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Thread: If (log x)/(y-z)=(log y)/(z-x)=(log z)/(x-y), then prove {(x^x).(y^y).(z^z)}=1

  1. #16
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    Re: If (log x)/(y-z)=(log y)/(z-x)=(log z)/(x-y), then prove {(x^x).(y^y).(z^z)}=1

    Quote Originally Posted by SlipEternal View Post
    Cross-multiplying gives the following equations:

    (z-x)\log x = (y-z)\log y (Equation 1)
    (x-y)\log x = (y-z)\log z (Equation 2)
    (x-y)\log y = (z-x)\log z (Equation 3)

    Adding the first two equations gives

    (z-y)\log x = (y-z)(\log y + \log z)

    Dividing both sides by y-z gives

    -\log x = \log y + \log z

    So, \log x + \log y + \log z = 0 (Equation 4)

    Next, from equation 3, adding (y-x)\log y to both sides gives:

    (y-x)\log y + (z-x)\log z = 0 (Equation 5)

    Now, consider

    \begin{align*}x\log x + y\log y + z\log z & = x\log x + (y-x+x)\log y + (z-x+x)\log z \\ & = x(\log x + \log y + \log z) + (y-x)\log y + (z-x)\log z \\ & = (y-x)\log y + (z-x)\log z \text{ by Equation 4} \\ & = 0 \text{ by Equation 5}\end{align*}

    Hence, x^x\cdot y^y\cdot z^z = 1 as desired.

    Thank you very much. I can't imagine this question would be so complex like this.
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  2. #17
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    Re: If (log x)/(y-z)=(log y)/(z-x)=(log z)/(x-y), then prove {(x^x).(y^y).(z^z)}=1

    can u tell me what mistake did i make in
    Spoiler:
    take z-x=a x-y=b and so y-z=-(a+b)
    substituting these values in the above eqn and cross multiplying u get
    x^(b)=z^-(a+b).......eq(1)
    x^(a)=y^-(a+b)......eq(2)
    y^(b)=z^(a)

    now substitute values of y^(b) and z^(a) in eq(1) and eq(2) we get
    x^(b)=(yz)^(-b)
    x^(a)=(yz)^(-a)
    now solving this u get
    x^(b-a)=(1/yz)^(b-a)
    so log(x)=log(1/yz)... eq(3)
    now coming to rhs take log on both sides u get
    xlog(x)+ylog(y)+zlog(z)=0
    using eq 3 and the fact that x=1/yz u get
    log(yz)^(-yz)+log(yz)^(yz)=0
    hence proved
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  3. #18
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    Re: If (log x)/(y-z)=(log y)/(z-x)=(log z)/(x-y), then prove {(x^x).(y^y).(z^z)}=1

    Quote Originally Posted by prasum View Post
    can u tell me what mistake did i make in
    Spoiler:
    take z-x=a x-y=b and so y-z=-(a+b)
    substituting these values in the above eqn and cross multiplying u get
    x^(b)=z^-(a+b).......eq(1)
    x^(a)=y^-(a+b)......eq(2)
    y^(b)=z^(a)

    now substitute values of y^(b) and z^(a) in eq(1) and eq(2) we get
    x^(b)=(yz)^(-b)
    x^(a)=(yz)^(-a)
    now solving this u get
    x^(b-a)=(1/yz)^(b-a)
    so log(x)=log(1/yz)... eq(3)
    now coming to rhs take log on both sides u get
    xlog(x)+ylog(y)+zlog(z)=0
    using eq 3 and the fact that x=1/yz u get
    log(yz)^(-yz)+log(yz)^(yz)=0
    hence proved
    I think you should not have assumed y or z = 1.
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  4. #19
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    Re: If (log x)/(y-z)=(log y)/(z-x)=(log z)/(x-y), then prove {(x^x).(y^y).(z^z)}=1

    Quote Originally Posted by prasum View Post
    can u tell me what mistake did i make in
    Spoiler:
    take z-x=a x-y=b and so y-z=-(a+b)
    substituting these values in the above eqn and cross multiplying u get
    x^(b)=z^-(a+b).......eq(1)
    x^(a)=y^-(a+b)......eq(2)
    y^(b)=z^(a)

    now substitute values of y^(b) and z^(a) in eq(1) and eq(2) we get
    x^(b)=(yz)^(-b)
    x^(a)=(yz)^(-a)
    now solving this u get
    x^(b-a)=(1/yz)^(b-a)
    so log(x)=log(1/yz)... eq(3)
    now coming to rhs take log on both sides u get
    xlog(x)+ylog(y)+zlog(z)=0
    using eq 3 and the fact that x=1/yz u get
    log(yz)^(-yz)+log(yz)^(yz)=0
    hence proved
    You gave essentially the same proof, but the OP had difficulty following it, so I proved it slightly differently.
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  5. #20
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    Re: If (log x)/(y-z)=(log y)/(z-x)=(log z)/(x-y), then prove {(x^x).(y^y).(z^z)}=1

    Use (log x)/(y-z)=(log y)/(z-x)=(log z)/(x-y) =k
    Then, (log x)=(y-z)*k... eq 1
    (log y)=(z-x)*k.... eq 2
    (log z)=(x-y)*k.... eq 3
    Now consider log(x^x*y^y*z^z)
    log(x^x*y^y*z^z) = xlog(x) + ylog(y) + zlog(z) =
    =x(y-z)*k + y( z-x)*k + z(x-y)*k from eqn 1,2,3
    =0
    I.e., log(x^x*y^y*z^z) =0 =log (1)
    I.e., (x^x*y^y*z^z) = 1.... hence proved.
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  6. #21
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    Re: If (log x)/(y-z)=(log y)/(z-x)=(log z)/(x-y), then prove {(x^x).(y^y).(z^z)}=1

    Quote Originally Posted by SlipEternal View Post
    You gave essentially the same proof, but the OP had difficulty following it, so I proved it slightly differently.
    hmm. i tried solving it and got x=y... which is a contradiction to the if statement at the beg... mind letting me know where my work is wrong?
    If (log x)/(y-z)=(log y)/(z-x)=(log z)/(x-y), then prove {(x^x).(y^y).(z^z)}=1-page-1.jpgIf (log x)/(y-z)=(log y)/(z-x)=(log z)/(x-y), then prove {(x^x).(y^y).(z^z)}=1-page-2.jpg
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