ln (e+ln (e+ln (e+...)))=?


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Question Number 71016 by mr W last updated on 10/Oct/19

$\ln (e + \ln (e + \ln (e + . . .))) = ?$
	Answered by mr W last updated on 11/Oct/19
	$ln (e+ln (e+ln (e+...)))=x ln (e+x)=x e+x=e^x (e+x)=e^(−e) e^(e+x) −(e+x)e^(−(e+x)) =−e^(−e) −(e+x)=W(−e^(−e) ) ⇒x=−[e+W(−e^(−e) )]= { ((−(e−0.070832)=−2.64745)),((−(e−4.138652)=1.42037)) :} but i think x>1, since x=ln (e+...)>ln e=1$
	$\ln (e + \ln (e + \ln (e + . . .))) = x$ $\ln (e + x) = x$ $e + x = e^{x}$ $(e + x) = e^{- e} e^{e + x}$ $- (e + x) e^{- (e + x)} = - e^{- e}$ $- (e + x) = W (- e^{- e})$ $\Rightarrow x = - [e + W (- e^{- e})] = {\begin{cases} - (e - 0.070832) = - 2.64745 \\ - (e - 4.138652) = 1.42037 \end{cases}$ $b u t i t h i n k x > 1, s i n c e x = \ln (e + . . .) > \ln e = 1$
	Answered by MJS last updated on 10/Oct/19

	$\ln (e + x) = x$ $approximating I get$ $x = - 2.64745 \lor x = 1.42037$
	Commented by mr W last updated on 11/Oct/19

	$t h a n k s s i r!$ $h o w t o e x p l a i n t h a t t w o v a l u e s a r e$ $p o s s i b l e f o r t h e l i m i t ?$
	Commented by MJS last updated on 11/Oct/19

	${that}^{'} s interesting . only the positive value is$ $valid, the negative value comes in the same$ $way as false solutions appear when we square$ $an equation .$
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