A) 0
B) 1
C) -1
D) 2
Correct Answer: B
Solution :
Given that, \[\frac{\log x}{a-b}=\frac{\log y}{b-c}=\frac{\log z}{c-a}\] Let each ratio be k and \[A=xyz\] then, \[\log x=k\,(a-b),\,\log \,y=k\,(b-c)\] and \[\log \,z=k\,(c-a)\] \[\therefore \] \[\log \,A=\log \,x+\log \,y+\log \,z\] \[=k\,(a-b)+k\,(b-c)+k\,(c-a)\] \[=k\,[a-b+b-c+c-a]\] \[=k[0]\] \[\therefore \] \[\log \,A=\log \,(xyz)=0\] \[[\because \,A=xyz]\] \[\Rightarrow \] \[xyz={{e}^{0}}\] \[\Rightarrow \] \[xyz=1\]
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