A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
Find the value of x and y in this system of equations. x^y=1/y^2 and y^x=1/sqrt(x).
anonymous
 3 years ago
Find the value of x and y in this system of equations. x^y=1/y^2 and y^x=1/sqrt(x).

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[x ^{y}=1\div y ^{2} and y ^{x}=1/\sqrt{x} \] u mean??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0From the second one, can I say that y^2 = 1/sqrt2...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So that gives you y and you can sub in the first to get x...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I solved this question before, but i can´t remember all the steps.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0May you show me, how?

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.2@viniterranova Let's continue, would you mind showing me your work?

ujjwal
 3 years ago
Best ResponseYou've already chosen the best response.0From the second i can say that y=1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So, if x^y=y I got the following expression. y= 1/y^2.

ujjwal
 3 years ago
Best ResponseYou've already chosen the best response.0I 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!!

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.2I 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If I mad x^y=y, and rewriting the expressoin, i can got y1/y^2=0. What do they think?

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.2How could you use the same variable "y"? it's defined differently. I have something. Let me show you

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.2\[x^y=\frac{1}{y^2}\] \[y^x=\frac{1}{\sqrt x}\] 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 \[x=\frac 12\]

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.2@viniterranova do you get this?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[y\times ( \log y)=2\log y\]gives y=1 and y=2 

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.2Yes @mukushla I missed that then y=1 and x=1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I got y=1 and x= 1 in another way.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think x=1/2 and y=2 from xy=1 if x=1/2 then y/2=1 y=2 ans

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0from 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 ....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks everyone for the help.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.