Question Number 172176 by Mikenice last updated on 23/Jun/22
$${solve}: \\ $$$$\mathrm{2}^{{n}} +{n}=\mathrm{4} \\ $$
Answered by mr W last updated on 24/Jun/22
![16×2^(n−4) =4−n 16 e^((n−4)ln 2) =4−n 16 ln 2=[(4−n) ln 2]e^((4−n)ln 2) (4−n)ln 2=W(16 ln 2) ⇒n=4−((W(16 ln 2))/(ln 2))≈4−((1.811771)/(ln 2))=1.386](https://www.tinkutara.com/question/Q172199.png)
$$\mathrm{16}×\mathrm{2}^{{n}−\mathrm{4}} =\mathrm{4}−{n} \\ $$$$\mathrm{16}\:{e}^{\left({n}−\mathrm{4}\right)\mathrm{ln}\:\mathrm{2}} =\mathrm{4}−{n} \\ $$$$\mathrm{16}\:\mathrm{ln}\:\mathrm{2}=\left[\left(\mathrm{4}−{n}\right)\:\mathrm{ln}\:\mathrm{2}\right]{e}^{\left(\mathrm{4}−{n}\right)\mathrm{ln}\:\mathrm{2}} \\ $$$$\left(\mathrm{4}−{n}\right)\mathrm{ln}\:\mathrm{2}={W}\left(\mathrm{16}\:\mathrm{ln}\:\mathrm{2}\right) \\ $$$$\Rightarrow{n}=\mathrm{4}−\frac{{W}\left(\mathrm{16}\:\mathrm{ln}\:\mathrm{2}\right)}{\mathrm{ln}\:\mathrm{2}}\approx\mathrm{4}−\frac{\mathrm{1}.\mathrm{811771}}{\mathrm{ln}\:\mathrm{2}}=\mathrm{1}.\mathrm{386} \\ $$