Question Number 150916 by tabata last updated on 16/Aug/21
$${x}^{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } =\mathrm{2}{x}+\mathrm{1}\: \\ $$$$ \\ $$$${how}\:{can}\:{it}\:{solve}\:{this}\:?{help}\:{me}\:{please} \\ $$
Answered by MJS_new last updated on 17/Aug/21
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{we}\:\mathrm{can}\:\mathrm{generally}\:\mathrm{solve} \\ $$$${x}^{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } ={ax}+{b} \\ $$$$\mathrm{but}\:\mathrm{in}\:\mathrm{this}\:\mathrm{special}\:\mathrm{case}\:\mathrm{we}\:\mathrm{can}: \\ $$$$\mathrm{let}\:{t}={x}−\mathrm{1}\:\Leftrightarrow\:{x}={t}+\mathrm{1} \\ $$$$\left({t}+\mathrm{1}\right)^{{t}^{\mathrm{2}} } =\mathrm{2}{t}+\mathrm{3} \\ $$$$\mathrm{let}\:{n}={t}^{\mathrm{2}} \:\Leftrightarrow\:{t}=\pm\sqrt{{n}} \\ $$$$\left(\mathrm{1}\pm\sqrt{{n}}\right)^{{n}} =\mathrm{3}\pm\mathrm{2}\sqrt{{n}} \\ $$$$\mathrm{obviously}\:{n}=\mathrm{2} \\ $$$$\Rightarrow\:{t}=\pm\sqrt{\mathrm{2}}\:\Rightarrow\:{x}=\mathrm{1}\pm\sqrt{\mathrm{2}} \\ $$
Commented by tabata last updated on 16/Aug/21
$${thank}\:{you}\:{sir} \\ $$
Commented by tabata last updated on 16/Aug/21
$${also}\:\mathrm{1}−\sqrt{\mathrm{2}}\:{is}\:{solution}\:{of}\:{this}\:{equation} \\ $$
Commented by MJS_new last updated on 17/Aug/21
$$\mathrm{yes},\:\mathrm{thank}\:\mathrm{you}. \\ $$
Commented by otchereabdullai@gmail.com last updated on 22/Aug/21
$$\mathrm{nice} \\ $$