If log x:log y:log z = (y - z),:,(z - x):(x - | vedclass

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Question

(b) $\log x:\log y:\log z = y - z:z - x:x - y$

$ \Rightarrow $ ${{\log x} \over {y - z}} = {{\log y} \over {z - x}} = {{\log z} \over {x - y}} = k\

Vedclass · Accepted Answer

${x^x}{y^y}{z^z} = 1$

Vedclass · Answer

(b) $\log x:\log y:\log z = y - z:z - x:x - y$

$ \Rightarrow $ ${{\log x} \over {y - z}} = {{\log y} \over {z - x}} = {{\log z} \over {x - y}} = k\,{m{(say)}}$

$ \Rightarrow $ $\log x = k(y - z),\,\log y = k(z - x),\,\log z = k(x - y)$

$	herefore \log x + \log y + \log z = 0$$ \Rightarrow $$x + y = 1$ $ \Rightarrow $$xyz = 1$.

$x\log x + y\log y + z\log z$

=$x.k.(y - z) + y.k.(z - x) + z.k(x - y) = 0$

$ \Rightarrow $$\log ({x^x}.{y^y}.{z^z}) = \log 1$

$	herefore x^xy^yz^z=1 $.

If $\log x:\log y:\log z = (y - z)\,:\,(z - x):(x - y)$ then

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