1

If

$$\frac{\log x}{y-z} = \frac{\log y}{z-x} = \frac{\log z}{x-y}$$ prove that $$xyz = 1 $$.

without using this method :-

$$ \mbox{Let }\frac{\log x}{y-z} = \frac{\log y}{z-x} = \frac{\log z}{x-y} = k\\ \mbox{this gives three equations :-}\\ \mbox{1.}\log x = k(y -z)\\ \mbox{2.}\log y = k(z - x)\\ \mbox{3.}\log z = k(x - y)\\ \mbox{Adding eq. 1, 2 and 3, we get}\\ \log x + \log y + \log z = k(y - z) + k(z-x)+k(x-y)\\ \implies \log xyz = ky - kz + kz - kx + kx - ky\\ \implies \log xyz = 0\\ \implies \log xyz = \log 1\\ \implies xyz = 1\\ \mbox{(which is the required proof)} $$

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

anonymous
  • 601
  • 1
    It's kind of a very bad question if someone asks you to do it, but not in the obvious and shortest and most sensible way. Why bother to answer such a thing, you have given a fine proof. – Luke Aug 27 '18 at 14:57
  • @Luke I want to know if there's any other way of proving this. – anonymous Aug 27 '18 at 14:59
  • Every other method is very similar, I'd expect. I've written something down using the componendo rule : on paper , it seems very different to your proof, but it really is the same. Can I post this? – Sarvesh Ravichandran Iyer Aug 27 '18 at 14:59
  • астон вілла олоф мэллбэрг Yes – anonymous Aug 27 '18 at 15:01
  • Do not repost questions that were put on hold, especially not without substantive modification. Instead [edit] the post to improve it. – quid Aug 27 '18 at 15:12
  • I now see that you at least tried to improve the post from the original version; I compared to the deleted one that was already improved. I thus undid my vote this one time; it is still not good practice to repost. Wait if the post is reopened and if it is not you can post in the dedicated thread on meta.. – quid Aug 27 '18 at 15:17

2 Answers2

2

$$\frac{\log x}{y-z} = \frac{\log y}{z-x}$$

Implies $$\frac{\log x}{y-z} = \frac{\log y}{z-x}=\frac{\log x+\log y}{(y-z)+(z-x)}$$

Therefore $$\frac{\log xy}{y-x)}= \frac{\log z}{x-y}= \frac{-\log z}{y-x} \Rightarrow \\ \log xy = -\log z \Rightarrow \\ \log xyz =0$$

N. S.
  • 132,525
1

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 $$\frac{\log x}{y-z} = \frac{\log y}{z-x} = \frac{\log z}{x-y}$$ 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 $\frac{\log x}{y-z}>0$. On the other hand, it implies $y>1$, which in turn implies $\frac{\log y}{z-x}<0$. Obviously it is not possible that a positive value equals a negative value.

If $y<z$, then $\frac{\log x}{y-z}<0$. But then $y<1$, and thus $\frac{\log y}{z-x}>0$. Again, this is not possible.

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

celtschk
  • 43,384