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Question Number 161096 by HongKing last updated on 11/Dec/21

if  x;y;z>0  and  a;b;c>0  different in pairs and  n;k∈N^∗   ((log x^n )/(b^k  - c^k )) = ((log y^n )/(c^k  - a^k )) = ((log z^n )/(a^k  - b^k ))  then find  (√(xyz))

$$\mathrm{if}\:\:\mathrm{x};\mathrm{y};\mathrm{z}>\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{a};\mathrm{b};\mathrm{c}>\mathrm{0} \\ $$ $$\mathrm{different}\:\mathrm{in}\:\mathrm{pairs}\:\mathrm{and}\:\:\mathrm{n};\mathrm{k}\in\mathbb{N}^{\ast} \\ $$ $$\frac{\mathrm{log}\:\mathrm{x}^{\boldsymbol{\mathrm{n}}} }{\mathrm{b}^{\boldsymbol{\mathrm{k}}} \:-\:\mathrm{c}^{\boldsymbol{\mathrm{k}}} }\:=\:\frac{\mathrm{log}\:\mathrm{y}^{\boldsymbol{\mathrm{n}}} }{\mathrm{c}^{\boldsymbol{\mathrm{k}}} \:-\:\mathrm{a}^{\boldsymbol{\mathrm{k}}} }\:=\:\frac{\mathrm{log}\:\mathrm{z}^{\boldsymbol{\mathrm{n}}} }{\mathrm{a}^{\boldsymbol{\mathrm{k}}} \:-\:\mathrm{b}^{\boldsymbol{\mathrm{k}}} } \\ $$ $$\mathrm{then}\:\mathrm{find}\:\:\sqrt{\boldsymbol{\mathrm{xyz}}} \\ $$

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