Question Number 160426 by alf123 last updated on 29/Nov/21
Commented by mr W last updated on 29/Nov/21
$${one}\:{equation}\:{for}\:{two}\:{unknowns} \\ $$$$\Rightarrow{no}\:{unique}\:{solution}! \\ $$
Commented by Rasheed.Sindhi last updated on 30/Nov/21
$$ \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{question}\:\mathrm{were}\:\mathrm{as}:\: \\ $$$$\blacktriangleright\mathrm{log}\left(\mathrm{log}\left(\mathrm{x}^{\mathrm{3}} \mathrm{y}−\mathrm{y}^{\mathrm{3}} \mathrm{x}\right)\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{log}\left(\mathrm{log}\left(\mathrm{x}^{\mathrm{2}} \mathrm{y}−\mathrm{y}^{\mathrm{2}} \mathrm{x}\right)=\mathrm{0}\right. \\ $$$$\mathrm{log}\left(\mathrm{log}\left(\mathrm{x}^{\mathrm{3}} \mathrm{y}−\mathrm{y}^{\mathrm{3}} \mathrm{x}\right)=\mathrm{log}\left(\mathrm{log}\left(\mathrm{x}^{\mathrm{2}} \mathrm{y}−\mathrm{y}^{\mathrm{2}} \mathrm{x}\right)\right.\right. \\ $$$$\mathrm{x}^{\mathrm{3}} \mathrm{y}−\mathrm{y}^{\mathrm{3}} \mathrm{x}=\mathrm{x}^{\mathrm{2}} \mathrm{y}−\mathrm{y}^{\mathrm{2}} \mathrm{x} \\ $$$${xy}\left({x}−{y}\right)\left({x}+{y}\right)={xy}\left({x}−{y}\right) \\ $$$$\:\:\:\:\:\:\:\:{x}+{y}=\mathrm{1} \\ $$