Question

Logx/y-z=logy/z-x=logz/x-y then prove that xyz=1

Answer 1

HELLO DEAR,

it seems there is some mistakes in QUESTIONS,

the correct question is If logx/y-z = logy/z-x = logz/x-y ,show that :- $\bold{x^x * y^y *z^z = 1}$

let $\bold{log\frac{ x}{y - z} =log \frac{ y}{ z - x} = log\frac{z}{x - y} = K}$

so,

logx = (y - z)k ,

logy = (z - x)k ,

logz = (x - y)k

now,

We have to prove,

$\bold{x^x*y^y*z^z = 1}$

let $\bold{x^x*y^y*z^z = P}$

$\bold{log(x^x *y^y*z^z) = log P}$

$\bold{xlogx + ylogy + zlogz = log P}$

now, using the above values,

$\bold{x(y - z)k + y(z - x)k + z(x - y)k = log P}$

$\bold{xky - xkz + ykz - xky + xkz - zky = log P}$

$\bold{0 = log P}$

$\bold{log1 = logP}$

HENCE, p = 1

Thus , $\bold{x^x*y^y*z^z = 1}$

I HOPE ITS HELP YOU DEAR,

THANKS

Answer 2

Logx/(y-z)=logy/(z-x)=logz/(x-y)=k ,. Then logx=k(y-z), logy=k(z-x), logz=k(x-y).  Then log(xyz)=logx+logy+logz, log(xyz)=ky-kz+kz-kx+kx-ky=0=log1. Xyz=1(proved)