Question Number 208263 by efronzo1 last updated on 09/Jun/24
Answered by Ghisom last updated on 09/Jun/24
$${a}=\mathrm{e}^{\alpha} \:\Rightarrow\:{b}=\mathrm{e}^{\lambda} {a}\wedge{c}=\mathrm{e}^{\mathrm{2}\lambda} {a} \\ $$$${A}=\mathrm{log}_{{c}} \:{a}\:=\frac{\alpha}{\alpha+\mathrm{2}\lambda} \\ $$$${B}=\mathrm{log}_{{b}} \:{c}\:=\frac{\alpha+\mathrm{2}\lambda}{\alpha+\lambda} \\ $$$${C}=\mathrm{log}_{{a}} \:{b}\:=\frac{\alpha+\lambda}{\alpha} \\ $$$${B}−{A}={C}−{B}=\delta\:\Rightarrow \\ $$$$\frac{\left(\mathrm{3}\alpha^{\mathrm{2}} +\mathrm{3}\alpha\lambda−\mathrm{2}\lambda^{\mathrm{2}} \right)\lambda}{\alpha\left(\alpha+\lambda\right)\left(\alpha+\mathrm{2}\lambda\right)}=\mathrm{0} \\ $$$$\lambda=\mathrm{0}\:\Rightarrow\:{a}={b}={c}\wedge\delta=\mathrm{1} \\ $$$$\lambda=\frac{\mathrm{3}\pm\sqrt{\mathrm{33}}}{\mathrm{4}}\alpha\:\Rightarrow\:\delta=\frac{\mathrm{3}}{\mathrm{2}} \\ $$