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

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

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

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(B) Given that $\frac{\log x}{y - z} = \frac{\log y}{z - x} = \frac{\log z}{x - y} = k$ (let). Then $\log x = k(y - z)$,$\log y = k(z - x)$,and $\log z = k(x - y)$. Now,consider the expression $x^y \cdot y^z \cdot z^x$. Taking the logarithm on both sides: $\log(x^y \cdot y^z \cdot z^x) = y \log x + z \log y + x \log z$. Substituting the values: $= y(k(y - z)) + z(k(z - x)) + x(k(x - y))$ $= k(y^2 - yz + z^2 - zx + x^2 - xy)$ $= k(x^2 + y^2 + z^2 - xy - yz - zx)$. This does not necessarily equal $0$. However,consider $x^x \cdot y^y \cdot z^z$: $\log(x^x \cdot y^y \cdot z^z) = x \log x + y \log y + z \log z$ $= x(k(y - z)) + y(k(z - x)) + z(k(x - y))$ $= k(xy - xz + yz - yx + zx - zy) = k(0) = 0$. Since $\log(x^x \cdot y^y \cdot z^z) = 0$,we have $x^x \cdot y^y \cdot z^z = 10^0 = 1$.

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

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