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

King Group TitleBest ResponseYou've already chosen the best response.0
ya ure rite !!!!!!!now pls solve!
 2 years ago

Tomas.A Group TitleBest ResponseYou've already chosen the best response.0
you can't solve because you have two variables and one equation
 2 years ago

slaaibak Group TitleBest ResponseYou've already chosen the best response.0
Chuck norris can
 2 years ago

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

King Group TitleBest ResponseYou've already chosen the best response.0
now can u solve?
 2 years ago

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

King Group TitleBest ResponseYou've already chosen the best response.0
so please solve.........
 2 years ago

slaaibak Group TitleBest 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
 2 years ago

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

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

King Group TitleBest ResponseYou've already chosen the best response.0
ummm...how do u get the answer??:(::(:(:(
 2 years ago

Tomas.A Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

King Group TitleBest ResponseYou've already chosen the best response.0
but the thing is i dont know how to go on furtheer.......
 2 years ago

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

Tomas.A Group TitleBest ResponseYou've already chosen the best response.0
\[\log_nm^p=p\cdot\log_nm\] use this and stop spamming
 2 years ago

Mertsj Group TitleBest ResponseYou've already chosen the best response.2
dw:1323440622455:dw
 2 years ago

Mertsj Group TitleBest ResponseYou've already chosen the best response.2
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
 2 years ago

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

King Group TitleBest ResponseYou've already chosen the best response.0
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...........
 2 years ago

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

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

King Group TitleBest ResponseYou've already chosen the best response.0
i tried i didnt get it.....
 2 years ago

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

Mertsj Group TitleBest ResponseYou've already chosen the best response.2
Good luck to you, King
 2 years ago

King Group TitleBest ResponseYou've already chosen the best response.0
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.........
 2 years ago

King Group TitleBest ResponseYou've already chosen the best response.0
thnx again mertsj Good luck to u 2!!!!!
 2 years ago

Mertsj Group TitleBest ResponseYou've already chosen the best response.2
yw and thanks
 2 years ago

King Group TitleBest ResponseYou've already chosen the best response.0
and thnx slaibaak u also helped me!!!!
 2 years ago

Mertsj Group TitleBest ResponseYou've already chosen the best response.2
You wished me good luck
 2 years ago

King Group TitleBest ResponseYou've already chosen the best response.0
oh ok !!bye now i gtg!!
 2 years ago
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