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

Mathematics
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\[\log_{3} (\log_{2} x) + \log_{1/3} (\log_{1/2} y)=1\]
ya ure rite !!!!!!!now pls solve!
you can't solve because you have two variables and one equation

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Other answers:

Chuck norris can
oh sorry wait also given xy^2=4
\[xy ^{2}=4\]
now can u solve?
Yep
\[\begin{cases}\log_{3} (\log_{2} x) + \log_{1/3} (\log_{1/2} y)=1\\xy ^{2}=4\end{cases}\]
so please solve.........
\[\log_3 (\log_2 x) - \log_3(-\log_2 y) = 1\] \[\log_3({ \log_2 x \over -\log_2 y }) = 1\] \[3 = { \log_2 x \over -\log_2 y }\] 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
Oops, there I see my mistake. x= 4/y^2 So just fix the last substitution
ummm.can u wait i shall try and get the answer and can u check if it is rite?
Yep
ummm...how do u get the answer??:(::(:(:(
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
but the thing is i dont know how to go on furtheer.......
ok guys i got this \[2-\log_{2} y ^{2}/\log_{2} y\]
\[\log_nm^p=p\cdot\log_nm\] use this and stop spamming
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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
well, if you don't try, I won't show you how to do it. answer is x=64 y=1/4
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...........
Sorry. Forgot about the 4. x = 64
Those were the basics. I practically gave you the answer, you just had to substitute and simplify
i tried i didnt get it.....
I don't think you are a fool, I only think you didn't try hard enough
Good luck to you, King
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.........
thnx again mertsj Good luck to u 2!!!!!
yw and thanks
and thnx slaibaak u also helped me!!!!
why thanks?
You wished me good luck
oh ok !!bye now i gtg!!
Bye

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