Question Number 7003 by Tawakalitu. last updated on 05/Aug/16 | ||
$${Solve}\:{for}\:{x} \\ $$$${x}^{{x}} \:+\:\mathrm{2}{x}\:=\:\mathrm{8} \\ $$ | ||
Commented by prakash jain last updated on 05/Aug/16 | ||
$$\mathrm{Put}\:{u}={x}−\mathrm{4}\Rightarrow{x}=\left({u}+\mathrm{4}\right) \\ $$$$\left({u}+\mathrm{4}\right)^{\left({u}+\mathrm{4}\right)} +\mathrm{2}{u}=\mathrm{0} \\ $$ | ||