King
Its a log question so i shall give it in the reply......
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King
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\[\log_{3} (\log_{2} x) + \log_{1/3} (\log_{2} y)=1.\]
Find x,y if given \[xy ^{2}=4\]
FoolForMath
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alright what have you tried ?
King
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um i took log base 1/3 as log base 3^-1 and that will be
\[-\log_{3} \]
Stom
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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
Lumenaire
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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
King
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can u please xplain how u got the 3rd equation?
Lumenaire
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log a - log b = log(a/b)
FoolForMath
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@King:From the first equation I got $\frac{x}{y} = 3$ can you do the rest ?
King
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no log x base y
King
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xplain
Lumenaire
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oh ok. log a / log b = log a base b
Lumenaire
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if they both have the same base. that is the rule
King
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FFM what have u written?$\frac{x}{y} = 3$
King
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oh ok
King
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so ok how to find x,y
Lumenaire
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read the rest. y^3 is x. and just plug that into your other equation
FoolForMath
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yes king
FoolForMath
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king its just substitution in the rest parts ..
Lumenaire
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xy^2 becomes y^5
so y^5 =4 so y = 4^0.2
and x = y^3 = 4^0.6
King
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ok...thnx guys can u solve 1 more question....shall i post it?
Lumenaire
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go ahead