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Question Number 91681 by hmamarques1994@gmail.com last updated on 02/May/20

Se f((√3^(x)^(1/3) ) + 3^(x)^(1/3) ) = (x)^(1/3) , calcule ((f(2))/(f(1)))∙

$$\: \\ $$$$\:\boldsymbol{\mathrm{Se}}\:\:\boldsymbol{\mathrm{f}}\left(\sqrt{\mathrm{3}^{\sqrt[{\mathrm{3}}]{\boldsymbol{\mathrm{x}}}} }\:+\:\mathrm{3}^{\sqrt[{\mathrm{3}}]{\boldsymbol{\mathrm{x}}}} \right)\:=\:\sqrt[{\mathrm{3}}]{\boldsymbol{\mathrm{x}}},\:\:\boldsymbol{\mathrm{calcule}}\:\:\frac{\boldsymbol{\mathrm{f}}\left(\mathrm{2}\right)}{\boldsymbol{\mathrm{f}}\left(\mathrm{1}\right)}\centerdot \\ $$$$\: \\ $$

Commented by john santu last updated on 02/May/20

set (x)^(1/(3 )) = t f(3^(t/2) +3^t ) = t (1) f(2) ; 2 = 3^(t/2) + 3^t , (3^(t/2) )^2 + 3^(t/2) −2=0 3^(t/2) = 1 ⇒ t = 0 , f(2) = 0 (2) f(1) ; (3^(t/2) )^2 + 3^(t/2) −1=0 3^(t/2) = ((−1 +(√5))/2) ⇒ (t/2) = log_3 ((((√5)−1)/2)) t = log_3 (((3−(√5))/2)) ⇒ f(1)=log_3 (((3−(√5))/2)) (3) ((f(2))/(f(1))) = 0

$${set}\:\sqrt[{\mathrm{3}\:\:}]{{x}}\:=\:{t}\: \\ $$$${f}\left(\mathrm{3}^{\frac{{t}}{\mathrm{2}}} +\mathrm{3}^{{t}} \:\right)\:=\:{t}\: \\ $$$$\left(\mathrm{1}\right)\:{f}\left(\mathrm{2}\right)\:;\:\mathrm{2}\:=\:\mathrm{3}^{\frac{{t}}{\mathrm{2}}} \:+\:\mathrm{3}^{{t}} \:,\:\left(\mathrm{3}^{\frac{{t}}{\mathrm{2}}} \right)^{\mathrm{2}} +\:\mathrm{3}^{\frac{{t}}{\mathrm{2}}} −\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{3}^{\frac{{t}}{\mathrm{2}}} =\:\mathrm{1}\:\Rightarrow\:{t}\:=\:\mathrm{0}\:,\:{f}\left(\mathrm{2}\right)\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:{f}\left(\mathrm{1}\right)\:;\:\left(\mathrm{3}^{\frac{{t}}{\mathrm{2}}} \right)^{\mathrm{2}} +\:\mathrm{3}^{\frac{{t}}{\mathrm{2}}} −\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{3}^{\frac{{t}}{\mathrm{2}}} \:=\:\frac{−\mathrm{1}\:+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow\:\frac{{t}}{\mathrm{2}}\:=\:\mathrm{log}_{\mathrm{3}} \:\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right) \\ $$$${t}\:=\:\mathrm{log}_{\mathrm{3}} \left(\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\:\Rightarrow\:{f}\left(\mathrm{1}\right)=\mathrm{log}_{\mathrm{3}} \left(\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$$$\left(\mathrm{3}\right)\:\frac{{f}\left(\mathrm{2}\right)}{{f}\left(\mathrm{1}\right)}\:=\:\mathrm{0}\: \\ $$

Commented by hmamarques1994@gmail.com last updated on 02/May/20

Show, man!

$${Show},\:{man}! \\ $$

Commented by john santu last updated on 02/May/20

man show

$${man}\:{show}\: \\ $$

Answered by mr W last updated on 02/May/20

say t=(√3^(x)^(1/3) )>0 t^2 =3^(x)^(1/3) f(t+t^2 )=log_3 t^2 say u=t+t^2 t^2 +t−u=0 t=(((√(1+4u))−1)/2) t^2 =u−t=u−(((√(1+4u))−1)/2)=((2u+1−(√(1+4u)))/2) f(u)=log_3 ((2u+1−(√(1+4u)))/2) ⇒f(x)=log_3 ((2x+1−(√(1+4x)))/2) ⇒((f(2))/(f(1)))=((log_3 ((5−3)/2))/(log_3 ((3−(√5))/2)))=(0/(log_3 ((3−(√5))/2)))=0

$${say}\:{t}=\sqrt{\mathrm{3}^{\sqrt[{\mathrm{3}}]{{x}}} }>\mathrm{0} \\ $$$${t}^{\mathrm{2}} =\mathrm{3}^{\sqrt[{\mathrm{3}}]{{x}}} \\ $$$${f}\left({t}+{t}^{\mathrm{2}} \right)=\mathrm{log}_{\mathrm{3}} \:{t}^{\mathrm{2}} \\ $$$${say}\:{u}={t}+{t}^{\mathrm{2}} \\ $$$${t}^{\mathrm{2}} +{t}−{u}=\mathrm{0} \\ $$$${t}=\frac{\sqrt{\mathrm{1}+\mathrm{4}{u}}−\mathrm{1}}{\mathrm{2}} \\ $$$${t}^{\mathrm{2}} ={u}−{t}={u}−\frac{\sqrt{\mathrm{1}+\mathrm{4}{u}}−\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{2}{u}+\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{4}{u}}}{\mathrm{2}} \\ $$$${f}\left({u}\right)=\mathrm{log}_{\mathrm{3}} \:\frac{\mathrm{2}{u}+\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{4}{u}}}{\mathrm{2}} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{log}_{\mathrm{3}} \:\frac{\mathrm{2}{x}+\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{4}{x}}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{f}\left(\mathrm{2}\right)}{{f}\left(\mathrm{1}\right)}=\frac{\mathrm{log}_{\mathrm{3}} \:\frac{\mathrm{5}−\mathrm{3}}{\mathrm{2}}}{\mathrm{log}_{\mathrm{3}} \:\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}}=\frac{\mathrm{0}}{\mathrm{log}_{\mathrm{3}} \:\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}}=\mathrm{0} \\ $$

Commented by hmamarques1994@gmail.com last updated on 02/May/20

Show, man!

$$\boldsymbol{{Show}},\:\boldsymbol{{man}}! \\ $$

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