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

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(A) 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 $\log z = k(a-b)$.Summing thes

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(A) 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 $\log z = k(a-b)$.Summing these: $\log x + \log y + \log z = k(b-c + c-a + a-b) = k(0) = 0$.Since $\log(xyz) = 0$,we have $xyz = 10^0 = 1$.Next,consider $a \log x + b \log y + c \log z = k[a(b-c) + b(c-a) + c(a-b)] = k(ab - ac + bc - ba + ca - cb) = k(0) = 0$.Thus,$\log(x^a \cdot y^b \cdot z^c) = 0$,which implies $x^a \cdot y^b \cdot z^c = 1$.Finally,consider $(b+c) \log x + (c+a) \log y + (a+b) \log z = k[(b+c)(b-c) + (c+a)(c-a) + (a+b)(a-b)] = k(b^2 - c^2 + c^2 - a^2 + a^2 - b^2) = k(0) = 0$.Thus,$\log(x^{b+c} \cdot y^{c+a} \cdot z^{a+b}) = 0$,which implies $x^{b+c} \cdot y^{c+a} \cdot z^{a+b} = 1$.Therefore,$xyz = x^a \cdot y^b \cdot z^c = x^{b+c} \cdot y^{c+a} \cdot z^{a+b} = 1$.

If $\frac{\log x}{b-c} = \frac{\log y}{c-a} = \frac{\log z}{a-b}$,then $xyz = x^a \cdot y^b \cdot z^c = x^{b+c} \cdot y^{c+a} \cdot z^{a+b} = $

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