what's the x and y?.....

Algebra Level 2

If x^y=y^x and x^2=y, where x and y are distinct positive real numbers, then the value of x+2y is

4 solutions

Vaibhav Agarwal
Mar 2, 2014

$xlog_{x}y=y => xlog_{x}x^{2}=x^{2} => 2x = x^{2} => x=2$

Thus y=4 and $\boxed{x+2y=10}$

great

Krishna Ar - 7 years, 3 months ago

thnx

Vaibhav Agarwal - 7 years, 3 months ago

I was too close.

Rhishikesh Dongre - 7 years, 1 month ago

Ganesh Banoth
Mar 20, 2014

2^4=4^2;compare with x^y=y^x then we get x=2;y=4; x+2y=10

Scott Wayne Husain
Mar 11, 2014

x^y=y^x multiplying by exp 1/y and we'll get x=y^(x/y) (1st eq)

substitute 1st eq to the second eq which is x^2=y we'll have y^2(x/y)=y

now in order to have this equal we need to have an exponent of 1 so 2x/y=1; 2x=y

substitute again in the 2nd eq x^2 = 2x

x=2 therefore y=4

so x+2y=10

x^y=y^x -----> eq'n 1 y=x^2 -----> eq'n 2

use eq'n 2 in 1

x^x^2=x^2x (x^2)lnx=(2x)lnx

cancel lnx

x^2=2x x^2-2x=0 x(x-2)=0 x=0 x=2

y=(2)^2 y=4

therefore: x+2y 2+2(4) 2+8 10

Marvin De Luna - 7 years, 2 months ago

Kowshik Das
Mar 18, 2014

x^y=y^x =>x^(x^2)=x^2x =>x^2=2x =>x=2 . so y=x^2 =2^2 =4. the ans is 2+(2*2)=10