If frac{log x}{b - c} = frac{log y}{c - a} = | vedclass

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(D) Let $\frac{\log x}{b - c} = \frac{\log y}{c - a} = \frac{\log z}{a - b} = k$.Then,$\log x = k(b - c)$,$\log y = k(c - a)$,and $\log z = k(a - b)$.

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All of these

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(D) Let $\frac{\log x}{b - c} = \frac{\log y}{c - a} = \frac{\log z}{a - b} = k$.Then,$\log x = k(b - c)$,$\log y = k(c - a)$,and $\log z = k(a - b)$.For option $A$: $xyz = e^{\log x + \log y + \log z} = e^{k(b - c + c - a + a - b)} = e^0 = 1$.For option $B$: $x^a y^b z^c = e^{a \log x + b \log y + c \log z} = e^{k(a(b - c) + b(c - a) + c(a - b))} = e^{k(ab - ac + bc - ba + ca - cb)} = e^0 = 1$.For option $C$: $x^{b + c} y^{c + a} z^{a + b} = e^{(b + c) \log x + (c + a) \log y + (a + b) \log z} = e^{k((b + c)(b - c) + (c + a)(c - a) + (a + b)(a - b))} = e^{k(b^2 - c^2 + c^2 - a^2 + a^2 - b^2)} = e^0 = 1$.Since all options are true,the correct answer is $D$.

If $\frac{\log x}{b - c} = \frac{\log y}{c - a} = \frac{\log z}{a - b}$,then which of the following is true?

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