Given-that-log-x-y-z-log-y-z-x-log-z-x-y-Show-that-x-x-y-y-z-z-1-

Question Number 17075 by tawa tawa last updated on 30/Jun/17

Given that : \log (\frac{x}{y - z}) = \log (\frac{y}{z - x}) = \log (\frac{z}{x - y})

Show that : x^{x} \times y^{y} \times z^{z} = 1

Commented by RasheedSoomro last updated on 30/Jun/17

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

\Rightarrow \frac{x}{y - z} = \frac{y}{z - x} = \frac{z}{x - y} = \frac{x + y + z}{y - z + z - x + x - y}

= \frac{x + y + z}{0} = \infty

? ? ?

Answered by 433 last updated on 30/Jun/17

\frac{x}{y - z} > 0 & \frac{y}{z - x} > 0 & \frac{z}{x - y} > 0

x (y - z) > 0 & y (z - x) > 0 & z (x - y) > 0

x y - x z > 0 & y z - y x > 0 & z x - z y > 0

x y > x z & y z > y x & z x > z y

x y > x z > z y > y x

x y > y x

Answered by mrW1 last updated on 30/Jun/17

This question can not be correct!

if \log (\frac{x}{y - z}) = \log (\frac{y}{z - x}) = \log (\frac{z}{x - y})

\Rightarrow \frac{x}{y - z} = \frac{y}{z - x} = \frac{z}{x - y} = \frac{1}{a} {let}^{'} s say

x \neq 0, y \neq 0, z \neq 0, a \neq 0

y - z = ax \dots (i)

z - x = ay \dots (ii)

x - y = az \dots (iii)

0 = a (x + y + z)

\Rightarrow x + y + z = 0

y + z = - x

y - z = ax

\Rightarrow y = \frac{- 1 + a}{2} x

\Rightarrow z = \frac{- 1 - a}{2} x

put this into (ii) or (iii)

x - \frac{- 1 + a}{2} x = a \frac{- 1 - a}{2} x

(2 + 1 - a) x = (- 1 - a) ax

(3 - a + a^{2} + a) x = 0

a^{2} + 3 = 0!

no such value for a is possible .

that means {it}^{'} s not true that

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

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