Question Number 183712 by Michaelfaraday last updated on 29/Dec/22
$${solve}: \\ $$$${W}\left({In}\left(\mathrm{4}{x}\right)\right)=\sqrt{\left({x}−\mathrm{1}\right)} \\ $$
Answered by mr W last updated on 29/Dec/22
$${e}^{\mathrm{ln}\:\left(\mathrm{4}{x}\right)} \mathrm{ln}\:\left(\mathrm{4}{x}\right)=\sqrt{{x}−\mathrm{1}} \\ $$$$\left(\mathrm{4}{x}\right)^{\mathrm{4}{x}} ={e}^{\sqrt{{x}−\mathrm{1}}} \\ $$$${let}\:{t}=\sqrt{{x}−\mathrm{1}}\geqslant\mathrm{0} \\ $$$$\left(\mathrm{4}{t}^{\mathrm{2}} +\mathrm{4}\right)^{\left(\mathrm{4}{t}^{\mathrm{2}} +\mathrm{4}\right)} ={e}^{{t}} \\ $$$${f}\left({t}\right)=\left(\mathrm{4}{t}^{\mathrm{2}} +\mathrm{4}\right)^{\left(\mathrm{4}{t}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$$${g}\left({t}\right)={e}^{{t}} \\ $$$${f}\left(\mathrm{0}\right)−{g}\left(\mathrm{0}\right)=\mathrm{4}^{\mathrm{4}} −{e}^{\mathrm{0}} =\mathrm{256}−\mathrm{1}=\mathrm{255} \\ $$$${f}'\left({t}\right)−{g}'\left({t}\right)=…\:>\mathrm{0} \\ $$$$\Rightarrow{f}\left({t}\right)−{g}\left({t}\right)\:{is}\:{strictly}\:{increasing}, \\ $$$${i}.{e}.\:{f}\left({t}\right)−{g}\left({t}\right)\geqslant\mathrm{255}\:{for}\:{t}\geqslant\mathrm{0} \\ $$$$\Rightarrow{f}\left({t}\right)−{g}\left({t}\right)=\mathrm{0}\:{has}\:{no}\:{roots}! \\ $$$$\Rightarrow{W}\left({In}\left(\mathrm{4}{x}\right)\right)=\sqrt{{x}−\mathrm{1}}\:{has}\:{no}\:{roots}! \\ $$
Commented by Michaelfaraday last updated on 29/Dec/22
$${wow},{thanks}\:{sir}\:{for}\:{your}\:{help} \\ $$$$ \\ $$