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    I just...

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    (Original post by The_Blade)
    I just...

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    What are you asking?
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    Can't do it

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    (Original post by The_Blade)
    Can't do it

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    put dy/dx=0 in the usual way, noting that we cannot have x=0
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    (Original post by The_Blade)
    Can't do it

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    Put the x^-1 in the denominator so that you have x^7

    Factorise the numerator and cancel appropriately

    Put the fraction =0

    Remember that if a fraction is 0 then it is the numerator that =0
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    (Original post by TenOfThem)
    Put the x^-1 in the denominator so that you have x^7

    Factorise the numerator and cancel appropriately

    Put the fraction =0

    Remember that if a fraction is 0 then it is the numerator that =0


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    the denominator is x^7

    it would be x^(-7) if you have it in the numerator
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    (Original post by TenOfThem)
    the denominator is x^7

    it would be x^(-7) if you have it in the numerator
    X^-1 ÷ X^6?

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    (Original post by The_Blade)
    X^-1 ÷ X^6?

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    yes - in the numerator

    but x^6 x x^1 if in the denominator
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    (Original post by TenOfThem)
    yes - in the numerator

    but x^6 x x^1 if in the denominator
    OK but I don't know how to go on from there. There answer is 1/3 e^-1

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    (Original post by TenOfThem)
    yes - in the numerator

    but x^6 x x^1 if in the denominator


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    When you cancel x^3 you are left with x^4 in the denominator

    Again though, the numerator = 0
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    (Original post by TenOfThem)
    When you cancel x^3 you are left with x^4 in the denominator

    Again though, the numerator = 0
    Why is the numerator 0

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    (Original post by TenOfThem)
    When you cancel x^3 you are left with x^4 in the denominator

    Again though, the numerator = 0
    I didn't get e^-1

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    You have done the last parts wrong. Personally I would divide the equation through by 3 to get:

    \frac{1}{3}=lnx

    Then take e of both sides, which would give you:

    e^{\frac{1}{3}}= e^{lnx}
    e^{\frac{1}{3}}= x

    Or:

    3lnx \equiv lnx^{3}

    So,

    e^1 = e^{lnx^3}

    e^1 = x^3

    x=e^{\frac{1}{3}}

    Then sub the e^{\frac{1}{3}} back into the original equation
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    (Original post by Phichi)
    You have done the last parts wrong. Personally I would divide the equation through by 3 to get:

    \frac{1}{3}=lnx

    Then take e of both sides, which would give you:

    e^{\frac{1}{3}}= e^{lnx}
    e^{\frac{1}{3}}= x

    Or:

    3lnx \equiv lnx^{3}

    So,

    e^1 = e^{lnx^3}

    e^1 = x^3

    x=e^{\frac{1}{3}}

    Then sub the e^{\frac{1}{3}} back into the original equation
    I don't see how that would get you 1/3 x e^-1

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    (Original post by The_Blade)
    I don't see how that would get you 1/3 x e^-1

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    We've found that:

    x=e^{\frac{1}{3}}

    Sub this into the original equation and find the value.

    So:


    \frac{\frac{1}{3}}{e^1}

    \frac{1}{3e^1} = \frac{1}{3}{e^{-1}}
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    (Original post by Phichi)
    We've found that:

    x=e^{\frac{1}{3}}

    Sub this into the original equation and find the value.

    So:


    \frac{\frac{1}{3}}{e^1}

    \frac{1}{3e^1} = \frac{1}{3}{e^{-1}}
    I see now. Thanks dude

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