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King

  • 3 years ago

\[\log_{3} (\log_{2} x) + \log_{1/3} (\log_{1/2} y)]=1

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  1. Tomas.A
    • 3 years ago
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    \[\log_{3} (\log_{2} x) + \log_{1/3} (\log_{1/2} y)=1\]

  2. King
    • 3 years ago
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    ya ure rite !!!!!!!now pls solve!

  3. Tomas.A
    • 3 years ago
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    you can't solve because you have two variables and one equation

  4. slaaibak
    • 3 years ago
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    Chuck norris can

  5. King
    • 3 years ago
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    oh sorry wait also given xy^2=4

  6. King
    • 3 years ago
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    \[xy ^{2}=4\]

  7. King
    • 3 years ago
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    now can u solve?

  8. slaaibak
    • 3 years ago
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    Yep

  9. Tomas.A
    • 3 years ago
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    \[\begin{cases}\log_{3} (\log_{2} x) + \log_{1/3} (\log_{1/2} y)=1\\xy ^{2}=4\end{cases}\]

  10. King
    • 3 years ago
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    so please solve.........

  11. slaaibak
    • 3 years ago
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    \[\log_3 (\log_2 x) - \log_3(-\log_2 y) = 1\] \[\log_3({ \log_2 x \over -\log_2 y }) = 1\] \[3 = { \log_2 x \over -\log_2 y }\] x=y^2/4 \[3 ={{ {\log_2{ y^2\over4}} } \over -\log_2 y}\] I'm not sure if I did everything correctly, but you can try to solve it further. If I made no mistakes, it's not too hard from here

  12. slaaibak
    • 3 years ago
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    Oops, there I see my mistake. x= 4/y^2 So just fix the last substitution

  13. King
    • 3 years ago
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    ummm.can u wait i shall try and get the answer and can u check if it is rite?

  14. slaaibak
    • 3 years ago
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    Yep

  15. King
    • 3 years ago
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    ummm...how do u get the answer??:(::(:(:(

  16. Tomas.A
    • 3 years ago
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    sorry but solving those logarithms is just waste of time, it's only technical thing, so you'd better show how you solve and then i can say what's wrong

  17. King
    • 3 years ago
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    but the thing is i dont know how to go on furtheer.......

  18. King
    • 3 years ago
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    ok guys i got this \[2-\log_{2} y ^{2}/\log_{2} y\]

  19. Tomas.A
    • 3 years ago
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    \[\log_nm^p=p\cdot\log_nm\] use this and stop spamming

  20. Mertsj
    • 3 years ago
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    |dw:1323440622455:dw|

  21. Mertsj
    • 3 years ago
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    So y^-3 = x but x = 4/y^2 So y^-3 = 4/y^2 and y^-1 = 4 so y = 1/4 and x = 16

  22. slaaibak
    • 3 years ago
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    well, if you don't try, I won't show you how to do it. answer is x=64 y=1/4

  23. King
    • 3 years ago
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    thnx a lot mertsj..... and slaibaak look my sir taught us only basics and u think i am a fool i would rather do it on my own but i tried i didn't get it so i asked u guys if u didn't wanna help u could have said no ok...........

  24. Mertsj
    • 3 years ago
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    Sorry. Forgot about the 4. x = 64

  25. slaaibak
    • 3 years ago
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    Those were the basics. I practically gave you the answer, you just had to substitute and simplify

  26. King
    • 3 years ago
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    i tried i didnt get it.....

  27. slaaibak
    • 3 years ago
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    I don't think you are a fool, I only think you didn't try hard enough

  28. Mertsj
    • 3 years ago
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    Good luck to you, King

  29. King
    • 3 years ago
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    first of all i hate it that i had to ask fr hep i always try as hard as i can then ask fr help......so if u think i dont try hard enough well den thats ure opinion...so frgt it.........

  30. King
    • 3 years ago
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    thnx again mertsj Good luck to u 2!!!!!

  31. Mertsj
    • 3 years ago
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    yw and thanks

  32. King
    • 3 years ago
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    and thnx slaibaak u also helped me!!!!

  33. King
    • 3 years ago
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    why thanks?

  34. Mertsj
    • 3 years ago
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    You wished me good luck

  35. King
    • 3 years ago
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    oh ok !!bye now i gtg!!

  36. Mertsj
    • 3 years ago
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    Bye

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