if $\displaystyle \frac{log x}{y-z}=\frac{log y}{z-x}=\frac{log z}{x-y}, $
then prove $\displaystyle x^x.y^y.z^z=1$
I tried very hard, using different methods. But I am unable to get the answer.
if $\displaystyle \frac{log x}{y-z}=\frac{log y}{z-x}=\frac{log z}{x-y}, $
then prove $\displaystyle x^x.y^y.z^z=1$
I tried very hard, using different methods. But I am unable to get the answer.
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 understand everything. Still I am not satisfied in the way you took [tex] $\displaystyle y\log y + z\log z$ [\tex] to $\displaystyle \log \left(yz\right)^{yz}$
I cant understand it fully.
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???
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.)
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.
Cross-multiplying gives the following equations:
$\displaystyle (z-x)\log x = (y-z)\log y$ (Equation 1)
$\displaystyle (x-y)\log x = (y-z)\log z$ (Equation 2)
$\displaystyle (x-y)\log y = (z-x)\log z$ (Equation 3)
Adding the first two equations gives
$\displaystyle (z-y)\log x = (y-z)(\log y + \log z)$
Dividing both sides by $\displaystyle y-z$ gives
$\displaystyle -\log x = \log y + \log z$
So, $\displaystyle \log x + \log y + \log z = 0$ (Equation 4)
Next, from equation 3, adding $\displaystyle (y-x)\log y$ to both sides gives:
$\displaystyle (y-x)\log y + (z-x)\log z = 0$ (Equation 5)
Now, consider
$\displaystyle \begin{align*}x\log x + y\log y + z\log z & = x\log x + (y-x+x)\log y + (z-x+x)\log z \\ & = x(\log x + \log y + \log z) + (y-x)\log y + (z-x)\log z \\ & = (y-x)\log y + (z-x)\log z \text{ by Equation 4} \\ & = 0 \text{ by Equation 5}\end{align*}$
Hence, $\displaystyle x^x\cdot y^y\cdot z^z = 1$ as desired.
By the way, if you solve for z, you get $\displaystyle z = \dfrac{x\log x + y\log y}{\log x + \log y}$. Using that to reduce the number of equations by one, you can get an equation of just $\displaystyle x$ and $\displaystyle y$:
$\displaystyle \left(\dfrac{x\log x + y\log y}{\log x + \log y} - x\right) \log\left(\dfrac{x \log x + y\log y}{\log x+\log y} \right) = (x-y)\log y$
Plugging in any random value for $\displaystyle y$ and solving for $\displaystyle x$, it seems you get $\displaystyle x=y$ 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).