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Question Number 160879 by amin96 last updated on 08/Dec/21

Commented by cortano last updated on 08/Dec/21

⇔ 27.x^(log _3 (x^(1/3) )) = x^((10)/3) ⇔ log _3 (27)+(1/3)(log _3 (x))^2 = ((10)/3) log _3 (x) ⇔ 3 +(1/3)t^2 −((10)/3)t = 0 ; [t = log _3 (x)] ⇔t^2 −10t+9=0 ⇔(t−1)(t−9)=0 → { ((t=1⇒x=3^1 =3)),((t=9⇒x=3^9 )) :}

$$\Leftrightarrow\:\mathrm{27}.\mathrm{x}^{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)} \:=\:\mathrm{x}^{\frac{\mathrm{10}}{\mathrm{3}}} \\ $$$$\Leftrightarrow\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{27}\right)+\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)\right)^{\mathrm{2}} =\:\frac{\mathrm{10}}{\mathrm{3}}\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right) \\ $$$$\Leftrightarrow\:\mathrm{3}\:+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{t}^{\mathrm{2}} −\frac{\mathrm{10}}{\mathrm{3}}\mathrm{t}\:=\:\mathrm{0}\:;\:\left[\mathrm{t}\:=\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)\right] \\ $$$$\Leftrightarrow\mathrm{t}^{\mathrm{2}} −\mathrm{10t}+\mathrm{9}=\mathrm{0} \\ $$$$\Leftrightarrow\left(\mathrm{t}−\mathrm{1}\right)\left(\mathrm{t}−\mathrm{9}\right)=\mathrm{0} \\ $$$$\rightarrow\begin{cases}{\mathrm{t}=\mathrm{1}\Rightarrow\mathrm{x}=\mathrm{3}^{\mathrm{1}} =\mathrm{3}}\\{\mathrm{t}=\mathrm{9}\Rightarrow\mathrm{x}=\mathrm{3}^{\mathrm{9}} \:}\end{cases} \\ $$

Answered by MathsFan last updated on 08/Dec/21

let log_(27) x=u → x=27^u 27•(27^u )^u =(((27^u )^(10) ))^(1/3) 27•27^u^2 =27^(10u/3) u^2 +1=((10u)/3) 3u^2 −10u+3=0 u_1 =3 and u_2 =(1/3) x_1 =27^3 =19 683 x_2 =3

$$\boldsymbol{\mathrm{let}}\:\:\boldsymbol{\mathrm{log}}_{\mathrm{27}} \boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{u}}\:\rightarrow\:\:\boldsymbol{\mathrm{x}}=\mathrm{27}^{\boldsymbol{\mathrm{u}}} \\ $$$$\:\mathrm{27}\bullet\left(\mathrm{27}^{\boldsymbol{\mathrm{u}}} \right)^{\boldsymbol{\mathrm{u}}} =\sqrt[{\mathrm{3}}]{\left(\mathrm{27}^{\boldsymbol{\mathrm{u}}} \right)^{\mathrm{10}} } \\ $$$$\:\mathrm{27}\bullet\mathrm{27}^{\boldsymbol{\mathrm{u}}^{\mathrm{2}} } =\mathrm{27}^{\mathrm{10}\boldsymbol{\mathrm{u}}/\mathrm{3}} \\ $$$$\:\boldsymbol{\mathrm{u}}^{\mathrm{2}} +\mathrm{1}=\frac{\mathrm{10}\boldsymbol{\mathrm{u}}}{\mathrm{3}} \\ $$$$\:\mathrm{3}\boldsymbol{\mathrm{u}}^{\mathrm{2}} −\mathrm{10}\boldsymbol{\mathrm{u}}+\mathrm{3}=\mathrm{0} \\ $$$$\:\boldsymbol{\mathrm{u}}_{\mathrm{1}} =\mathrm{3}\:\:\:\:\boldsymbol{\mathrm{and}}\:\:\:\:\:\boldsymbol{\mathrm{u}}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\boldsymbol{\mathrm{x}}_{\mathrm{1}} =\mathrm{27}^{\mathrm{3}} =\mathrm{19}\:\mathrm{683} \\ $$$$\boldsymbol{\mathrm{x}}_{\mathrm{2}} =\mathrm{3} \\ $$

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