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Find the value of x and y in this system of equations. x^y=1/y^2 and y^x=1/sqrt(x).

Mathematics
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\[x ^{y}=1\div y ^{2} and y ^{x}=1/\sqrt{x} \] u mean??
From the second one, can I say that y^2 = 1/sqrt2...
ok. Keep going.

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Other answers:

It´s right.
So that gives you y and you can sub in the first to get x...
I solved this question before, but i can´t remember all the steps.
May you show me, how?
@viniterranova Let's continue, would you mind showing me your work?
From the second i can say that y=1
I mad x^y = y.
So, if x^y=y I got the following expression. y= 1/y^2.
I don't know how you got these.. But i am sure they (your above expressions) will still give correct answer since i get y=1 from 2nd expression!!
I know you are substituting but don't use the same variable \[x^y=y\] This will be troublesome, use another variable, if you want to but I don't see how it'd help @viniterranova
Thanks ash2326.
If I mad x^y=y, and rewriting the expressoin, i can got y-1/y^2=0. What do they think?
How could you use the same variable "y"? it's defined differently. I have something. Let me show you
ok.
\[x^y=\frac{1}{y^2}\] \[y^x=\frac{1}{\sqrt x}\] Take log both sides, for both the equations \[y\log x=-2\log y\] \[x\log y=-\frac 1 2\log x\] multiply both the equations, we would get \[xy\log x \times \log y=\log x \times\log y\] or \[xy=1\] We have the equation \[y\log x=-2\log y\] put x=1/y \[y\log{\frac 1y}=-2\log y\] we get \[y\times ( -\log y)=-2\log y\] \[y=2\] so \[x=\frac 12\]
@viniterranova do you get this?
\[y\times ( -\log y)=-2\log y\]gives y=1 and y=2 ------
Yes @mukushla I missed that then y=1 and x=1
neat sol
Great.
It´s rigth.
I got y=1 and x= 1 in another way.
i think x=1/2 and y=2 from xy=1 if x=1/2 then y/2=1 y=2 ans
from y log x =-2log y sub x=1/y y log(1/y)=-2log y y(-log y)=-2log y dividing both sides by -log y gives y=-2log y/(-log y) y=2 ans..... and from xy=1 2x=1 then x=1/2 ans ....
Thanks everyone for the help.
ok yw, have fun....

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