Question Number 71016 by mr W last updated on 10/Oct/19
$$\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+…\right)\right)\right)=? \\ $$
Answered by mr W last updated on 11/Oct/19
$$\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+…\right)\right)\right)={x} \\ $$$$\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+{x}\right)={x} \\ $$$${e}+{x}={e}^{{x}} \\ $$$$\left({e}+{x}\right)={e}^{−{e}} {e}^{{e}+{x}} \\ $$$$−\left({e}+{x}\right){e}^{−\left({e}+{x}\right)} =−{e}^{−{e}} \\ $$$$−\left({e}+{x}\right)={W}\left(−{e}^{−{e}} \right) \\ $$$$\Rightarrow{x}=−\left[{e}+{W}\left(−{e}^{−{e}} \right)\right]=\begin{cases}{−\left({e}−\mathrm{0}.\mathrm{070832}\right)=−\mathrm{2}.\mathrm{64745}}\\{−\left({e}−\mathrm{4}.\mathrm{138652}\right)=\mathrm{1}.\mathrm{42037}}\end{cases} \\ $$$${but}\:{i}\:{think}\:{x}>\mathrm{1},\:{since}\:{x}=\mathrm{ln}\:\left({e}+…\right)>\mathrm{ln}\:{e}=\mathrm{1} \\ $$
Answered by MJS last updated on 10/Oct/19
$$\mathrm{ln}\:\left({e}+{x}\right)\:={x} \\ $$$$\mathrm{approximating}\:\mathrm{I}\:\mathrm{get} \\ $$$${x}=−\mathrm{2}.\mathrm{64745}\vee{x}=\mathrm{1}.\mathrm{42037} \\ $$
Commented by mr W last updated on 11/Oct/19
$${thanks}\:{sir}! \\ $$$${how}\:{to}\:{explain}\:{that}\:{two}\:{values}\:{are} \\ $$$${possible}\:{for}\:{the}\:{limit}? \\ $$
Commented by MJS last updated on 11/Oct/19
$$\mathrm{that}'\mathrm{s}\:\mathrm{interesting}.\:\mathrm{only}\:\mathrm{the}\:\mathrm{positive}\:\mathrm{value}\:\mathrm{is} \\ $$$$\mathrm{valid},\:\mathrm{the}\:\mathrm{negative}\:\mathrm{value}\:\mathrm{comes}\:\mathrm{in}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{way}\:\mathrm{as}\:\mathrm{false}\:\mathrm{solutions}\:\mathrm{appear}\:\mathrm{when}\:\mathrm{we}\:\mathrm{square} \\ $$$$\mathrm{an}\:\mathrm{equation}. \\ $$