If {a^x} = bc,{b^y} = ca,and {c^z} = ab,then | vedclass

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(C) Given equations are: $a^x = bc$,$b^y = ca$,$c^z = ab$ Taking logarithm on both sides for each equation (assuming base $a, b, c$ are consistent): $

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$x + y + z + 2$

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(C) Given equations are: $a^x = bc$,$b^y = ca$,$c^z = ab$ Taking logarithm on both sides for each equation (assuming base $a, b, c$ are consistent): $x = \log_a(bc) = \log_a b + \log_a c$ $y = \log_b(ca) = \log_b c + \log_b a$ $z = \log_c(ab) = \log_c a + \log_c b$ Adding $1$ to each side: $x+1 = \log_a b + \log_a c + 1 = \log_a(abc)$ $y+1 = \log_b c + \log_b a + 1 = \log_b(abc)$ $z+1 = \log_c a + \log_c b + 1 = \log_c(abc)$ Taking reciprocals: $\frac{1}{x+1} = \log_{abc} a$,$\frac{1}{y+1} = \log_{abc} b$,$\frac{1}{z+1} = \log_{abc} c$ Summing these: $\frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} = \log_{abc} a + \log_{abc} b + \log_{abc} c = \log_{abc}(abc) = 1$ Alternatively,if $a=b=c=2$,then $2^x = 4 \Rightarrow x=2$,$2^y = 4 \Rightarrow y=2$,$2^z = 4 \Rightarrow z=2$. Then $xyz = 2 	imes 2 	imes 2 = 8$. Checking options: $x+y+z+2 = 2+2+2+2 = 8$. Thus,$xyz = x+y+z+2$.

If ${a^x} = bc$,${b^y} = ca$,and ${c^z} = ab$,then the value of $xyz$ is equal to:

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