Question

What is the solution of x^y=y^x and y=2x?

Answer 1

x = 2, y = 4.

Explanation:

x^y = y^x

Substitute y = 2x in the above:

x^2x = (2x)^x

x^2x = 2*x * x^x Divide both sides by x^x:

x^2x / x^x = 2*x

x^(2x-x) = 2^x

x^x = 2^x

So x = 2.

and y = 2x = 4.

We can also make x = 0 and y = 0 but

I'm not sure if this result is valid because 0^0 is undefined.

Looking further into this there seems to be different opinions on this with some mathematicians say 0^0 = 1, so x=0, y = 0 may be acceptable.

Answer 2

x=0, y=0

x=2, y=4

Explanation:

• x^y= y^x
• y= 2x

• x^(2x)= (2x)^x
• (x^2)^x= (2x)^x
• x^2=2x
• x(x-2)=0
• x=0 ⇒ y=0
• x=2 ⇒ y=4