Question

If $\frac{\log x}{l+m-2n}=\frac{\log y}{m+n-2l}=\frac{\log z}{n+l-2m}$, then the value of $x^2{y}^2{z}^2$ will be

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

$Let\frac{\log x}{l+m-2n}=\frac{\log y}{m+n-2l}=\frac{\log z}{n+l-2m}=P\text{Let the log be base 'a'}\therefore {\log }_{a}x={P(l+m-2n)\Rightarrow x=a}^{P(l+m-2n)}lo{g}_{a}y=P(m+n-2l)\Rightarrow y=a^{P(m+n-2l)}lo{g}_{a}z={P(n+l-2m)\Rightarrow z=a}^{P(n+l-2m)}$

$x^2{y}^2{z}^2=a^{2P(l+m-2n)}\cdot a^{2P(m+n-2l)}\cdot a^{2P(n+l-2m)}=a^{2P(l+m-2n+m+n-2l+n+l-2m)}=a^{2P\left(0\right)}=a^0=1$