Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

Chuck norris can

Answer 5

oh sorry wait
also given
xy^2=4

Answer 6

\[xy ^{2}=4\]

Answer 7

now can u solve?

Answer 8

Yep

Answer 9

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

Answer 10

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

Answer 11

\[\log_3 (\log_2 x) - \log_3(-\log_2 y) = 1\]

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


\[3 = { \log_2 x \over -\log_2 y }\]


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

Answer 12

Oops, there I see my mistake. x= 4/y^2

So just fix the last substitution

Answer 13

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

Answer 14

Yep

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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