If x=loga⁡bc,y=logh⁡c⋅a,z=logc⁡ab then xyz is

x=loga⁡bc⇒ax=bc⇒ax+1=abc⇒logabc⁡a=1x+1similarlylogabc⁡b=1y+1logabc⁡c=1z+1now, logabc⁡abc=1x+1+1y+1+1z+1⇒(x+1)(y+1)(z+1)=(y+1)(z+1)+(x+1)(z+1)+(x+1)(y+1)Simplifying xyz = x+y+z+2

If x=loga⁡bc,y=logh⁡c⋅a,z=logc⁡ab then xyz is

x=loga⁡bc⇒ax=bc⇒ax+1=abc⇒logabc⁡a=1x+1similarlylogabc⁡b=1y+1logabc⁡c=1z+1now, logabc⁡abc=1x+1+1y+1+1z+1⇒(x+1)(y+1)(z+1)=(y+1)(z+1)+(x+1)(z+1)+(x+1)(y+1)Simplifying xyz = x+y+z+2

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If $x = \log_{a} b c, y = \log_{h} c \cdot a, z = \log_{c} a b$ then xyz is

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x+y+z+1

x+y+z+3

x+y+z

x+y+z+2

answer is C.

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Detailed Solution

$\begin{aligned} x = \log_{a} b c \Rightarrow a^{x} = b c \Rightarrow a^{x + 1} = a b c \\ \Rightarrow \log_{a b c} a = \frac{1}{x + 1} \end{aligned}$
similarly
$\begin{aligned} \log_{a b c} b = \frac{1}{y + 1} \\ \log_{a b c} c = \frac{1}{z + 1} \end{aligned}$
now, $\log_{a b c} a b c = \frac{1}{x + 1} + \frac{1}{y + 1} + \frac{1}{z + 1}$
$\begin{array}{l} \Rightarrow (x + 1) (y + 1) (z + 1) \\ = (y + 1) (z + 1) + (x + 1) (z + 1) + (x + 1) (y + 1) \end{array}$
Simplifying xyz = x+y+z+2