Question

If $\frac{\log x}{l+m-2n}=\frac{\log y}{m+n-2l}=\frac{\log z}{n+l-2m}$, then find $x^2{y}^2{z}^2$.

Answer 1

Given $\frac{\log x}{l+m-2n}=\frac{\log y}{m+n-2l}=\frac{\log z}{n+l-2m}$

Let $\frac{\log x}{l+m-2n}=\frac{\log y}{m+n-2l}=\frac{\log z}{n+l-2m}=K$

$\Rightarrow \frac{\log x}{l+m-2n}=K$

$\Rightarrow \log x=K\times (l+m-2n)$

multiply by 2

$\Rightarrow 2\log x=2K\times (l+m-2n)$ $\dots \left(I\right)$

Similarly,

$\Rightarrow \frac{\log y}{m+n+-2l}=K$ $\Rightarrow \frac{\log z}{n+l-2m}=K$

$\Rightarrow \log y=K\times (m+n-2l)$ $\Rightarrow \log z=K\times (n+l-2m)$

multiply by 2

$\Rightarrow 2\log y=2K\times (m+n-2l)$ $\dots \left(II\right)$ $\Rightarrow 2\log z=2K\times (n+l-2m)$ $\dots \left(III\right)$

Add $\left(I\right),\left(II\right),\left(III\right)$

$\Rightarrow 2\log x+2\log y+2\log z=2K(l+m-2n)+2(m+n-2l)+2(n+l-2m)$

$\Rightarrow \log x^2+\log y^2+\log z^2=2K[l+m-2n+m+n-2l+n+l-2m]$ $\dots [\because a\log b=\log b^a]$

$\Rightarrow \log x^2+\log y^2+\log z^2=2K\left[0\right]$

$\Rightarrow \log (x^2\times y^2\times z^2)=0$ $\dots [\because \log a+\log b=\log (a.b\left)\right]$

$\Rightarrow \log \left(x^2{y}^2{z}^2\right)=0$

$\Rightarrow x^2{y}^2{z}^2={10}^0$

$\therefore x^2{y}^2{z}^2=1$