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  4. If ( log 3/x-y)=( log 5/y-z)=(1og7/z-x) ,then 3x+y.5y+z.7z+x is equal to

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Q. If $ \frac{\log 3}{x-y}=\frac{\log 5}{y-z}=\frac{1og7}{z-x} $ ,then $ {{3}^{x+y}}{{.5}^{y+z}}{{.7}^{z+x}} $ is equal to

Bihar CECEBihar CECE 2015

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 $