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King

  • 3 years ago

Its a log question so i shall give it in the reply......

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  1. King
    • 3 years ago
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    \[\log_{3} (\log_{2} x) + \log_{1/3} (\log_{2} y)=1.\] Find x,y if given \[xy ^{2}=4\]

  2. FoolForMath
    • 3 years ago
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    alright what have you tried ?

  3. King
    • 3 years ago
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    um i took log base 1/3 as log base 3^-1 and that will be \[-\log_{3} \]

  4. Stom
    • 3 years ago
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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

  5. Lumenaire
    • 3 years ago
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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

  6. King
    • 3 years ago
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    can u please xplain how u got the 3rd equation?

  7. Lumenaire
    • 3 years ago
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    log a - log b = log(a/b)

  8. FoolForMath
    • 3 years ago
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    @King:From the first equation I got $\frac{x}{y} = 3$ can you do the rest ?

  9. King
    • 3 years ago
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    no log x base y

  10. King
    • 3 years ago
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    xplain

  11. Lumenaire
    • 3 years ago
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    oh ok. log a / log b = log a base b

  12. Lumenaire
    • 3 years ago
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    if they both have the same base. that is the rule

  13. King
    • 3 years ago
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    FFM what have u written?$\frac{x}{y} = 3$

  14. King
    • 3 years ago
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    oh ok

  15. King
    • 3 years ago
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    so ok how to find x,y

  16. Lumenaire
    • 3 years ago
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    read the rest. y^3 is x. and just plug that into your other equation

  17. FoolForMath
    • 3 years ago
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    yes king

  18. FoolForMath
    • 3 years ago
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    king its just substitution in the rest parts ..

  19. Lumenaire
    • 3 years ago
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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

  20. King
    • 3 years ago
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    ok...thnx guys can u solve 1 more question....shall i post it?

  21. Lumenaire
    • 3 years ago
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    go ahead

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