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$\color{maroon}{pqr \neq 0}$

$\color{maroon}{p^{-x} = \dfrac{1}{q}}$

Or,$\log(p^{-x}) = \log(\dfrac{1}{q})$

Or, $-x \log (p) = \log(1)-\log(q)$ $\qquad\Big[∵\color{blue}{\log_b(m^n)= n. \log_b(m)\\\qquad\qquad \log_b(\dfrac{m}{n})= \log_b(m)-\log_b(n)} \Big]$

Or, $x\log(p) = \log (q)$ $\qquad[∵\color{blue}{\log(1)=0}]$

Or, $\color{green}{x= \dfrac{\log(q)}{\log(p)}}$

$\color{maroon}{q^{-y} = \dfrac{1}{r}}$

Or,$\log(q^{-y}) = \log(\dfrac{1}{r})$

Or, $-y \log (q) = \log(1)-\log(r)$ $\qquad\Big[∵\color{blue}{\log_b(m^n)= n. \log_b(m)\\\qquad\qquad \log_b(\dfrac{m}{n})= \log_b(m)-\log_b(n)}\Big]$

Or, $y\log(q) = \log (r)$ $\qquad[∵\color{blue}{\log(1)=0}]$

Or, $\color{green}{y= \dfrac{\log(r)}{\log(q)}}$

$\color{maroon}{r^{-z} = \dfrac{1}{p}}$

Or,$\log(r^{-z}) = \log(\dfrac{1}{p})$

Or, $-z \log (r) = \log(1)-\log(p)$ $\qquad\Big[∵\color{blue}{\log_b(m^n)= n. \log_b(m)\\\qquad\qquad \log_b(\dfrac{m}{n})= \log_b(m)-\log_b(n)}\Big]$

Or, $z\log(r) = \log (p)$ $\qquad[∵\color{blue}{\log(1)=0}]$

Or, $\color{green}{z= \dfrac{\log(p)}{\log(r)}}$

∴ $\color{black}{x \times y \times z}$ = $ \dfrac{\log(q)}{\log(p)} \times \dfrac{\log(r)}{\log(q)} \times \dfrac{\log(p)}{\log(r)}$

$\qquad \qquad = \color{black}{1}$

Correct Answer: $C$

$\color{maroon}{p^{-x} = \dfrac{1}{q}}$

Or,$\log(p^{-x}) = \log(\dfrac{1}{q})$

Or, $-x \log (p) = \log(1)-\log(q)$ $\qquad\Big[∵\color{blue}{\log_b(m^n)= n. \log_b(m)\\\qquad\qquad \log_b(\dfrac{m}{n})= \log_b(m)-\log_b(n)} \Big]$

Or, $x\log(p) = \log (q)$ $\qquad[∵\color{blue}{\log(1)=0}]$

Or, $\color{green}{x= \dfrac{\log(q)}{\log(p)}}$

$\color{maroon}{q^{-y} = \dfrac{1}{r}}$

Or,$\log(q^{-y}) = \log(\dfrac{1}{r})$

Or, $-y \log (q) = \log(1)-\log(r)$ $\qquad\Big[∵\color{blue}{\log_b(m^n)= n. \log_b(m)\\\qquad\qquad \log_b(\dfrac{m}{n})= \log_b(m)-\log_b(n)}\Big]$

Or, $y\log(q) = \log (r)$ $\qquad[∵\color{blue}{\log(1)=0}]$

Or, $\color{green}{y= \dfrac{\log(r)}{\log(q)}}$

$\color{maroon}{r^{-z} = \dfrac{1}{p}}$

Or,$\log(r^{-z}) = \log(\dfrac{1}{p})$

Or, $-z \log (r) = \log(1)-\log(p)$ $\qquad\Big[∵\color{blue}{\log_b(m^n)= n. \log_b(m)\\\qquad\qquad \log_b(\dfrac{m}{n})= \log_b(m)-\log_b(n)}\Big]$

Or, $z\log(r) = \log (p)$ $\qquad[∵\color{blue}{\log(1)=0}]$

Or, $\color{green}{z= \dfrac{\log(p)}{\log(r)}}$

∴ $\color{black}{x \times y \times z}$ = $ \dfrac{\log(q)}{\log(p)} \times \dfrac{\log(r)}{\log(q)} \times \dfrac{\log(p)}{\log(r)}$

$\qquad \qquad = \color{black}{1}$

Correct Answer: $C$

$p^{-x}=\dfrac{1}{q},q^{-y}=\dfrac{1}{r},r^{-z}=\dfrac{1}{p}$

We can write like this $\dfrac{1}{p^{x}}=\dfrac{1}{q},\dfrac{1}{q^{y}}=\dfrac{1}{r},\dfrac{1}{r^{z}}=\dfrac{1}{p}$

$\implies \dfrac{1}{p^{x}}\times \dfrac{1}{q^{y}}\times \dfrac{1}{r^{z}} = \dfrac{1}{q}\times\dfrac{1}{r} \times \dfrac{1}{p}$

$\implies \dfrac{1}{p^{x}}\times \dfrac{1}{q^{y}}\times \dfrac{1}{r^{z}} = \dfrac{1}{p}\times\dfrac{1}{q} \times \dfrac{1}{r}$

Compare the coefficients of both sides, and we get,

$\implies x=y=z=1$

$\therefore xyz = 1\times 1 \times 1 = 1$

So, the correct answer is $(C).$

We can write like this $\dfrac{1}{p^{x}}=\dfrac{1}{q},\dfrac{1}{q^{y}}=\dfrac{1}{r},\dfrac{1}{r^{z}}=\dfrac{1}{p}$

$\implies \dfrac{1}{p^{x}}\times \dfrac{1}{q^{y}}\times \dfrac{1}{r^{z}} = \dfrac{1}{q}\times\dfrac{1}{r} \times \dfrac{1}{p}$

$\implies \dfrac{1}{p^{x}}\times \dfrac{1}{q^{y}}\times \dfrac{1}{r^{z}} = \dfrac{1}{p}\times\dfrac{1}{q} \times \dfrac{1}{r}$

Compare the coefficients of both sides, and we get,

$\implies x=y=z=1$

$\therefore xyz = 1\times 1 \times 1 = 1$

So, the correct answer is $(C).$

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