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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

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.........

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??:(::(:(:(

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

|dw:1323440622455:dw|

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

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

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