Question Number 208852 by Tawa11 last updated on 25/Jun/24
Answered by Frix last updated on 25/Jun/24
![3^(−3(x^2 −x)) =x^2 −x ⇒ x^2 −x>0 ⇔ x<0∨x>1 t=x^2 −x ⇔ x=(1/2)±((√(4t+1))/2) 3^(−3t) =t [ _(but let′s continue)^(It′s obvious to me that t=(1/3)) ] −3tln 3 =ln t t=e^(−u) −3e^(−u) ln 3 =−u ue^u =3ln 3 u=ln 3 t=(1/3) x=((3±(√(21)))/6)](https://www.tinkutara.com/question/Q208854.png)
$$\mathrm{3}^{−\mathrm{3}\left({x}^{\mathrm{2}} −{x}\right)} ={x}^{\mathrm{2}} −{x}\:\Rightarrow\:{x}^{\mathrm{2}} −{x}>\mathrm{0}\:\Leftrightarrow\:{x}<\mathrm{0}\vee{x}>\mathrm{1} \\ $$$${t}={x}^{\mathrm{2}} −{x}\:\Leftrightarrow\:{x}=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{4}{t}+\mathrm{1}}}{\mathrm{2}} \\ $$$$\mathrm{3}^{−\mathrm{3}{t}} ={t}\:\:\:\:\:\left[\:_{\mathrm{but}\:\mathrm{let}'\mathrm{s}\:\mathrm{continue}} ^{\mathrm{It}'\mathrm{s}\:\mathrm{obvious}\:\mathrm{to}\:\mathrm{me}\:\mathrm{that}\:{t}=\frac{\mathrm{1}}{\mathrm{3}}} \right] \\ $$$$−\mathrm{3}{t}\mathrm{ln}\:\mathrm{3}\:=\mathrm{ln}\:{t} \\ $$$${t}=\mathrm{e}^{−{u}} \\ $$$$−\mathrm{3e}^{−{u}} \mathrm{ln}\:\mathrm{3}\:=−{u} \\ $$$${u}\mathrm{e}^{{u}} =\mathrm{3ln}\:\mathrm{3} \\ $$$${u}=\mathrm{ln}\:\mathrm{3} \\ $$$${t}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${x}=\frac{\mathrm{3}\pm\sqrt{\mathrm{21}}}{\mathrm{6}} \\ $$
Commented by Tawa11 last updated on 25/Jun/24
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$