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ln-x-x-a-x-




Question Number 6055 by 1771727373 last updated on 11/Jun/16
ln(x)+x=a  x=?
$${ln}\left({x}\right)+{x}={a} \\ $$$${x}=? \\ $$
Answered by Yozzii last updated on 11/Jun/16
lnx=a−x⇒x=e^(a−x) ⇒xe^x =e^a   ⇒W(xe^x )=W(e^a )  x=W(e^a )  W(z)=Σ_(n=1) ^∞ (((−1)^(n−1) n^(n−2) )/((n−1)!))z^n    (ref. Wolfram)  ∴W(e^a )=Σ_(n=1) ^∞ (((−1)^(n−1) n^(n−2) e^(na) )/((n−1)!))  x=Σ_(n=1) ^∞ (((−1)^(n−1) n^(n−2) e^(na) )/((n−1)!))
$${lnx}={a}−{x}\Rightarrow{x}={e}^{{a}−{x}} \Rightarrow{xe}^{{x}} ={e}^{{a}} \\ $$$$\Rightarrow{W}\left({xe}^{{x}} \right)={W}\left({e}^{{a}} \right) \\ $$$${x}={W}\left({e}^{{a}} \right) \\ $$$${W}\left({z}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}^{{n}−\mathrm{2}} }{\left({n}−\mathrm{1}\right)!}{z}^{{n}} \:\:\:\left({ref}.\:{Wolfram}\right) \\ $$$$\therefore{W}\left({e}^{{a}} \right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}^{{n}−\mathrm{2}} {e}^{{na}} }{\left({n}−\mathrm{1}\right)!} \\ $$$${x}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}^{{n}−\mathrm{2}} {e}^{{na}} }{\left({n}−\mathrm{1}\right)!} \\ $$

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