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Question Number 208852 by Tawa11 last updated on 25/Jun/24

Answered by Frix last updated on 25/Jun/24

$$3^{-3(x^2-x)}=x^2-x\Rightarrow x^2-x>0\iff x<0\vee x>1$$$$t=x^2-x\iff x=\frac{1}{2}\pm \frac{\sqrt{4t+1}}{2}$$$$3^{-3t}=t\left[{}_{\mathrm{but}{\mathrm{let}}^{\prime }\mathrm{s}\mathrm{continue}}^{{\mathrm{It}}^{\prime }\mathrm{s}\mathrm{obvious}\mathrm{to}\mathrm{me}\mathrm{that}t=\frac{1}{3}}\right]$$$$-3t\ln 3=\ln t$$$$t={\mathrm{e}}^{-u}$$$$-{3\mathrm{e}}^{-u}\ln 3=-u$$$$u{\mathrm{e}}^u=3\ln 3$$$$u=\ln 3$$$$t=\frac{1}{3}$$$$x=\frac{3\pm \sqrt{21}}{6}$$

Commented by Tawa11 last updated on 25/Jun/24

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}.$$

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