Question Number 175121 by Michaelfaraday last updated on 20/Aug/22
Answered by BaliramKumar last updated on 20/Aug/22
$$\left(\mathrm{3}^{−\mathrm{1}} \right)^{\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}}{\mathrm{16}−\mathrm{2}{x}^{\mathrm{3}} }} \:=\:\left(\mathrm{3}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}{x}}} \\ $$$$\mathrm{3}^{\left(\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}}{\mathrm{2}{x}^{\mathrm{3}} −\mathrm{16}}\right)} \:=\:\mathrm{3}^{\left(\frac{\mathrm{2}}{\mathrm{4}{x}}\right)} \\ $$$$\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}}{\mathrm{2}\left({x}^{\mathrm{3}} −\mathrm{8}\right)}\:=\:\frac{\mathrm{2}}{\mathrm{4}{x}} \\ $$$$\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}}{{x}^{\mathrm{3}} −\mathrm{8}}\:=\:\frac{\mathrm{1}}{{x}} \\ $$$${x}^{\mathrm{3}} −\mathrm{8}\:=\:{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} \:=\:\mathrm{8} \\ $$$${x}\:=\:\pm\mathrm{2} \\ $$
Commented by Frix last updated on 20/Aug/22
$$+\mathrm{2}\:\mathrm{is}\:\mathrm{wrong}\:\mathrm{because}\:\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}}{\mathrm{16}−\mathrm{2}{x}^{\mathrm{3}} }\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined} \\ $$
Commented by Tawa11 last updated on 20/Aug/22
$$\mathrm{Great}\:\mathrm{sirs} \\ $$
Commented by peter frank last updated on 24/Aug/22
$$\mathrm{good} \\ $$
Answered by Frix last updated on 20/Aug/22
$$\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}}{\mathrm{16}−\mathrm{2}{x}^{\mathrm{3}} }=−\frac{{x}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)} \\ $$$$\mathrm{3}^{\frac{{x}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)}} =\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}{x}}} \\ $$$$\frac{{x}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)}=\frac{\mathrm{1}}{\mathrm{2}{x}} \\ $$$$\mathrm{4}{x}+\mathrm{8}=\mathrm{0} \\ $$$${x}=−\mathrm{2} \\ $$