Question Number 24865 by kosarrr last updated on 27/Nov/17
$$\frac{\mathrm{27}^{{x}−\mathrm{1}} +\mathrm{81}^{{x}} }{\mathrm{3}^{\mathrm{3}{x}} }=\frac{\mathrm{4}}{\mathrm{27}} \\ $$
Answered by iv@0uja last updated on 27/Nov/17
$$\frac{\mathrm{3}^{\mathrm{3}\left({x}−\mathrm{1}\right)} +\mathrm{3}^{\mathrm{4}{x}} }{\mathrm{3}^{\mathrm{3}{x}} }=\frac{\mathrm{4}}{\mathrm{3}^{\mathrm{3}} } \\ $$$$\mathrm{3}^{−\mathrm{3}} +\mathrm{3}^{{x}} =\mathrm{4}\centerdot\mathrm{3}^{−\mathrm{3}} \\ $$$$\mathrm{3}^{{x}} =\mathrm{3}\centerdot\mathrm{3}^{−\mathrm{3}} =\mathrm{3}^{−\mathrm{2}} \\ $$$${x}=−\mathrm{2} \\ $$