Question Number 191589 by MATHEMATICSAM last updated on 28/Apr/23
$${a}^{{x}} \:=\:{bc},\:{b}^{{y}} \:=\:{ca},\:{c}^{{z}} \:=\:{ab}. \\ $$$$\mathrm{Prove}\:\mathrm{that},\:\frac{{x}}{\mathrm{1}\:+\:{x}}\:+\:\frac{{y}}{\mathrm{1}\:+\:{y}}\:+\:\frac{{z}}{\mathrm{1}\:+\:{z}}\:=\:\mathrm{2}. \\ $$$$\left(\mathrm{Without}\:\mathrm{using}\:\mathrm{log}\right) \\ $$$${a}\:\neq\:{b}\:\neq\:{c} \\ $$
Commented by mr W last updated on 27/Apr/23
$${not}\:{true}\:{if} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{2}},\:{b}=\mathrm{1},\:{c}=\mathrm{2} \\ $$$${x}={y}={z}=−\mathrm{1} \\ $$
Answered by Rasheed.Sindhi last updated on 26/Apr/23
$${a}^{{x}} \:=\:{bc},\:{b}^{{y}} \:=\:{ca},\:{c}^{{z}} \:=\:{ab}. \\ $$$$\mathrm{Prove}\:\mathrm{that},\:\frac{{x}}{\mathrm{1}\:+\:{x}}\:+\:\frac{{y}}{\mathrm{1}\:+\:{y}}\:+\:\frac{{z}}{\mathrm{1}\:+\:{z}}\:=\:\mathrm{2}. \\ $$$${a}^{{x}} {b}^{{y}} {c}^{{z}} =\left({bc}\right)\left({ca}\right)\left({ab}\right)={a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} \\ $$$${x}=\mathrm{2},{y}=\mathrm{2},{z}=\mathrm{2} \\ $$$$\:\:\:\:\frac{{x}}{\mathrm{1}\:+\:{x}}\:+\:\frac{{y}}{\mathrm{1}\:+\:{y}}\:+\:\frac{{z}}{\mathrm{1}\:+\:{z}}\: \\ $$$$\:=\frac{\mathrm{2}}{\mathrm{1}\:+\:\mathrm{2}}\:+\:\frac{\mathrm{2}}{\mathrm{1}\:+\:\mathrm{2}}\:+\:\frac{\mathrm{2}}{\mathrm{1}\:+\:\mathrm{2}}\: \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{3}=\mathrm{2} \\ $$