Question-210855

Question Number 210855 by zhou0429 last updated on 20/Aug/24

Answered by Frix last updated on 20/Aug/24

x, y >0 ∧ x≠y xln x =yln y Let y=px∧p>0∧p≠1 xln x =pxln px ln x =pln p +pln x (1−p)ln x =pln p ((ln x)/(ln p))=(p/(1−p)) ln_p x =(p/(1−p)) x=p^(p/(1−p)) y=p^(1/(1−p))

$${x},\:{y}\:>\mathrm{0}\:\wedge\:{x}\neq{y} \\ $$$${x}\mathrm{ln}\:{x}\:={y}\mathrm{ln}\:{y} \\ $$$$\mathrm{Let}\:{y}={px}\wedge{p}>\mathrm{0}\wedge{p}\neq\mathrm{1} \\ $$$${x}\mathrm{ln}\:{x}\:={px}\mathrm{ln}\:{px} \\ $$$$\mathrm{ln}\:{x}\:={p}\mathrm{ln}\:{p}\:+{p}\mathrm{ln}\:{x} \\ $$$$\left(\mathrm{1}−{p}\right)\mathrm{ln}\:{x}\:={p}\mathrm{ln}\:{p} \\ $$$$\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:{p}}=\frac{{p}}{\mathrm{1}−{p}} \\ $$$$\mathrm{ln}_{{p}} \:{x}\:=\frac{{p}}{\mathrm{1}−{p}} \\ $$$${x}={p}^{\frac{{p}}{\mathrm{1}−{p}}} \\ $$$${y}={p}^{\frac{\mathrm{1}}{\mathrm{1}−{p}}} \\ $$

Commented by zhou0429 last updated on 20/Aug/24