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

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(B) Given $x = \log_a(bc)$,adding $1$ to both sides gives $1 + x = 1 + \log_a(bc) = \log_a(a) + \log_a(bc) = \log_a(abc)$.Taking the reciprocal,$(1 +

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$(1 + x)^{-1} + (1 + y)^{-1} + (1 + z)^{-1}$

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(B) Given $x = \log_a(bc)$,adding $1$ to both sides gives $1 + x = 1 + \log_a(bc) = \log_a(a) + \log_a(bc) = \log_a(abc)$.Taking the reciprocal,$(1 + x)^{-1} = \frac{1}{\log_a(abc)} = \log_{abc}(a)$.Similarly,$(1 + y)^{-1} = \log_{abc}(b)$ and $(1 + z)^{-1} = \log_{abc}(c)$.Adding these expressions,we get $(1 + x)^{-1} + (1 + y)^{-1} + (1 + z)^{-1} = \log_{abc}(a) + \log_{abc}(b) + \log_{abc}(c)$.Using the property $\log_n(m) + \log_n(p) = \log_n(mp)$,we have $\log_{abc}(a 	imes b 	imes c) = \log_{abc}(abc) = 1$.

If $x = \log_a(bc)$,$y = \log_b(ca)$,and $z = \log_c(ab)$,then which of the following is equal to $1$?

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