Question Number 151164 by puissant last updated on 18/Aug/21
Commented by puissant last updated on 18/Aug/21
$${solve}\:{in}\:\mathbb{R} \\ $$
Answered by MJS_new last updated on 18/Aug/21
$$\mathrm{let}\:{t}=\mathrm{e}^{\pi{x}} \:\Leftrightarrow\:{x}=\frac{\mathrm{ln}\:{t}}{\pi} \\ $$$${t}+\frac{\sqrt{\mathrm{2}}}{{t}}−\mathrm{1}−\sqrt{\mathrm{2}}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} −\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){t}+\sqrt{\mathrm{2}}=\mathrm{0} \\ $$$$\left({t}−\mathrm{1}\right)\left({t}−\sqrt{\mathrm{2}}\right)=\mathrm{0} \\ $$$${t}=\mathrm{1}\vee{t}=\sqrt{\mathrm{2}} \\ $$$$\Leftrightarrow \\ $$$${x}=\mathrm{0}\vee{x}=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}\pi} \\ $$