find all ordered pairs of positive integers (x,y) that satisfy the following equation: x^y=2^64. please answer and explain how you got it!!!!

2^64 can be written as a power in the following ways

2^64 , 4^32 , 16^16, 256^8 , 65536^4, 4294967296^2 and (4294967296^2 )^1

2^64 = 2x2x2x2...x2x2 --- 64 of them

(2x2)x(2x2)x2...x2)x(2x2) --- 32 pairs of 2's
= 4^32

(2x2x2x2)x(2x2...x2x(2x2x2x2) --- 16 groups of 4 2's
= 16^16
etc

e.g. I can only split the 64 into equal groups,
I could not use three 2's, since 64 does not divide evenly by 3

so ordered pairs of whole positive numbers are
(2,64) , (4,32), (16,16) , (256,8) , (655336, 4)
(4294967296,2) and (4294967296^2 , 1)

To find all ordered pairs of positive integers (x, y) that satisfy the equation x^y = 2^64, we need to determine the values of x and y that make the equation true.

First, we can represent 2^64 as 2^(2^6) since 64 is 2^6. Simplifying further, we have 2^(2^6) = 2^2^6 = 2^(2, 4, 8, 16, 32, 64), where each comma-separated value represents another exponentiation of 2.

Now, let's examine the equation x^y = 2^(2^6).

We know that x and y must be positive integers. Since 2^(2^6) is 2 raised to a power of 2, 4, 8, 16, 32, or 64, we can deduce that x must equal 2 and, therefore, y must be one of the powers mentioned above.

So, the ordered pairs (x, y) that satisfy the given equation are:
(2, 2),
(2, 4),
(2, 8),
(2, 16),
(2, 32),
(2, 64).

Therefore, the six ordered pairs of positive integers (x, y) that satisfy the equation x^y = 2^64 are (2, 2), (2, 4), (2, 8), (2, 16), (2, 32), and (2, 64).