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Question Number 7003 by Tawakalitu. last updated on 05/Aug/16

Solve for x  x^x  + 2x = 8

$${Solve}\:{for}\:{x} \\ $$$${x}^{{x}} \:+\:\mathrm{2}{x}\:=\:\mathrm{8} \\ $$

Commented by prakash jain last updated on 05/Aug/16

Put u=x−4⇒x=(u+4)  (u+4)^((u+4)) +2u=0

$$\mathrm{Put}\:{u}={x}−\mathrm{4}\Rightarrow{x}=\left({u}+\mathrm{4}\right) \\ $$$$\left({u}+\mathrm{4}\right)^{\left({u}+\mathrm{4}\right)} +\mathrm{2}{u}=\mathrm{0} \\ $$

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