If frac{log x}{l+m-2n} = frac{log y}{m+n-2l} | vedclass

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(D) Let each ratio be equal to a constant $k$.Then,$\log x = k(l+m-2n)$,$\log y = k(m+n-2l)$,and $\log z = k(n+l-2m)$.Adding these three equations:$\l

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(D) Let each ratio be equal to a constant $k$.Then,$\log x = k(l+m-2n)$,$\log y = k(m+n-2l)$,and $\log z = k(n+l-2m)$.Adding these three equations:$\log x + \log y + \log z = k(l+m-2n + m+n-2l + n+l-2m)$$\log(xyz) = k(2l - 2l + 2m - 2m + 2n - 2n) = k(0) = 0$.Since $\log(xyz) = 0$,we have $xyz = e^0 = 1$.Therefore,$x^2 y^2 z^2 = (xyz)^2 = 1^2 = 1$.

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

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