Question Number 37414 by hari321 last updated on 12/Jun/18
$${x}^{{x}} =\mathrm{2} \\ $$$$ \\ $$$${find}\:{the}\:{value}\:{of}\:{x}. \\ $$
Answered by MrW3 last updated on 12/Jun/18
$${x}^{{x}} =\mathrm{2} \\ $$$${x}=\mathrm{2}^{\frac{\mathrm{1}}{{x}}} \\ $$$${x}={e}^{\frac{\mathrm{ln}\:\mathrm{2}}{{x}}} \\ $$$$\mathrm{ln}\:\mathrm{2}=\frac{\mathrm{ln}\:\mathrm{2}}{{x}}{e}^{\frac{\mathrm{ln}\:\mathrm{2}}{{x}}} \\ $$$$\Rightarrow\frac{\mathrm{ln}\:\mathrm{2}}{{x}}={W}\left(\mathrm{ln}\:\mathrm{2}\right) \\ $$$$\Rightarrow{x}=\frac{\mathrm{ln}\:\mathrm{2}}{{W}\left(\mathrm{ln}\:\mathrm{2}\right)}\approx\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{0}.\mathrm{4444}}=\mathrm{1}.\mathrm{5596} \\ $$$$ \\ $$$${in}\:{general},\:{the}\:{solution}\:{of}\:{eqn}. \\ $$$${x}^{{x}} ={a}\:{with}\:{a}\geqslant\frac{\mathrm{1}}{\:^{{e}} \sqrt{{e}}}\approx\mathrm{0}.\mathrm{6922} \\ $$$${is}\:{x}=\frac{\mathrm{ln}\:{a}}{{W}\left(\mathrm{ln}\:{a}\right)} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 13/Jun/18
$${what}\:{is}\:{the}\:{meaning}\:{of}\:{W}\left({ln}\mathrm{2}\right)\:{pls}\:{explain} \\ $$
Commented by MrW3 last updated on 13/Jun/18
$${W}\left({x}\right)\:{is}\:{the}\:{Lambert}\:{W}\:{function}\:{which} \\ $$$${is}\:{the}\:{inverse}\:{function}\:{of}\:{y}={xe}^{{x}} . \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 13/Jun/18
$${i}\:{have}\:{collected}\:{information}\:{as}\:{sir}\:{MRW}\mathrm{3}\:{guided} \\ $$$$ \\ $$$${here}\:{i}\:{am}\:{posting}.... \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 13/Jun/18
Commented by tanmay.chaudhury50@gmail.com last updated on 13/Jun/18
Commented by tanmay.chaudhury50@gmail.com last updated on 13/Jun/18
Commented by tanmay.chaudhury50@gmail.com last updated on 13/Jun/18
Commented by Rasheed.Sindhi last updated on 14/Jun/18
$$\mathcal{H}{elpful}!\:\mathcal{E}{asy}\:{approach}! \\ $$