Question

If $\frac{\log x}{l+m-2n}=\frac{\log y}{m+n-2l}=\frac{\log x}{n+l-2m}$, then $xyz$ is equal to

• A. $0$
• B. $lmn$
• C. $1$
• D. $2$

Answer 1

The correct option is C $1$

Let $\frac{\log x}{l+m-2n}=\frac{\log y}{m+n-2l}=\frac{\log z}{n+l-2m}=k\left(say\right)$

So, we get

$\log x=k(l+m-2n)$ ....... $\left(i\right)$

$\log y=k(m+n-2l)$ ....... $\left(ii\right)$

$\log z=k(n+l-2m)$ ....... $\left(iii\right)$

$\therefore \log x+\log y+\log z=k(l+m-2n)+k(m+n-2l)+k(n+l-2m)$

$\Rightarrow \log x+\log y+\log z=kl+km-2kn+km+kn-2kl+kn+kl-2km$

$\Rightarrow \log \left(xyz\right)=0$

$\Rightarrow \log xyz=\log 1$

$\Rightarrow xyz=1$