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

- King

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

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- anonymous

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

- King

ya ure rite !!!!!!!now pls solve!

- anonymous

you can't solve because you have two variables and one equation

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## More answers

- slaaibak

Chuck norris can

- King

oh sorry wait
also given
xy^2=4

- King

\[xy ^{2}=4\]

- King

now can u solve?

- slaaibak

Yep

- anonymous

\[\begin{cases}\log_{3} (\log_{2} x) + \log_{1/3} (\log_{1/2} y)=1\\xy ^{2}=4\end{cases}\]

- King

so please solve.........

- slaaibak

\[\log_3 (\log_2 x) - \log_3(-\log_2 y) = 1\]
\[\log_3({ \log_2 x \over -\log_2 y }) = 1\]
\[3 = { \log_2 x \over -\log_2 y }\]
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

Oops, there I see my mistake. x= 4/y^2
So just fix the last substitution

- King

ummm.can u wait i shall try and get the answer and can u check if it is rite?

- slaaibak

Yep

- King

ummm...how do u get the answer??:(::(:(:(

- anonymous

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

but the thing is i dont know how to go on furtheer.......

- King

ok guys i got this
\[2-\log_{2} y ^{2}/\log_{2} y\]

- anonymous

\[\log_nm^p=p\cdot\log_nm\]
use this
and stop spamming

- Mertsj

|dw:1323440622455:dw|

- Mertsj

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

well, if you don't try, I won't show you how to do it. answer is
x=64
y=1/4

- King

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

Sorry. Forgot about the 4. x = 64

- slaaibak

Those were the basics. I practically gave you the answer, you just had to substitute and simplify

- King

i tried i didnt get it.....

- slaaibak

I don't think you are a fool, I only think you didn't try hard enough

- Mertsj

Good luck to you, King

- 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.........

- King

thnx again mertsj Good luck to u 2!!!!!

- Mertsj

yw and thanks

- King

and thnx slaibaak u also helped me!!!!

- King

why thanks?

- Mertsj

You wished me good luck

- King

oh ok !!bye now i gtg!!

- Mertsj

Bye

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