x-x-1-2-2x-1-how-can-it-solve-this-help-me-please-

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

I don′t think we can generally solve x^((x−1)^2 ) =ax+b but in this special case we can: let t=x−1 ⇔ x=t+1 (t+1)^t^2 =2t+3 let n=t^2 ⇔ t=±(√n) (1±(√n))^n =3±2(√n) obviously n=2 ⇒ t=±(√2) ⇒ x=1±(√2)

$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{we}\:\mathrm{can}\:\mathrm{generally}\:\mathrm{solve} \\ $$$${x}^{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } ={ax}+{b} \\ $$$$\mathrm{but}\:\mathrm{in}\:\mathrm{this}\:\mathrm{special}\:\mathrm{case}\:\mathrm{we}\:\mathrm{can}: \\ $$$$\mathrm{let}\:{t}={x}−\mathrm{1}\:\Leftrightarrow\:{x}={t}+\mathrm{1} \\ $$$$\left({t}+\mathrm{1}\right)^{{t}^{\mathrm{2}} } =\mathrm{2}{t}+\mathrm{3} \\ $$$$\mathrm{let}\:{n}={t}^{\mathrm{2}} \:\Leftrightarrow\:{t}=\pm\sqrt{{n}} \\ $$$$\left(\mathrm{1}\pm\sqrt{{n}}\right)^{{n}} =\mathrm{3}\pm\mathrm{2}\sqrt{{n}} \\ $$$$\mathrm{obviously}\:{n}=\mathrm{2} \\ $$$$\Rightarrow\:{t}=\pm\sqrt{\mathrm{2}}\:\Rightarrow\:{x}=\mathrm{1}\pm\sqrt{\mathrm{2}} \\ $$

Commented by tabata last updated on 16/Aug/21