Hi,
Given information: x,y are none-zero integers, is x^y<y^x?
1. As x = y^2, then I'll replace all "x" with "y" in the above inequality.
x^y = (y^2)^y = y^(2*y)
y^x = y^(y^2) = y^(y*y)
Try to plug in some easy values:
If y = 1 then y^(2y) = y^(y^2) = 1, or x^y = y^x
If y = 3 then
y^(2y) = 3^6
y^(y^2) = 3^9
and 3^6<3^9, or x^y<y^x
Hence 1 is insufficient.
2. y>2, there're two unknown here, but only one condition that y>2, hence it's insufficient to confirm the inequality.
Example:
x = 1, y = 3, then 1^3<3^1
x = -2, y =4, then -2^4>4^(-2)
1&2.
x = y^2 and y>2
x^y = (y^2)^y = y^(2*y)
y^x = y^(y^2) = y^(y*y)
To compare A^B with A^C: if A>1, C>B>0, then A^C>A^B
Since:
y>2
y^2>2y>0 (as y>2)
Then y^(2*y) is always smaller than y^(y*y), or x^y<y^x
=>Suff.
Pick C.
The answer is C.
"There is nothing either good or bad - but thinking makes it so" - Shakespeare.
