Question Number 114795 by Algoritm last updated on 21/Sep/20
Answered by MJS_new last updated on 21/Sep/20
$$\mathrm{3}^{{x}} =\mathrm{27}−\mathrm{9}^{\mathrm{2}{y}} \\ $$$$\mathrm{2}^{{x}} =\frac{\mathrm{1}}{\mathrm{8}}−\mathrm{4}^{−{y}} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{1}\right)\:\mathrm{27}−\mathrm{9}^{\mathrm{2}{y}} >\mathrm{0}\:\Rightarrow\:{y}<\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\left(\mathrm{2}\right)\:\frac{\mathrm{1}}{\mathrm{8}}−\mathrm{4}^{−{y}} >\mathrm{0}\:\Rightarrow\:{y}>\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{but}\:{y}<\frac{\mathrm{3}}{\mathrm{4}}\wedge{y}>\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{is}\:\mathrm{impossible} \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{solution}\:\mathrm{in}\:\mathbb{R} \\ $$