Question Number 148914 by naka3546 last updated on 01/Aug/21
$${Solve}\:\:{the}\:\:{equation} \\ $$$$\:\:\:\mathrm{2}^{{x}} \:+\:{x}\:=\:\mathrm{11} \\ $$$${with}\:\:{Omega}\:\:{Function}\:. \\ $$
Commented by naka3546 last updated on 01/Aug/21
$${thank}\:{you},\:{sir}. \\ $$
Commented by mr W last updated on 01/Aug/21
$$\mathrm{2}^{\mathrm{11}} =\left(\mathrm{11}−{x}\right)\mathrm{2}^{\mathrm{11}−{x}} \\ $$$$\mathrm{2}^{\mathrm{11}} \mathrm{ln}\:\mathrm{2}=\left(\mathrm{11}−{x}\right)\mathrm{ln}\:\mathrm{2}{e}^{\left(\mathrm{11}−{x}\right)\mathrm{ln}\:\mathrm{2}} \\ $$$$\left(\mathrm{11}−{x}\right)\mathrm{ln}\:\mathrm{2}={W}\left(\mathrm{2}^{\mathrm{11}} \mathrm{ln}\:\mathrm{2}\right) \\ $$$${x}=\mathrm{11}−\frac{{W}\left(\mathrm{2}^{\mathrm{11}} \mathrm{ln}\:\mathrm{2}\right)}{\mathrm{ln}\:\mathrm{2}}=\mathrm{3} \\ $$