Question Number 215639 by Wuji last updated on 12/Jan/25
$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{z}}\right)^{\left(\mathrm{1}+\mathrm{z}\right)} =\mathrm{2} \\ $$
Answered by mr W last updated on 12/Jan/25
$$\left(\mathrm{1}+\frac{\mathrm{1}}{{z}}\right)^{\mathrm{1}+{z}} =\mathrm{2}=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{−\mathrm{1}} =\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{1}−\mathrm{2}} \\ $$$$\Rightarrow{z}=−\mathrm{2} \\ $$
Commented by Wuji last updated on 13/Jan/25
$$\mathrm{2}=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{−\mathrm{1}} =\left(\mathrm{1}+\frac{\mathrm{1}}{−\mathrm{2}}\right)^{\mathrm{1}−\mathrm{2}} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{z}}\right)^{\mathrm{1}+\mathrm{z}} =\left(\mathrm{1}+\frac{\mathrm{1}}{−\mathrm{2}}\right)^{\mathrm{1}−\mathrm{2}} \\ $$$$\mathrm{now}\:\mathrm{the}\:\mathrm{comparison}\:\mathrm{is}\:\mathrm{obvious} \\ $$
Answered by Frix last updated on 13/Jan/25
$$\mathrm{Let}\:{z}=\frac{{x}}{\mathrm{1}−{x}}\wedge{x}\neq\mathrm{1}\:\Leftrightarrow\:{x}=\frac{{z}}{\mathrm{1}+{z}} \\ $$$${x}^{\frac{\mathrm{1}}{{x}−\mathrm{1}}} =\mathrm{2} \\ $$$${x}=\mathrm{2}^{{x}−\mathrm{1}} \\ $$$${x}=\frac{\mathrm{2}^{{x}} }{\mathrm{2}} \\ $$$$\mathrm{2}{x}=\mathrm{2}^{{x}} \\ $$$${x}=\mathrm{2} \\ $$$${z}=−\mathrm{2} \\ $$