SOLUTION: Please help me solve this question. I don't know what operation to do: {{{ If x^y = 16, where x and y are positive integers and x is less than y, what is the value of x-y ?

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Question 415790: Please help me solve this question. I don't know what operation to do:
Please+help+me+solve+this+question.+I+don%27t+know+what+operation+to+do%3A+%0D%0A If x^y = 16, where x and y are positive integers and x is less than y, what is the value of x-y ?
Found 2 solutions by ewatrrr, htmentor:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

x^y = 16

for example:  2^4 = 16   | x and y are positive integers and x < y

   x-y = 2 - 4 = -2

Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
If x%5Ey+=+16, there are only a few possible solutions for x and y.
16^1 = 16, but the problem states that x < y.
4^2 = 16, but again x < y.
So the only possibility is 2^4 = 16, so x=2, y=4 -> x-y = 2-4 = -2.