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KingBest ResponseYou've already chosen the best response.0
\[\log_{3} (\log_{2} x) + \log_{1/3} (\log_{2} y)=1.\] Find x,y if given \[xy ^{2}=4\]
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.2
alright what have you tried ?
 2 years ago

KingBest ResponseYou've already chosen the best response.0
um i took log base 1/3 as log base 3^1 and that will be \[\log_{3} \]
 2 years ago

StomBest ResponseYou've already chosen the best response.0
log3(log2,x) + log1/3(log2,y) =1 log3(log2,x)  log3(log2,y) =1 log3(log2,x/log2,y) =1 log3(x/y) =1 x/y=3
 2 years ago

LumenaireBest ResponseYou've already chosen the best response.0
replace 1 with log base 3 of 3 then: \[\log_{3}(\log_{2}x)  \log_{3}(\log_{2}y) = \log_{3}3 \] then:\[\log_{2}x / \log_{2}y = 3\] \[\log_{y}x = 3 \] y^3 = x so y is 4^0.2 x = 4^0.6
 2 years ago

KingBest ResponseYou've already chosen the best response.0
can u please xplain how u got the 3rd equation?
 2 years ago

LumenaireBest ResponseYou've already chosen the best response.0
log a  log b = log(a/b)
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.2
@King:From the first equation I got $\frac{x}{y} = 3$ can you do the rest ?
 2 years ago

LumenaireBest ResponseYou've already chosen the best response.0
oh ok. log a / log b = log a base b
 2 years ago

LumenaireBest ResponseYou've already chosen the best response.0
if they both have the same base. that is the rule
 2 years ago

KingBest ResponseYou've already chosen the best response.0
FFM what have u written?$\frac{x}{y} = 3$
 2 years ago

LumenaireBest ResponseYou've already chosen the best response.0
read the rest. y^3 is x. and just plug that into your other equation
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.2
king its just substitution in the rest parts ..
 2 years ago

LumenaireBest ResponseYou've already chosen the best response.0
xy^2 becomes y^5 so y^5 =4 so y = 4^0.2 and x = y^3 = 4^0.6
 2 years ago

KingBest ResponseYou've already chosen the best response.0
ok...thnx guys can u solve 1 more question....shall i post it?
 2 years ago
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