Logx/y-z=logy/z-x=logz/x-y find x^x.y^y.z^z=?

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Logx/y-z=logy/z-x=logz/x-y find x^x.y^y.z^z=?

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Answer 1 · 2018-04-04T19:11:09

$x^{x} . y^{y} . z^{z} = 1$ $x^x.y^y.z^z=1$

Explanation:

$\frac{log x}{y - z} = \frac{log y}{z - x} = \frac{log z}{x - y} = k$ $logx/(y-z)=logy/(z-x)=logz/(x-y) =k$

Therefore,

$log x = (y - z) k$ $logx=(y-z)k$ $d d d d d$ $color(white)(ddddd$ $log y = (z - x) k$ $logy=(z-x)k$ $d d d d d$ $color(white)(ddddd$ $log z = (x - y) k$ $logz=(x-y)k$

Assume, $A = x^{x} . y^{y} . z^{z}$ $A=x^x.y^y.z^z$

Taking log,

$log A = log (x^{x} . y^{y} . z^{z})$ $logA=log(x^x.y^y.z^z)$

$\Rightarrow {log x}^{x} + {log y}^{y} + {log z}^{z}$ $=>logx^x+logy^y+logz^z$

$\Rightarrow x log x + y log y + z log z$ $=>xlogx+ylogy+zlogz$

$\Rightarrow x (y - z) k + y (z - x) k + z (x - y) k$ $=>x(y-z)k+y(z-x)k+z(x-y)k$

$\Rightarrow x k y - x k z + y k z - y k x + z k x - z k y$ $=>xky-xkz+ykz-ykx+zkx-zky$

$\Rightarrow 0$ $=>0$

$\Rightarrow log A = 0$ $=> logA=0$

$\Rightarrow x^{x} . y^{y} . z^{z} = 1$ $=>x^x.y^y.z^z=1$

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Logx/y-z=logy/z-x=logz/x-y find x^x.y^y.z^z=?

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