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

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Professor PC Thomas Classes · Answer

$ \begin{aligned} & \begin{array}{l} \frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b} \ \log x=k(b-c) \end{array} \ & \qquad \log y=k(c-a) \ & \log z=k(a-b) \ & 	ext { let. } \end{aligned} $ \ (A) $T=x y z$ $ \log T=\log x+\log y+\log z $ substuting the above values we get $\mathrm{T}=1$ $$ 	herefore x y z=1 $$ similarly we can prove $$ \begin{aligned} & x^a, y^b, c^z=1 \ & x^{b+c} y^{c+a} z^{a+b}=1 \ & x y z=x^a y^b z^c \end{aligned} $$ (B) Let $x^a y^b z^c=p$ $$ \begin{aligned} & \Rightarrow \log P=a \log x+b \log y+c \log z \ & \Rightarrow \log P=\{a(b-c)+b(c-a)+c(a-b)\} \ & \Rightarrow \log P=0 \Rightarrow P=1 \end{aligned} $$ (C) Let $x^{b+c} \cdot y^{c+a} \cdot z^{a+b}=Q$ $$ \begin{aligned} \log Q & =(b+c) \log x+(c+a) \log y+(a+b) \log z \ \Rightarrow \quad \log Q & =k(b+c)(b-c)+(c+a)(c-a)+(a+b)(a-b) \ \log Q & =0 \ & Q=1 \end{aligned} $$