Question Number 91681 by hmamarques1994@gmail.com last updated on 02/May/20
$$\: \\ $$$$\:\boldsymbol{\mathrm{Se}}\:\:\boldsymbol{\mathrm{f}}\left(\sqrt{\mathrm{3}^{\sqrt[{\mathrm{3}}]{\boldsymbol{\mathrm{x}}}} }\:+\:\mathrm{3}^{\sqrt[{\mathrm{3}}]{\boldsymbol{\mathrm{x}}}} \right)\:=\:\sqrt[{\mathrm{3}}]{\boldsymbol{\mathrm{x}}},\:\:\boldsymbol{\mathrm{calcule}}\:\:\frac{\boldsymbol{\mathrm{f}}\left(\mathrm{2}\right)}{\boldsymbol{\mathrm{f}}\left(\mathrm{1}\right)}\centerdot \\ $$$$\: \\ $$
Commented by john santu last updated on 02/May/20
$${set}\:\sqrt[{\mathrm{3}\:\:}]{{x}}\:=\:{t}\: \\ $$$${f}\left(\mathrm{3}^{\frac{{t}}{\mathrm{2}}} +\mathrm{3}^{{t}} \:\right)\:=\:{t}\: \\ $$$$\left(\mathrm{1}\right)\:{f}\left(\mathrm{2}\right)\:;\:\mathrm{2}\:=\:\mathrm{3}^{\frac{{t}}{\mathrm{2}}} \:+\:\mathrm{3}^{{t}} \:,\:\left(\mathrm{3}^{\frac{{t}}{\mathrm{2}}} \right)^{\mathrm{2}} +\:\mathrm{3}^{\frac{{t}}{\mathrm{2}}} −\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{3}^{\frac{{t}}{\mathrm{2}}} =\:\mathrm{1}\:\Rightarrow\:{t}\:=\:\mathrm{0}\:,\:{f}\left(\mathrm{2}\right)\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:{f}\left(\mathrm{1}\right)\:;\:\left(\mathrm{3}^{\frac{{t}}{\mathrm{2}}} \right)^{\mathrm{2}} +\:\mathrm{3}^{\frac{{t}}{\mathrm{2}}} −\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{3}^{\frac{{t}}{\mathrm{2}}} \:=\:\frac{−\mathrm{1}\:+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow\:\frac{{t}}{\mathrm{2}}\:=\:\mathrm{log}_{\mathrm{3}} \:\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right) \\ $$$${t}\:=\:\mathrm{log}_{\mathrm{3}} \left(\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\:\Rightarrow\:{f}\left(\mathrm{1}\right)=\mathrm{log}_{\mathrm{3}} \left(\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$$$\left(\mathrm{3}\right)\:\frac{{f}\left(\mathrm{2}\right)}{{f}\left(\mathrm{1}\right)}\:=\:\mathrm{0}\: \\ $$
Commented by hmamarques1994@gmail.com last updated on 02/May/20
$${Show},\:{man}! \\ $$
Commented by john santu last updated on 02/May/20
$${man}\:{show}\: \\ $$
Answered by mr W last updated on 02/May/20
$${say}\:{t}=\sqrt{\mathrm{3}^{\sqrt[{\mathrm{3}}]{{x}}} }>\mathrm{0} \\ $$$${t}^{\mathrm{2}} =\mathrm{3}^{\sqrt[{\mathrm{3}}]{{x}}} \\ $$$${f}\left({t}+{t}^{\mathrm{2}} \right)=\mathrm{log}_{\mathrm{3}} \:{t}^{\mathrm{2}} \\ $$$${say}\:{u}={t}+{t}^{\mathrm{2}} \\ $$$${t}^{\mathrm{2}} +{t}−{u}=\mathrm{0} \\ $$$${t}=\frac{\sqrt{\mathrm{1}+\mathrm{4}{u}}−\mathrm{1}}{\mathrm{2}} \\ $$$${t}^{\mathrm{2}} ={u}−{t}={u}−\frac{\sqrt{\mathrm{1}+\mathrm{4}{u}}−\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{2}{u}+\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{4}{u}}}{\mathrm{2}} \\ $$$${f}\left({u}\right)=\mathrm{log}_{\mathrm{3}} \:\frac{\mathrm{2}{u}+\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{4}{u}}}{\mathrm{2}} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{log}_{\mathrm{3}} \:\frac{\mathrm{2}{x}+\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{4}{x}}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{f}\left(\mathrm{2}\right)}{{f}\left(\mathrm{1}\right)}=\frac{\mathrm{log}_{\mathrm{3}} \:\frac{\mathrm{5}−\mathrm{3}}{\mathrm{2}}}{\mathrm{log}_{\mathrm{3}} \:\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}}=\frac{\mathrm{0}}{\mathrm{log}_{\mathrm{3}} \:\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}}=\mathrm{0} \\ $$
Commented by hmamarques1994@gmail.com last updated on 02/May/20
$$\boldsymbol{{Show}},\:\boldsymbol{{man}}! \\ $$