Question Number 187731 by Rupesh123 last updated on 21/Feb/23
Answered by Frix last updated on 21/Feb/23
$$\mathrm{Very}\:\mathrm{obviously}\:{x}=\mathrm{0} \\ $$
Answered by Rasheed.Sindhi last updated on 21/Feb/23
$$\mathrm{2}^{{x}} +\mathrm{2}^{−{x}} =\mathrm{2} \\ $$$$\mathrm{2}^{{x}} ={y}\Rightarrow\mathrm{2}^{−{x}} =\frac{\mathrm{1}}{{y}} \\ $$$${y}+\frac{\mathrm{1}}{{y}}=\mathrm{2} \\ $$$${y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{1}=\mathrm{0} \\ $$$$\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${y}=\mathrm{1} \\ $$$$\mathrm{2}^{{x}} =\mathrm{1} \\ $$$$\mathrm{2}^{{x}} =\mathrm{2}^{\mathrm{0}} \\ $$$${x}=\mathrm{0} \\ $$
Answered by CElcedricjunior last updated on 21/Feb/23
$$\mathrm{2}^{\mathrm{2}\boldsymbol{{x}}} −\mathrm{2}^{\boldsymbol{{x}}} +\mathrm{1}=\mathrm{0}=>\left(\mathrm{2}^{\boldsymbol{{x}}} −\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\:\blacksquare{Moivre} \\ $$$$=>\mathrm{2}^{\boldsymbol{{x}}} =\mathrm{1}=>\boldsymbol{{xln}}\mathrm{2}=\boldsymbol{{ln}}\mathrm{1}=>\boldsymbol{{x}}=\frac{\boldsymbol{{ln}}\mathrm{1}}{\boldsymbol{{ln}}\mathrm{2}}=\frac{\mathrm{0}}{\boldsymbol{{ln}}\mathrm{2}}=\mathrm{0} \\ $$$$=>\boldsymbol{{x}}=\mathrm{0}\:\bigstar\mathscr{C}{edric}\:{junior} \\ $$
Answered by manxsol last updated on 21/Feb/23
$${a}+\frac{\mathrm{1}}{{a}}\gg\mathrm{2}\Rightarrow{a}=\mathrm{1}\:\:\:\mathrm{1}+\mathrm{1}=\mathrm{2} \\ $$$$\mathrm{2}^{{z}} =\mathrm{1}\Rightarrow{x}=\mathrm{0} \\ $$