If x = log_a (bc), y = log_b (ca), and z = lo | vedclass

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(B) Given $x = \log_a (bc)$, $y = \log_b (ca)$, and $z = \log_c (ab)$.Adding $1$ to each term:$1+x = 1 + \log_a (bc) = \log_a a + \log_a (bc) = \log_a

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$1$

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(B) Given $x = \log_a (bc)$, $y = \log_b (ca)$, and $z = \log_c (ab)$.Adding $1$ to each term:$1+x = 1 + \log_a (bc) = \log_a a + \log_a (bc) = \log_a (abc)$.Similarly, $1+y = \log_b (abc)$ and $1+z = \log_c (abc)$.Now, taking the reciprocal:$\frac{1}{1+x} = \frac{1}{\log_a (abc)} = \log_{abc} a$.$\frac{1}{1+y} = \frac{1}{\log_b (abc)} = \log_{abc} b$.$\frac{1}{1+z} = \frac{1}{\log_c (abc)} = \log_{abc} c$.Adding these expressions:$\frac{1}{1+x} + \frac{1}{1+y} + \frac{1}{1+z} = \log_{abc} a + \log_{abc} b + \log_{abc} c = \log_{abc} (abc) = 1$.

If $x = \log_a (bc)$, $y = \log_b (ca)$, and $z = \log_c (ab)$, then the value of $\frac{1}{1+x} + \frac{1}{1+y} + \frac{1}{1+z}$ is

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