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

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

    if \frac{log x}{y-z}=\frac{log y}{z-x}=\frac{log z}{x-y},

    then prove x^x.y^y.z^z=1

    I tried very hard, using different methods. But I am unable to get the answer.
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  2. #2
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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

    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
    Thanks from NameIsHidden
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  3. #3
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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
    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 actually get

    <br />
x^{-(a+b)}=y^a=z^b

    How did you take it to x^b=z^{-(a+b)}?

    If p^m=q^n, can I write  p^n=q^m ?????
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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

    u have to crossmultiply logx/-(a+b)=log y/a and log y/a=log z/b also log x/(-(a+b))=log z/b

    try u will get same thing
    Thanks from NameIsHidden
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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
    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
    Thank you very much. I got it.
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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

    I understand everything. Still I am not satisfied in the way you took [tex] y\log y + z\log z [\tex] to \log \left(yz\right)^{yz}

    I cant understand it fully.
    Last edited by NameIsHidden; May 20th 2014 at 05:31 AM.
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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

    yeah thats the catch here (yz)^(yz) is not equal to (y^y)(z^z)
    but if y=some no and z=1 we can arrive at this conclusion
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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
    yeah thats the catch here (yz)^(yz) is not equal to (y^y)(z^z)
    but if y=some no and z=1 we can arrive at this conclusion
    I forgot to say that in the question it was stated that x, y, z are real numbers such that x,y,z \neq 1 and  x,y,z > 0.

    Then how can you still prove it?
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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

    Was there anything else at all that you forgot to tell us? This habit of not giving the entire original problem results in a huge waste of time for all concerned. If I see it again, I for one will not even look at your future posts.
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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 JeffM View Post
    Was there anything else at all that you forgot to tell us? This habit of not giving the entire original problem results in a huge waste of time for all concerned. If I see it again, I for one will not even look at your future posts.
    I thought it was given x,y, z is not equal to 1, because log 1 is useless and the same way log -10 or any negative number is useless. Am I right? Okay dont look at my any posts, no cost.

    I GAVE MY FULL QUESTION IN MY FIRST THREAD. JUST I GAVE THE SIMPLIFIED FORM IN THE TITLE. YOU FELT LAZY TO READ THE ENTIRE BODY AND **YOU DID NOT SEE THE ORIGINAL PROBLEM**.

    IN MY SECOND POST, I TYPED WRONG IN ONE LINE "+" INSTEAD OF A "-" AND COPIED PASTED IT. THAT'S WRONG . OKAY I ACCEPT IT. I DID **NOT HIDE ANYTHING THERE**

    I JUST HID THE NOT EQUALS TO ONE AND ARE POSITIVE REAL NUMBERS BECAUSE I HOPE YOU KNOW LOG OF A NEGATIVE NUMBER IS UNDEFINED AND LOG OF 1 IS USELESS.

    Even if it was not given in question, one can't assume a variable to be always 0. I am not saying any fault at prasum. He answers to me. I am grateful to him. Knowing the full question JeffM, will you answer my question???
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  11. #11
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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

    Actually, I did see and solve the original problem, but we ended up wasting a lot of time because people were not sure what you were asking.

    Furthermore, log(1) is not useless; it is zero, one of India's great contributions to mathematics.

    And, yes, there are things deducible about x, y, and z, such as they are not equal to each other.

    I shall see whether I can find a proof, but before I go looking I wanted to be 100% positive that I had all the information specified. (I'll look; I don't guarantee I'll find one. I have a busy day. I cannot look right now.)
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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 NameIsHidden View Post
    if \frac{log x}{y-z}=\frac{log y}{z-x}=\frac{log z}{x-y},
    then prove x^x.y^y.z^z=1
    I tried very hard, using different methods. But I am unable to get the answer.
    I have stayed out of this thread because I did not think the statement could be true.
    I had hoped someone had a clever way to prove it. I would welcome a solution.

    But my gut feeling is that it is not true as stated. I have tried to reverse engineer the result: NO LUCK!

    If $x^xy^yz^z=1$ then $x\log(x)+y\log(y)+z\log(z)=0$.
    That means that at least one of $x,~y,~\text{or}~z$ is between $0~\&~1$

    Now given that necessary fact, I have not been able to construct either that works with the given period(i.e. one way or the other).
    I hope someone here can do that.
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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

    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.
    Last edited by SlipEternal; May 20th 2014 at 06:32 PM.
    Thanks from prasum, NameIsHidden and Plato
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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

    I fixed it.
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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

    By the way, if you solve for z, you get z = \dfrac{x\log x + y\log y}{\log x + \log y}. Using that to reduce the number of equations by one, you can get an equation of just x and y:

    \left(\dfrac{x\log x + y\log y}{\log x + \log y} - x\right) \log\left(\dfrac{x \log x + y\log y}{\log x+\log y} \right) = (x-y)\log y

    Plugging in any random value for y and solving for x, it seems you get x=y is the only solution over the reals. However, I believe there is an implicit solution to this system of equations (but there is no known constructive solution for it).
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