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viniterranova Group Title

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

  • one year ago
  • one year ago

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  1. sriramkumar Group Title
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    \[x ^{y}=1\div y ^{2} and y ^{x}=1/\sqrt{x} \] u mean??

    • one year ago
  2. estudier Group Title
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    From the second one, can I say that y^2 = 1/sqrt2...

    • one year ago
  3. viniterranova Group Title
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    ok. Keep going.

    • one year ago
  4. viniterranova Group Title
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    It´s right.

    • one year ago
  5. estudier Group Title
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    So that gives you y and you can sub in the first to get x...

    • one year ago
  6. viniterranova Group Title
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    I solved this question before, but i can´t remember all the steps.

    • one year ago
  7. viniterranova Group Title
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    May you show me, how?

    • one year ago
  8. ash2326 Group Title
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    @viniterranova Let's continue, would you mind showing me your work?

    • one year ago
  9. ujjwal Group Title
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    From the second i can say that y=1

    • one year ago
  10. viniterranova Group Title
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    I mad x^y = y.

    • one year ago
  11. viniterranova Group Title
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    So, if x^y=y I got the following expression. y= 1/y^2.

    • one year ago
  12. ujjwal Group Title
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    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!!

    • one year ago
  13. ash2326 Group Title
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    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

    • one year ago
  14. viniterranova Group Title
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    Thanks ash2326.

    • one year ago
  15. viniterranova Group Title
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    If I mad x^y=y, and rewriting the expressoin, i can got y-1/y^2=0. What do they think?

    • one year ago
  16. ash2326 Group Title
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    How could you use the same variable "y"? it's defined differently. I have something. Let me show you

    • one year ago
  17. viniterranova Group Title
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    ok.

    • one year ago
  18. ash2326 Group Title
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    \[x^y=\frac{1}{y^2}\] \[y^x=\frac{1}{\sqrt x}\] 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 \[x=\frac 12\]

    • one year ago
  19. ash2326 Group Title
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    @viniterranova do you get this?

    • one year ago
  20. mukushla Group Title
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    \[y\times ( -\log y)=-2\log y\]gives y=1 and y=2 ------

    • one year ago
  21. ash2326 Group Title
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    Yes @mukushla I missed that then y=1 and x=1

    • one year ago
  22. mukushla Group Title
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    neat sol

    • one year ago
  23. viniterranova Group Title
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    Great.

    • one year ago
  24. viniterranova Group Title
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    It´s rigth.

    • one year ago
  25. viniterranova Group Title
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    I got y=1 and x= 1 in another way.

    • one year ago
  26. mark_o. Group Title
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    i think x=1/2 and y=2 from xy=1 if x=1/2 then y/2=1 y=2 ans

    • one year ago
  27. mark_o. Group Title
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    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 ....

    • one year ago
  28. viniterranova Group Title
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    Thanks everyone for the help.

    • one year ago
  29. mark_o. Group Title
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    ok yw, have fun....

    • one year ago
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