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

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

Originally Posted by SlipEternal
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.

Thank you very much. I can't imagine this question would be so complex like this.

2. ## 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

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

Originally Posted by prasum
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.

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

Originally Posted by prasum
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.

5. ## 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.

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

Originally Posted by SlipEternal
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?

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# Log x/y-z=Log y/z-x=Log z/x-y=x power x*y power y*z power z

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