Question Number 30340 by mondodotto@gmail.com last updated on 20/Feb/18
Answered by mrW2 last updated on 20/Feb/18
$$\mathrm{4}^{{x}} =\mathrm{16}{x} \\ $$$${e}^{{x}\mathrm{ln}\:\mathrm{4}} =\mathrm{16}{x} \\ $$$${xe}^{−{x}\mathrm{ln}\:\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{16}} \\ $$$$\left(−{x}\mathrm{ln}\:\mathrm{4}\right){e}^{−{x}\mathrm{ln}\:\mathrm{4}} =−\frac{\mathrm{ln}\:\mathrm{4}}{\mathrm{16}} \\ $$$$\Rightarrow−{x}\mathrm{ln}\:\mathrm{4}={W}\left(−\frac{\mathrm{ln}\:\mathrm{4}}{\mathrm{16}}\right) \\ $$$$\Rightarrow{x}=−\frac{{W}\left(−\frac{\mathrm{ln}\:\mathrm{4}}{\mathrm{16}}\right)}{\mathrm{ln}\:\mathrm{4}}\approx\begin{cases}{−\frac{−\mathrm{0}.\mathrm{0953}}{\mathrm{ln}\:\mathrm{4}}=\mathrm{0}.\mathrm{0687}}\\{−\frac{−\mathrm{3}.\mathrm{7741}}{\mathrm{ln}\:\mathrm{4}}=\mathrm{2}.\mathrm{7224}}\end{cases} \\ $$
Commented by mondodotto@gmail.com last updated on 21/Feb/18
$$\mathrm{sir}\:\mathrm{thanx}\:\mathrm{a}\:\mathrm{lot} \\ $$