Vtyjkyuk

If

$$fraclog xy-z = fraclog yz-x = fraclog zx-y$$
prove that $$xyz = 1
$$.

without using this method :-

$$
mboxLet fraclog xy-z = fraclog yz-x = fraclog zx-y = k\
mboxthis gives three equations :-\
mbox1.log x = k(y -z)\
mbox2.log y = k(z - x)\
mbox3.log z = k(x - y)\
mboxAdding eq. 1, 2 and 3, we get\
log x + log y + log z = k(y - z) + k(z-x)+k(x-y)\
implies log xyz = ky - kz + kz - kx + kx - ky\
implies log xyz = 0\
implies log xyz = log 1\
implies xyz = 1\
mbox(which is the required proof)
$$

I am trying my best but not able to do it.

$$fraclog xy-z = fraclog yz-x$$

Implies
$$fraclog xy-z = fraclog yz-x=fraclog x+log y(y-z)+(z-x)$$

Therefore
$$fraclog xyy-x)= fraclog zx-y= frac-log zy-x
Rightarrow \
log xy = -log z Rightarrow \
log xyz =0$$

Note: I'm assuming that the equation is to be interpreted as equation of real values, and using the logarithm as real-valued function.

The equation
$$fraclog xy-z = fraclog yz-x = fraclog zx-y$$
has no solution, therefore any statement about its solutions is vacuously true, including the statement that $xyz=1$.

Proof that there is no solution to this equation:

First, we see that all of $x,y,z$ must be positive, as otherwise the logarithm is not defined. Moreover, no two of those values can be equal, otherwise we would have a division by zero, which again is not defined. Moreover, none of them can equal to $1$, as then the logarithm would be zero and thus the other two logarithms would need to be zero, too, in contradiction to all three values being different.

Now we notice that the equation is invariant under cyclic permutation of $(x,y,z)$. Therefore we can without loss of generality assume that $x$ is the largest of the three.

Then $z-x$ is negative, and $x-y$ is positive. This means that exactly one of $y$ and $z$ must be larger than $1$. Since $x$ is by assumption the largest number, this implies that $x>1$ as well.

Now we have either $y>z$ or $y<z$. If $y>z$, then $fraclog xy-z>0$. On the other hand, it implies $y>1$, which in turn implies $fraclog yz-x<0$. Obviously it is not possible that a positive value equals a negative value.

If $y<z$, then $fraclog xy-z<0$. But then $y<1$, and thus $fraclog yz-x>0$. Again, this is not possible.

Thus for any solution, we have neither $y=z$ nor $y>z$ nor $y<z$. Therefore no solution exists. $square$