agarwalaayush2007 wrote:

Bunuel wrote:

Is y^x > x^y?

(1) y > x

(2) x is a positive number

Hi Bunuel. Is there a conceptual approach to solving this problem?

Hello

I personally dont think there is any conceptual approach, we have to try various cases and see for ourselves. But yes, if we do try and test various possible cases for x and y (and check in which cases x^y is greater, & in which cases y^x is greater), we can definitely use that knowledge in other questions. That knowledge could be useful.

But since you asked, there is a small related concept which comes to mind, and I will share it here, it might be useful some day who knows:

If both x and y are positive integers greater than 3, and if x < y, then:

x^y is always greater than y^x (smaller number raised to a bigger power will be greater than the bigger number raised to smaller power)

If we try to apply this concept in this particular question, we can see that even after combining the two statements, if x=4 & y=5; then x^y > y^x

But we can check that if x=2 & y=3; then x^y < y^x (thats why the concept I gave is applicable for numbers greater than 3).

So in this question, the data is not sufficient.