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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\log_{3} (\log_{2} x) + \log_{1/3} (\log_{1/2} y)=1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ya ure rite !!!!!!!now pls solve!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you can't solve because you have two variables and one equation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh sorry wait also given xy^2=4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\begin{cases}\log_{3} (\log_{2} x) + \log_{1/3} (\log_{1/2} y)=1\\xy ^{2}=4\end{cases}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so please solve.........

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.0\[\log_3 (\log_2 x)  \log_3(\log_2 y) = 1\] \[\log_3({ \log_2 x \over \log_2 y }) = 1\] \[3 = { \log_2 x \over \log_2 y }\] 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

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.0Oops, there I see my mistake. x= 4/y^2 So just fix the last substitution

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ummm.can u wait i shall try and get the answer and can u check if it is rite?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ummm...how do u get the answer??:(::(:(:(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but the thing is i dont know how to go on furtheer.......

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok guys i got this \[2\log_{2} y ^{2}/\log_{2} y\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\log_nm^p=p\cdot\log_nm\] use this and stop spamming

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.2So 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

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.0well, if you don't try, I won't show you how to do it. answer is x=64 y=1/4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thnx 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...........

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.2Sorry. Forgot about the 4. x = 64

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.0Those were the basics. I practically gave you the answer, you just had to substitute and simplify

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i tried i didnt get it.....

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.0I don't think you are a fool, I only think you didn't try hard enough

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0first 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.........

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thnx again mertsj Good luck to u 2!!!!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and thnx slaibaak u also helped me!!!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh ok !!bye now i gtg!!
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