Estude! Com Prof. Caju

(IEZZI) x<sup>(log₂x2)2- log₂(2x) -2 + (x+2)log_(x+2)2 4 = 3

S: { 1 , 2 , 2^-3/4}</sup>

Felipe22,

[tex3]x^{(log^2_2x)^2- log_ 2(2x) -2 }+ {(x+2)}^{{log_{(x+2)^2}} 4} = 3\implies x^{(log^2_2x)^2- [log_ 22)+log_2x)] -2 }+ {(x+2)}^{{log_{(x+2)^2}} 4} = 3 \\
(x+2)^{{log_{(x+2)^2}} 4} =(x+2)^{\frac{1}{2}log_{(x+2)}4}=4^{\frac{1}{2}}=2\\

x^{(2log_2x)^2- 1-log_2x -2 }+2 = 3 \implies x^{(2log_2x)^2-log_2x -3 } = 1\\
\therefore \boxed{x = 1 }\vee (2log_2x)^2 -log_2x-3=0\\
log_2x = y \implies 4y^2-y-3= 0 \therefore y =-\frac{3}{4} \vee y = 1\\
log_2x=-\frac{3}{4} \implies \boxed{x=2^{-\frac{3}{4}}}\\
log_2x = 1 \implies \boxed{x = 2} [/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]x^{(log^2_2x)^2- log_ 2(2x) -2 }+ {(x+2)}^{{log_{(x+2)^2}} 4} = 3\implies x^{(log^2_2x)^2- [log_ 22)+log_2x)] -2 }+ {(x+2)}^{{log_{(x+2)^2}} 4} = 3 \\
(x+2)^{{log_{(x+2)^2}} 4} =(x+2)^{\frac{1}{2}log_{(x+2)}4}=4^{\frac{1}{2}}=2\\

x^{(2log_2x)^2- 1-log_2x -2 }+2 = 3 \implies x^{(2log_2x)^2-log_2x -3 } = 1\\
\therefore \boxed{x = 1 }\vee (2log_2x)^2 -log_2x-3=0\\
log_2x = y \implies 4y^2-y-3= 0 \therefore y =-\frac{3}{4} \vee y = 1\\
log_2x=-\frac{3}{4} \implies \boxed{x=2^{-\frac{3}{4}}}\\
log_2x = 1 \implies \boxed{x = 2} [/tex3]

Mais uma vez te agradeço Petras.
Que Deus te ilumine sempre.
Abç!