x-2-16-x-find-x-

Question Number 138627 by KwesiDerek last updated on 15/Apr/21

$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} =\mathrm{16}^{\boldsymbol{\mathrm{x}}} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{x}} \\ $$

Commented by soudo last updated on 15/Apr/21

$$\mathrm{nice} \\ $$

Answered by mr W last updated on 15/Apr/21

x=±16^(x/2) =±4^x x=±e^(xln 4) (−xln 4)e^(−xln 4) =±ln 4 −xln 4=W(±ln 4) ⇒x=−((W(±ln 4))/(ln 4))= { ((−((W(ln 4))/(ln 4))=−(1/2))),((−((W(−ln 4))/(ln 4))⇒no real)) :}

$${x}=\pm\mathrm{16}^{\frac{{x}}{\mathrm{2}}} =\pm\mathrm{4}^{{x}} \\ $$$${x}=\pm{e}^{{x}\mathrm{ln}\:\mathrm{4}} \\ $$$$\left(−{x}\mathrm{ln}\:\mathrm{4}\right){e}^{−{x}\mathrm{ln}\:\mathrm{4}} =\pm\mathrm{ln}\:\mathrm{4} \\ $$$$−{x}\mathrm{ln}\:\mathrm{4}=\mathbb{W}\left(\pm\mathrm{ln}\:\mathrm{4}\right) \\ $$$$\Rightarrow{x}=−\frac{\mathbb{W}\left(\pm\mathrm{ln}\:\mathrm{4}\right)}{\mathrm{ln}\:\mathrm{4}}=\begin{cases}{−\frac{\mathbb{W}\left(\mathrm{ln}\:\mathrm{4}\right)}{\mathrm{ln}\:\mathrm{4}}=−\frac{\mathrm{1}}{\mathrm{2}}}\\{−\frac{\mathbb{W}\left(−\mathrm{ln}\:\mathrm{4}\right)}{\mathrm{ln}\:\mathrm{4}}\Rightarrow{no}\:{real}}\end{cases} \\ $$

Commented by KwesiDerek last updated on 15/Apr/21

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by MJS_new last updated on 16/Apr/21

x^2 −16^x =0 (x−4^x )(x+4^x )=0 x−4^x =0∨x+4^x =0 f_1 (x)=x−4^x f_1 ′(x)=1−4^x ln 4 =0 ⇒ x_0 =−((ln ln 4)/(ln 4)) f_1 ′′(x)<0 ∀x∈R ⇒ max at x_0 with f_1 (x_0 )<0 ⇒ no solution for f_1 (x)=0 f_2 (x)=x+4^x f_2 ′(x)=1+4^x ln 4 >0 ∀x∈R ∧ −∞<f_2 (x)<+∞ ⇒ ⇒ exactly one solution for f_2 (x)=0 x=−4^x ⇔ x=−2^(2x) ⇒ obviously x=−(1/2)

$${x}^{\mathrm{2}} −\mathrm{16}^{{x}} =\mathrm{0} \\ $$$$\left({x}−\mathrm{4}^{{x}} \right)\left({x}+\mathrm{4}^{{x}} \right)=\mathrm{0} \\ $$$${x}−\mathrm{4}^{{x}} =\mathrm{0}\vee{x}+\mathrm{4}^{{x}} =\mathrm{0} \\ $$$${f}_{\mathrm{1}} \left({x}\right)={x}−\mathrm{4}^{{x}} \\ $$$${f}_{\mathrm{1}} '\left({x}\right)=\mathrm{1}−\mathrm{4}^{{x}} \mathrm{ln}\:\mathrm{4}\:=\mathrm{0}\:\Rightarrow\:{x}_{\mathrm{0}} =−\frac{\mathrm{ln}\:\mathrm{ln}\:\mathrm{4}}{\mathrm{ln}\:\mathrm{4}} \\ $$$${f}_{\mathrm{1}} ''\left({x}\right)<\mathrm{0}\:\forall{x}\in\mathbb{R}\:\Rightarrow\:\mathrm{max}\:\mathrm{at}\:{x}_{\mathrm{0}} \:\mathrm{with}\:{f}_{\mathrm{1}} \left({x}_{\mathrm{0}} \right)<\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{f}_{\mathrm{1}} \left({x}\right)=\mathrm{0} \\ $$$${f}_{\mathrm{2}} \left({x}\right)={x}+\mathrm{4}^{{x}} \\ $$$${f}_{\mathrm{2}} '\left({x}\right)=\mathrm{1}+\mathrm{4}^{{x}} \mathrm{ln}\:\mathrm{4}\:>\mathrm{0}\:\forall{x}\in\mathbb{R}\:\wedge\:−\infty<{f}_{\mathrm{2}} \left({x}\right)<+\infty\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{exactly}\:\mathrm{one}\:\mathrm{solution}\:\mathrm{for}\:{f}_{\mathrm{2}} \left({x}\right)=\mathrm{0} \\ $$$${x}=−\mathrm{4}^{{x}} \:\Leftrightarrow\:{x}=−\mathrm{2}^{\mathrm{2}{x}} \:\Rightarrow\:\mathrm{obviously}\:{x}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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