A) 0
B) 2
C) 1
D) None of these
Correct Answer: C
Solution :
We have, \[\frac{\log 3}{x-y}=\frac{\log 5}{y-z}=\frac{\log 7}{z-x}=\lambda \][say] \[\Rightarrow \]\[\log 3=\lambda (x-y),\log 5=\lambda (y-z),\log 7\] \[=\lambda (z-x)\] \[\Rightarrow \]\[3={{10}^{\lambda (x-y)}},5={{10}^{\lambda (y-z)}},7={{10}^{\lambda (z-x)}}\] \[\Rightarrow \]\[{{3}^{x+y}}{{.5}^{y+z}}{{.7}^{z+x}}={{10}^{\lambda ({{x}^{2}}-{{y}^{2}})}}\] \[{{.10}^{\lambda ({{y}^{2}}-{{z}^{2}})}}{{.10}^{\lambda ({{z}^{2}}-{{x}^{2}})}}\] \[\Rightarrow \]\[{{3}^{x+y}}{{.5}^{y+z}}{{.7}^{z+x}}\] \[={{10}^{\lambda ({{x}^{2}}-{{y}^{2}}+{{y}^{2}}-{{z}^{2}}+{{z}^{2}}-{{x}^{2}})}}\] \[\Rightarrow \]\[{{3}^{x+y}}{{.5}^{y+z}}{{.7}^{z+x}}={{10}^{0}}\] \[\Rightarrow \]\[{{3}^{x+y}}{{.5}^{y+z}}{{.7}^{z+x}}=1\]
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