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    if ln( x^6 y) - 2 ln (xy) + ln (3xy^2) = 2

    prove that y = kx^n where k and n are constants

    find the exact values of y = kx ^n

    1) i've separated the equation:

    ln x^6 + lny - 2(lnx + lny) + ln 3x + lny^2 =2

    where do I go from here?
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    (Original post by Custardcream000)
    if ln( x^6 y) - 2 ln (xy) + ln (3xy^2) = 2

    prove that y = kx^n where k and n are constants

    find the exact values of y = kx ^n

    1) i've separated the equation:

    ln x^6 + lny - 2(lnx + lny) + ln 3x + lny^2 =2

    where do I go from here?
    No need to separate them. Get them all under the same logarithm instead.
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    (Original post by Custardcream000)
    if ln( x^6 y) - 2 ln (xy) + ln (3xy^2) = 2

    prove that y = kx^n where k and n are constants

    find the exact values of y = kx ^n

    1) i've separated the equation:

    ln x^6 + lny - 2(lnx + lny) + ln 3x + lny^2 =2

    where do I go from here?
    I think you've headed in the wrong direction to start with... kind of. Doesn't mean you can't go on from there, but you've kind of made... 1 more step... I dunno. Maybe that doesn't matter

    Think about somehow combining the logs.
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    (Original post by RDKGames)
    No need to separate them. Get them all under the same logarithm instead.
    so do you get ln( x^6y 3xy^2 / x^2 y^2 ) =2

    ln(3x^5y)=2

    where do i go from here??
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    (Original post by Custardcream000)
    so do you get ln( x^6y 3xy^2 / x^2 y^2 ) =2

    ln(3x^5y)=2

    where do i go from here??
    ln is log with base e. So anti-log both sides.
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    (Original post by RDKGames)
    ln is log with base e. So anti-log both sides.
    3x^5 y = e^ 2

    y= e^2 / 3x^5
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    (Original post by Custardcream000)
    3x^5 y = e^ 2

    y= e^2 / 3x^5
    Yep. You can take the x out of the denominator as \displaystyle \frac{1}{x}=x^{-1}
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    (Original post by RDKGames)
    Yep. You can take the x out of the denominator as \frac{1}{x}=x^{-1}
    so y = (e^2/3) x^-5
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    (Original post by Custardcream000)
    so y = (e^2/3) x^-5
    Yep.
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    (Original post by RDKGames)
    Yep.
    thanks for your help
 
 
 
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