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Question Number 131697 by mathlove last updated on 07/Feb/21

x^x =6 x=?

$${x}^{{x}} =\mathrm{6}\:\:\:\:\:\:\:\:\:\:{x}=? \\ $$

Answered by mr W last updated on 07/Feb/21

x=6^(1/x) =e^((ln 6)/x) (1/x)e^((ln 6)/x) =1 ((ln 6)/x)e^((ln 6)/x) =ln 6 ((ln 6)/x)=W(ln 6) ⇒x=((ln 6)/(W(ln 6)))≈2.231829

$${x}=\mathrm{6}^{\frac{\mathrm{1}}{{x}}} ={e}^{\frac{\mathrm{ln}\:\mathrm{6}}{{x}}} \\ $$$$\frac{\mathrm{1}}{{x}}{e}^{\frac{\mathrm{ln}\:\mathrm{6}}{{x}}} =\mathrm{1} \\ $$$$\frac{\mathrm{ln}\:\mathrm{6}}{{x}}{e}^{\frac{\mathrm{ln}\:\mathrm{6}}{{x}}} =\mathrm{ln}\:\mathrm{6} \\ $$$$\frac{\mathrm{ln}\:\mathrm{6}}{{x}}={W}\left(\mathrm{ln}\:\mathrm{6}\right) \\ $$$$\Rightarrow{x}=\frac{\mathrm{ln}\:\mathrm{6}}{{W}\left(\mathrm{ln}\:\mathrm{6}\right)}\approx\mathrm{2}.\mathrm{231829} \\ $$

Commented by mathlove last updated on 07/Feb/21

thanks sir

$${thanks}\:{sir} \\ $$

Commented by mathlove last updated on 07/Feb/21

what W(ln6)=?

$${what}\:\:{W}\left({ln}\mathrm{6}\right)=? \\ $$

Commented by mr W last updated on 07/Feb/21

Lambert W function W(ln 6)≈0.802821260392

$${Lambert}\:{W}\:{function} \\ $$$${W}\left(\mathrm{ln}\:\mathrm{6}\right)\approx\mathrm{0}.\mathrm{802821260392} \\ $$

Answered by Dwaipayan Shikari last updated on 07/Feb/21

x^x =6 ⇒xlog(x)=log(6)⇒log(x)e^(log(x)) =log(6)⇒log(x)=W_0 (log(6)) x=e^(W_0 (log(6))) =((log(6))/(W_0 (log(6)))) x=((log(6))/(log(6)−log^2 (6)+(3/2)log^3 (6)−(8/3)log^4 (6)+((125)/(24))log^5 (6)−..)) Exact solution for x =(1/(1−log(6)+(3/2)log^2 (6)−(8/3)log^3 (6)+...))

$${x}^{{x}} =\mathrm{6} \\ $$$$\Rightarrow{xlog}\left({x}\right)={log}\left(\mathrm{6}\right)\Rightarrow{log}\left({x}\right){e}^{{log}\left({x}\right)} ={log}\left(\mathrm{6}\right)\Rightarrow{log}\left({x}\right)={W}_{\mathrm{0}} \left({log}\left(\mathrm{6}\right)\right) \\ $$$${x}={e}^{{W}_{\mathrm{0}} \left({log}\left(\mathrm{6}\right)\right)} =\frac{{log}\left(\mathrm{6}\right)}{{W}_{\mathrm{0}} \left({log}\left(\mathrm{6}\right)\right)} \\ $$$${x}=\frac{{log}\left(\mathrm{6}\right)}{{log}\left(\mathrm{6}\right)−{log}^{\mathrm{2}} \left(\mathrm{6}\right)+\frac{\mathrm{3}}{\mathrm{2}}{log}^{\mathrm{3}} \left(\mathrm{6}\right)−\frac{\mathrm{8}}{\mathrm{3}}{log}^{\mathrm{4}} \left(\mathrm{6}\right)+\frac{\mathrm{125}}{\mathrm{24}}{log}^{\mathrm{5}} \left(\mathrm{6}\right)−..} \\ $$$${Exact}\:{solution}\:{for}\:{x} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−{log}\left(\mathrm{6}\right)+\frac{\mathrm{3}}{\mathrm{2}}{log}^{\mathrm{2}} \left(\mathrm{6}\right)−\frac{\mathrm{8}}{\mathrm{3}}{log}^{\mathrm{3}} \left(\mathrm{6}\right)+...} \\ $$

Commented by I want to learn more last updated on 07/Feb/21

Sir, please what is the formular for the W_0 (log 6) for the exact form.

$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{formular}\:\mathrm{for}\:\mathrm{the}\:\mathrm{W}_{\mathrm{0}} \left(\mathrm{log}\:\mathrm{6}\right)\:\:\mathrm{for}\:\mathrm{the}\:\mathrm{exact}\:\mathrm{form}. \\ $$

Commented by Dwaipayan Shikari last updated on 07/Feb/21

Asymtotic relation W_0 (x)=Σ_(n=1) ^∞ (((−n)^(n−1) )/(n!))x^n

$${Asymtotic}\:{relation} \\ $$$${W}_{\mathrm{0}} \left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−{n}\right)^{{n}−\mathrm{1}} }{{n}!}{x}^{{n}} \\ $$

Commented by I want to learn more last updated on 07/Feb/21

Thanks sir, i appreciate.

$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate}. \\ $$

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