Question Number 203035 by Frix last updated on 07/Jan/24
$$\mathrm{e}=\mathrm{2} \\ $$$$\mathrm{Proof}: \\ $$$$\mathrm{Let}\:{x}=\frac{\mathrm{e}+\mathrm{2}}{\mathrm{2}} \\ $$$$\mathrm{2}{x}=\mathrm{e}+\mathrm{2} \\ $$$$\mathrm{2}{x}\left(\mathrm{e}−\mathrm{2}\right)=\left(\mathrm{e}+\mathrm{2}\right)\left(\mathrm{e}−\mathrm{2}\right) \\ $$$$\mathrm{2e}{x}−\mathrm{4}{x}=\mathrm{e}^{\mathrm{2}} −\mathrm{4} \\ $$$$−\mathrm{4}{x}+\mathrm{4}=−\mathrm{2e}{x}+\mathrm{e}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}={x}^{\mathrm{2}} −\mathrm{2e}{x}+\mathrm{e}^{\mathrm{2}} \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{2}} =\left({x}−\mathrm{e}\right)^{\mathrm{2}} \\ $$$$\sqrt{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }=\sqrt{\left({x}−\mathrm{e}\right)^{\mathrm{2}} } \\ $$$${x}−\mathrm{2}={x}−\mathrm{e} \\ $$$$−\mathrm{2}=−\mathrm{e} \\ $$$$\mathrm{e}=\mathrm{2} \\ $$
Commented by Calculusboy last updated on 07/Jan/24
$$\boldsymbol{{nice}}\:\boldsymbol{{sir}} \\ $$
Commented by AST last updated on 07/Jan/24
$$\mathrm{2}{x}={e}+\mathrm{2}\Leftrightarrow\mathrm{2}{x}\left({e}−\mathrm{2}\right)=\left({e}+\mathrm{2}\right)\left({e}−\mathrm{2}\right)\:{is}\:{only}\:{valid} \\ $$$${when}\:{e}\neq\mathrm{2} \\ $$$$\sqrt{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }=\sqrt{\left({x}−{e}\right)^{\mathrm{2}} }\Rightarrow{x}−\mathrm{2}={x}−{e}\:{is}\:{only}\:{valid} \\ $$$${when}\:\:{x}\geqslant\mathrm{2}\:\wedge\:{x}\geqslant{e} \\ $$$$\mathrm{e}\:{here}\:{also}\:{has}\:{nothing}\:{to}\:{do}\:{with}\:{Euler}'{s}\: \\ $$$${number}\left(\approx\mathrm{2}.\mathrm{718281}\right).\:{So},\:{it}\:{is}\:{acting}\:{like}\:{a} \\ $$$${random}\:{variable}.\:{It}\:{could}\:{also}\:{be}\:{a},{b},{s},{t},{x}… \\ $$
Commented by Frix last updated on 08/Jan/24
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