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Question Number 143293 by Huy last updated on 12/Jun/21

$$f\left(x\right)=ln(1+ln\left(x\right)).$$$$f\left(x\right)=ln\left(f^{\prime }\left(x\right)\right).Findx$$

Answered by Olaf_Thorendsen last updated on 12/Jun/21

$$f\left(x\right)=\ln (1+\ln x)$$$$f^{\prime }\left(x\right)=\frac{\frac{1}{x}}{1+\ln x}$$$$f\left(x\right)=\ln \left(f^{\prime }\left(x\right)\right)$$$$\ln (1+\ln x)=\ln \left(\frac{1}{x(1+\ln x)}\right)$$$$1+\ln x=\frac{1}{x(1+\ln x)}$$$$x{(1+\ln x)}^2=1$$$$x=1\mathrm{is}\mathrm{an}\mathrm{evident}\mathrm{solution}$$

Answered by mathmax by abdo last updated on 12/Jun/21

$$\mathrm{f}\left(\mathrm{x}\right)=\log (1+\mathrm{logx})\mathrm{we}\mathrm{have}1+\mathrm{logx}>0\Rightarrow \mathrm{logx}>-1\Rightarrow \mathrm{x}>{\mathrm{e}}^{-1}$$$$\mathrm{f}\mathrm{is}\mathrm{defined}\mathrm{on}]{\mathrm{e}}^{-1},+\mathrm{\infty }[$$$$\mathrm{f}\left(\mathrm{x}\right)=\log \left({\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)\right)\Rightarrow \log (1+\mathrm{logx})=\log \left(\frac{1}{\mathrm{x}(1+\mathrm{logx})}\right)\Rightarrow$$$$1+\mathrm{logx}=\frac{1}{\mathrm{x}(1+\mathrm{logx})}\Rightarrow \mathrm{x}{(1+\mathrm{logx})}^2=1\Rightarrow$$$$\mathrm{x}(1+2\mathrm{logx}+{\log }^2\mathrm{x})-1=0\Rightarrow {\mathrm{xlog}}^2\mathrm{x}+2\mathrm{xlogx}+\mathrm{x}-1=0=\mathrm{\Psi }\left(\mathrm{x}\right)$$$$\mathrm{\Psi }\left(\mathrm{x}\right)={\mathrm{xlog}}^2\mathrm{x}+2\mathrm{xlogx}+\mathrm{x}-1\Rightarrow$$$${\mathrm{\Psi }}^{{}^{\prime }}\left(\mathrm{x}\right)={\log }^2\mathrm{x}+2\mathrm{logx}+2\mathrm{logx}+2+1={\log }^2\mathrm{x}+4\mathrm{logx}+3$$$$={\mathrm{u}}^2+4\mathrm{u}+3(\mathrm{u}=\mathrm{logx})$$$${\mathrm{\Delta }}^{{}^{\prime }}=2^2-3=1\Rightarrow \mathrm{logx}=-2+1\mathrm{or}\mathrm{logx}=-2-1\Rightarrow \mathrm{x}={\mathrm{e}}^{-1}\mathrm{or}\mathrm{x}={\mathrm{e}}^{-3}$$$$\mathrm{x}{\mathrm{e}}^{-3}{\mathrm{e}}^{-1}+\mathrm{\infty }$$$${\mathrm{\Psi }}^{{}^{\prime }}0-0+$$$$\mathrm{\Psi }\mathrm{decr}\mathrm{incr}+\mathrm{\infty }$$$$\mathrm{\Psi }\mathrm{is}\mathrm{decreazing}\mathrm{onI}=]\frac{1}{{\mathrm{e}}^3},\frac{1}{\mathrm{e}}[\Rightarrow \mathrm{\exists }!{\mathrm{x}}_0\in \mathrm{I}/\mathrm{\Psi }\left({\mathrm{x}}_0\right)=0$$$$\mathrm{\Psi }\mathrm{is}\mathrm{increazing}\mathrm{onJ}=]\frac{1}{\mathrm{e}},+\mathrm{\infty }[\Rightarrow \mathrm{\exists }!{\mathrm{y}}_0\in \mathrm{J}/\mathrm{\Psi }\left({\mathrm{y}}_0\right)=0$$$$\mathrm{we}\mathrm{see}\mathrm{that}\mathrm{\Psi }\left(1\right)=0\Rightarrow {\mathrm{y}}_0=1$$$$\mathrm{we}\mathrm{can}\mathrm{determine}{\mathrm{x}}_0\mathrm{by}\mathrm{newton}\mathrm{method}...$$

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