King
\[\log_{3} (\log_{2} x) + \log_{1/3} (\log_{1/2} y)]=1
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Tomas.A
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\[\log_{3} (\log_{2} x) + \log_{1/3} (\log_{1/2} y)=1\]
King
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ya ure rite !!!!!!!now pls solve!
Tomas.A
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you can't solve because you have two variables and one equation
slaaibak
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Chuck norris can
King
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oh sorry wait
also given
xy^2=4
King
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\[xy ^{2}=4\]
King
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now can u solve?
slaaibak
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Yep
Tomas.A
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\[\begin{cases}\log_{3} (\log_{2} x) + \log_{1/3} (\log_{1/2} y)=1\\xy ^{2}=4\end{cases}\]
King
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so please solve.........
slaaibak
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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
slaaibak
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Oops, there I see my mistake. x= 4/y^2
So just fix the last substitution
King
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ummm.can u wait i shall try and get the answer and can u check if it is rite?
slaaibak
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Yep
King
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ummm...how do u get the answer??:(::(:(:(
Tomas.A
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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
King
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but the thing is i dont know how to go on furtheer.......
King
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ok guys i got this
\[2-\log_{2} y ^{2}/\log_{2} y\]
Tomas.A
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\[\log_nm^p=p\cdot\log_nm\]
use this
and stop spamming
Mertsj
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|dw:1323440622455:dw|
Mertsj
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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
slaaibak
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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
King
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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...........
Mertsj
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Sorry. Forgot about the 4. x = 64
slaaibak
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Those were the basics. I practically gave you the answer, you just had to substitute and simplify
King
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i tried i didnt get it.....
slaaibak
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I don't think you are a fool, I only think you didn't try hard enough
Mertsj
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Good luck to you, King
King
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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.........
King
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thnx again mertsj Good luck to u 2!!!!!
Mertsj
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yw and thanks
King
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and thnx slaibaak u also helped me!!!!
King
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why thanks?
Mertsj
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You wished me good luck
King
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oh ok !!bye now i gtg!!
Mertsj
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Bye