If x=log _{a} b c, y=log _{b} c a, z=log _{c} | vedclass

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(A) Given $x=\log _{a} b c$,adding $1$ to both sides gives $x+1=\log _{a} b c + 1 = \log _{a} b c + \log _{a} a = \log _{a} (a b c)$.Taking the recipr

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$x y z=x+y+z+2$

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(A) Given $x=\log _{a} b c$,adding $1$ to both sides gives $x+1=\log _{a} b c + 1 = \log _{a} b c + \log _{a} a = \log _{a} (a b c)$.Taking the reciprocal,$\frac{1}{x+1} = \log _{a b c} a$.Similarly,$\frac{1}{y+1} = \log _{a b c} b$ and $\frac{1}{z+1} = \log _{a b c} c$.Adding these three equations: $\frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} = \log _{a b c} a + \log _{a b c} b + \log _{a b c} c = \log _{a b c} (a b c) = 1$.Now,$\frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} = 1$ implies $(y+1)(z+1) + (x+1)(z+1) + (x+1)(y+1) = (x+1)(y+1)(z+1)$.Expanding both sides: $(yz+y+z+1) + (xz+x+z+1) + (xy+x+y+1) = xyz + xy + yz + zx + x + y + z + 1$.Simplifying: $xy + yz + zx + 2(x+y+z) + 3 = xyz + xy + yz + zx + x + y + z + 1$.Subtracting $xy + yz + zx + x + y + z + 1$ from both sides gives $x+y+z+2 = xyz$.

If $x=\log _{a} b c, y=\log _{b} c a, z=\log _{c} a b,$ then

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