Question

If $\frac{\log 3}{x-y}=\frac{\log 5}{y-z}=\frac{\log 7}{z-x}$ the $3^{x+y}\cdot 5^{y+z}\cdot 7^{z+x}$ is equal to

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

Answer 1

Answer: (C)

$1$

We have, $\frac{\log 3}{x-y}=\frac{\log 5}{y-z}=\frac{\log 7}{z-x}=\lambda$(say)

$\Rightarrow \log 3=\lambda (x-y),\log 5=\lambda (y-z),\log 7=\lambda (z-x)$

$\Rightarrow 3={10}^{\lambda (x-y)},5={10}^{\lambda (y-z)},7={10}^{\lambda (z-x)}$

$\Rightarrow 3^{x+y}\cdot 5^{y+z}\cdot 7^{z+x}={10}^{\lambda (x^2-y^2)}\cdot {10}^{\lambda (y^2-z^2)}\cdot {10}^{\lambda (z^2-x^2)}$

$\Rightarrow 3^{x+y}\cdot 5^{y+z}\cdot 7^{z+x}={10}^{\lambda (x^2-y^2+y^2-z^2+z^2-x^2)}$

$\Rightarrow 3^{x+y}\cdot 5^{y+z}\cdot 7^{z+x}={10}^0$

$\Rightarrow 3^{x+y}\cdot 5^{y+z}\cdot 7^{z+x}=1$.