Question

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

• A. 0
• B. 2
• C. 1
• D. None of these

Answer 1

Answer: (C)

We have, $\frac{\log 3}{x-y}=\frac{\log 5}{y-z}=\frac{\log 7}{z-x}=\lambda$ [say] $\Rightarrow$ $\log 3=\lambda (x-y),\log 5=\lambda (y-z),\log 7$ $=\lambda (z-x)$ $\Rightarrow$ $3={10}^{\lambda (x-y)},5={10}^{\lambda (y-z)},7={10}^{\lambda (z-x)}$ $\Rightarrow$ $3^{x+y}{.5}^{y+z}{.7}^{z+x}={10}^{\lambda (x^2-y^2)}$ ${.10}^{\lambda (y^2-z^2)}{.10}^{\lambda (z^2-x^2)}$ $\Rightarrow$ $3^{x+y}{.5}^{y+z}{.7}^{z+x}$ $={10}^{\lambda (x^2-y^2+y^2-z^2+z^2-x^2)}$ $\Rightarrow$ $3^{x+y}{.5}^{y+z}{.7}^{z+x}={10}^0$ $\Rightarrow$ $3^{x+y}{.5}^{y+z}{.7}^{z+x}=1$