Question Number 35743 by mondodotto@gmail.com last updated on 22/May/18
$$\boldsymbol{\mathrm{find}}\:\boldsymbol{{x}} \\ $$$$\boldsymbol{\mathrm{log}}\left(\boldsymbol{\mathrm{log}}\mathrm{2}+\boldsymbol{\mathrm{log}}_{\mathrm{2}} \left(\boldsymbol{{x}}+\mathrm{1}\right)\right)=\mathrm{0} \\ $$
Commented by prof Abdo imad last updated on 25/May/18
$${if}\:{log}\:{mean}\:{ln}\:\:\left({e}\right)\Leftrightarrow\:{ln}\left(\mathrm{2}\right)\:+{ln}_{\mathrm{2}} \left({x}+\mathrm{1}\right)\:=\mathrm{1} \\ $$$$\Leftrightarrow\:{ln}\left(\mathrm{2}\right)\:+\:\frac{{ln}\left({x}+\mathrm{1}\right)}{{ln}\left(\mathrm{2}\right)}\:=\mathrm{1}\:\:\:{with}\:{x}>−\mathrm{1} \\ $$$$\Leftrightarrow\left\{{ln}\left(\mathrm{2}\right)\right\}^{\mathrm{2}} \:+{ln}\left({x}+\mathrm{1}\right)\:=\:{ln}\left(\mathrm{2}\right) \\ $$$$\Leftrightarrow{ln}\left({x}+\mathrm{1}\right)={ln}\left(\mathrm{2}\right)\:−\left\{{ln}\left(\mathrm{2}\right)\right\}^{\mathrm{2}} \:\Leftrightarrow \\ $$$${x}+\mathrm{1}\:={e}^{{ln}\left(\mathrm{2}\right)\:−\left({ln}\left(\mathrm{2}\right)\right)^{\mathrm{2}} } \:=\mathrm{2}\:{e}^{−\left\{{ln}\left(\mathrm{2}\right)\right\}^{\mathrm{2}} } \:\Leftrightarrow \\ $$$${x}\:=\:\mathrm{2}\:{e}^{−\left\{{ln}\left(\mathrm{2}\right)\right\}^{\mathrm{2}} } \:−\mathrm{1}\:\:. \\ $$
Answered by MJS last updated on 23/May/18
$$\mathrm{log}=\mathrm{ln}\:\mathrm{I}\:\mathrm{hope}… \\ $$$$\mathrm{ln}\left(\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{log}_{\mathrm{2}} \left({x}+\mathrm{1}\right)\right)=\mathrm{0} \\ $$$$\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{log}_{\mathrm{2}} \left({x}+\mathrm{1}\right)=\mathrm{1} \\ $$$$\mathrm{log}_{\mathrm{2}} \left({x}+\mathrm{1}\right)=\mathrm{1}−\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\left({x}+\mathrm{1}\right)=\mathrm{2}^{\mathrm{1}−\mathrm{ln}\left(\mathrm{2}\right)} \\ $$$${x}=\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{ln}\left(\mathrm{2}\right)} }−\mathrm{1}\approx.\mathrm{237} \\ $$
Commented by mondodotto@gmail.com last updated on 23/May/18
$$\mathrm{thanx} \\ $$