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

x=0, y=0

x=2, y=4

Step-by-step explanation:

x = 2, y = 4.

Step-by-step 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.

32/80  

91/140

the last one 54 doesn't go into 45. 54/100 ?

step-by-step explanation:

ask the math teacher she or he would know lol.

step-by-step explanation: