Question

If $\frac{logx}{l+m-2n}=\frac{logy}{m+n-2l}=\frac{logz}{n+l-2m}$, then $xyz$ is equal to:

• A. $0$
• B. $1$
• C. $lmn$
• D. $2$

Answer 1

Answer: (B)

$1$

Let $\frac{\log x}{l+m-2n}=\frac{\log y}{m+n-2l}=\frac{\log z}{n+l-2m}=k$ (say)
$\Rightarrow \log x=k(l+m-2n)$ .............(1)
$\log y=k(m+n-2l)$ .............(2)
$\log z=k(n+l-2m)$ ................(3)
$\therefore \log x+\log y+\log z$
$=k(l+m-2n)+k(m+n-2l)+k(n+l-2m)$
$\Rightarrow \log \left(xyz\right)=0$
$\Rightarrow \log xyz=\log 1$
$\Rightarrow xyz=1$