Question Number 32912 by mondodotto@gmail.com last updated on 06/Apr/18
$$\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{x}} \\ $$$$\:\boldsymbol{\mathrm{log}}\sqrt{\boldsymbol{{x}}}=\sqrt{\boldsymbol{\mathrm{log}{x}}} \\ $$
Answered by mrW2 last updated on 06/Apr/18
$${x}\geqslant\mathrm{0} \\ $$$$\mathrm{log}\:{x}\geqslant\mathrm{0}\Rightarrow{x}\geqslant\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:{x}=\sqrt{\mathrm{log}\:{x}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{log}\:{x}\right)^{\mathrm{2}} =\mathrm{log}\:{x} \\ $$$$\Rightarrow\left(\frac{\mathrm{1}}{\mathrm{4}}\mathrm{log}\:{x}−\mathrm{1}\right)\mathrm{log}\:{x}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{log}\:{x}=\mathrm{0}\Rightarrow{x}=\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}\mathrm{log}\:{x}−\mathrm{1}=\mathrm{0}\Rightarrow\mathrm{log}\:{x}=\mathrm{4}\Rightarrow{x}=\mathrm{10}^{\mathrm{4}} \\ $$
Commented by mondodotto@gmail.com last updated on 06/Apr/18
$$\mathrm{thanks} \\ $$
Commented by MJS last updated on 06/Apr/18
$$\mathrm{sometimes}\:\mathrm{log}\:\mathrm{means}\:\mathrm{ln},\:\mathrm{everything} \\ $$$$\mathrm{stays}\:\mathrm{the}\:\mathrm{same}\:\mathrm{in}\:\mathrm{this}\:\mathrm{case},\:\mathrm{but} \\ $$$$\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{solution}\:\mathrm{would}\:\mathrm{be}\:{x}={e}^{\mathrm{4}} \\ $$
Commented by mondodotto@gmail.com last updated on 06/Apr/18
$$\boldsymbol{\mathrm{still}}\:\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{don}}'\boldsymbol{\mathrm{t}}\:\boldsymbol{\mathrm{get}}\:\boldsymbol{\mathrm{you}},\boldsymbol{\mathrm{more}}\:\boldsymbol{\mathrm{explaination}}\:\boldsymbol{\mathrm{please}} \\ $$
Commented by MJS last updated on 06/Apr/18
$$\mathrm{everything}\:\mathrm{stays}\:\mathrm{the}\:\mathrm{same}\:\mathrm{with} \\ $$$$\mathrm{logarithm}\:\mathrm{to}\:\mathrm{base}\:{b},\:\mathrm{except}\:\mathrm{the} \\ $$$$\mathrm{last}\:\mathrm{line}: \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\mathrm{log}_{{b}} \:{x}−\mathrm{1}=\mathrm{0}\:\Rightarrow\:\mathrm{log}_{{b}} \:{x}=\mathrm{4}\:\Rightarrow \\ $$$$\Rightarrow\:{x}={b}^{\mathrm{4}} \\ $$$$\mathrm{if}\:\mathrm{log}=\mathrm{log}_{\mathrm{10}} \:\Rightarrow\:{x}=\mathrm{10}^{\mathrm{4}} \\ $$$$\mathrm{if}\:\mathrm{log}=\mathrm{ln}=\mathrm{log}_{{e}} \:\Rightarrow\:{x}={e}^{\mathrm{4}} \\ $$
Commented by mondodotto@gmail.com last updated on 07/Apr/18
$$\mathrm{thanx} \\ $$