)... Those with even power can have x as NEGATIVE too.. So values can be \(2^8....(2)^8....4^4....(4)^4.....16^2....(16)^2.....256^1\).. So \(y^x\) can be 8^2=64 8^{2}=1/64 4^4=256.. 4^(4)=1/256 2^16 2^(16) 1^256=1 So 7 values.. D
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Re: If x and y are integers and x^y=256, what is the number of the differe
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01 Jul 2017, 23:38
aazt wrote: 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 This is a very good question! \(x^y = 256\) Lets check all the values of x and y that satisfy the equation. \(2^8 = 256\) \(4^4 = 256\) \(16^2 = 256\) \(256^1 = 256\) For all those numbers which are raised to even powers we know that we can get the same answer with a negative number as well. \(2^8 = 256\) \({2}^8 = 256\) \(4^4 = 256\) \({4}^4 = 256\) \(16^2 = 256\) \({16}^2 = 256\) \(256^1 = 256\) We are looking for different values for \(y^x\) with unique values: \(8^2 = 16\) \(8^{2} = \frac{1}{8^2} = \frac{1}{16}\) \(4^4 = 256\) \(4^{4} = \frac{1}{4^4} = \frac{1}{256}\) \(2^16 =\) no need to calculate as you know the value is going to be greater than 256 \(2^{16} =\)\(\frac{1}{2^{16}}\) = Once again no need to calculate as you know the value is going to be different from all above values \(1^256 = 1\) As we can see that we are getting\(7\) different values for \(y^x\) Hence, Answer is D
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Re: If x and y are integers and x^y=256, what is the number of the differe
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26 Jul 2017, 18:56
Can someone explain what happens if we account for negative values of y? For example, let x = 65536 (=256^2), and y = 2. Then 2^65536 gives another unique value for y^x, does it not?



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Re: If x and y are integers and x^y=256, what is the number of the differe
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26 Jul 2017, 20:39
aazt wrote: 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 \(x^y = 256\) \(256 = 2^8 = 4^4 = 16^2 = 256^1\), but since \(x\) and \(y\) can be both positive and negative integers, \((2)^8, (4)^4, (16)^2\) also fits the condition. So, \(y^x\) can be \(1^{256}, 2^{16}, 8^{2}, 4^{4}, 2^{16}, 4^{4}, 8^{2}\) = 7 values. Ans  D.
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Re: If x and y are integers and x^y=256, what is the number of the differe
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26 Jul 2017, 20:49
rocheng wrote: Can someone explain what happens if we account for negative values of y? For example, let x = 65536 (=256^2), and y = 2. Then 2^65536 gives another unique value for y^x, does it not? Hi, It's given that \(x^y = 256\), but if we assume negative values for \(y\), then \(x^y\) will not equal \(256\), for e.g., if \(x^y = 256\), then \(x\) and \(y\) satisfy the value \(256\) and \(1\), respectively, but when we assign \(x = 256\) and \(y = 1\), we get \(x^y = 256^{1} = \frac{1}{256}\), which breaches the initial condition that \(x^y = 256\). The given conditions must hold true throughout, for this reason \(y\) cannot be a negative integer.
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Re: If x and y are integers and x^y=256, what is the number of the differe
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31 Jul 2017, 20:22
TimeTraveller wrote: rocheng wrote: Can someone explain what happens if we account for negative values of y? For example, let x = 65536 (=256^2), and y = 2. Then 2^65536 gives another unique value for y^x, does it not? Hi, It's given that \(x^y = 256\), but if we assume negative values for \(y\), then \(x^y\) will not equal \(256\), for e.g., if \(x^y = 256\), then \(x\) and \(y\) satisfy the value \(256\) and \(1\), respectively, but when we assign \(x = 256\) and \(y = 1\), we get \(x^y = 256^{1} = \frac{1}{256}\), which breaches the initial condition that \(x^y = 256\). The given conditions must hold true throughout, for this reason \(y\) cannot be a negative integer. Hi  thanks for the response. You said that "if we assume negative values for \(y\), then \(x^y\) will not equal \(256\)." What about my example? \(x = 65536\), \(y = 2\), \(x^y = 256\)



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Re: If x and y are integers and x^y=256, what is the number of the differe
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Re: If x and y are integers and x^y=256, what is the number of the differe
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