If log x : log y : log z = (y - z) : (z - x) | vedclass

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(B) Given $\frac{\log x}{y - z} = \frac{\log y}{z - x} = \frac{\log z}{x - y} = k$ (say).Then $\log x = k(y - z)$,$\log y = k(z - x)$,and $\log z = k(

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$x^x \cdot y^y \cdot z^z = 1$

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(B) Given $\frac{\log x}{y - z} = \frac{\log y}{z - x} = \frac{\log z}{x - y} = k$ (say).Then $\log x = k(y - z)$,$\log y = k(z - x)$,and $\log z = k(x - y)$.Consider the expression $x \log x + y \log y + z \log z$:$= x \cdot k(y - z) + y \cdot k(z - x) + z \cdot k(x - y)$$= k(xy - xz + yz - yx + zx - zy)$$= k(0) = 0$.Thus,$\log(x^x) + \log(y^y) + \log(z^z) = 0$.$\log(x^x \cdot y^y \cdot z^z) = \log(1)$.Therefore,$x^x \cdot y^y \cdot z^z = 1$.

If $\log x : \log y : \log z = (y - z) : (z - x) : (x - y)$,then

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